Assignment 2

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course Mth 1631

1/21/13 around 5p.m.

002.

Note that there are 12 questions in this exercise.

We begin with a short exercise intended to introduce you to the 'basic-points' idea, a very important idea which you will use throughout the course to quickly construct graphs of basic functions and their transformations and combinations. This is a main theme which runs throughout the course.

We then continue with a series of exercises on solving three simultaneous equations in three unknowns, using the process of elimination.

Note on graphs:

Graphs in this course aren't something you get from a graphing calculator. The graphing calculator has its uses and its benefits, but is not needed for most of this course. Graphs are generally constructed using transformations and other techniques, applied to a few simple functions. You will soon understand what this means.

You aren't expected to take the time to do meticulous artwork in this exercise. You will be asked to sketch points on some graphs, and a few curves (parabolas in this case). These graphs are for your own reference, and only need to be neat enough that you can tell what you're seeing. Nobody else needs to see them. Some you might well be able to imagine in your head, with no need to put anything on paper.

You are welcome to use graph paper, but it's probably easier to just sketch a pair of coordinate axes on a piece of paper and mark off an appropriate scale. A set of x and y axes, with each axis running from -8 to 8, will be fine for this exercise. Don't waste time labeling every point on the axes. If you just label the points for -8, -4, 4 and 8 you'll be able to tell what the other coordinates are, and you might not even need to label them (you could just make a larger 'tick mark' at each of these points). You should be able to sketch a usable set of coordinate axes in a minute or less.

It is entirely possible that you'll complete at least some of the first five and understand everything in them by simply following the instructions, then reading the given solutions. If this is the case on one or more of the first five exercises, you can simply put 'OK' for the confidence rating and self-critique rating. By doing so you certify that you have done everything correctly and understand the exercise.

Question `q001: Sketch a set of coordinate axes, with the x axis horizontal and directed to the right, the y axis vertical and directed upwards.

Sketch the point P = (-3, -1) on a set of coordinate axes.

Sketch the point you get if you shift this point -1 units in the horizontal direction. What are the coordinates of your point?

Sketch the point you get if you shift the original point 3 units in the vertical direction. What are the coordinates of your point?

Sketch the point you get if you move the original point 4 times as far from the x axis. What are the coordinates of your point?

If you move the original point 4 times as far from the x axis, then shift the resulting point -1 units in the horizontal direction, and finally shift the point 3 units in the vertical direction, what are the coordinates of the final point?

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Your solution:

Shifting the point -1 units in the horizontal direction the coordinates are (-4,-1)

Shifting the original point 3 units in the vertical direction the coordinates are (-3,2)

Shifting the original point 4 times as far from the x axis the coordinates are (1,1)

Shifting the original point 4 times as far from the x axis then shift the resulting point -1 units in the horizontal direction, and finally shift the point 3 units in the vertical direction the coordinates are (0,2).

confidence rating #$&*:

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Given Solution:

Shifting the point -1 units in the horizontal direction we end up at the point (-3 + (-1), -1) = (-4, -1).

Shifting the point 3 units in the vertical direction we end up at the point (-3, -1 + ( 3)) = (-3, 2).

The point (-3, -1) is -1 units from the x axis.

• If the point is moves 4 times further from the x axis, the y coordinate will become 4 * -1 = -4.

• The x coordinate will not change.

• So the coordinates of the new point will be (-3, -4).

If you then shift the resulting point -1 units in the horizontal direction, it will end up at (-3 + (-1), -4) = (-4, -4).

If you shift this new point 3 units in the vertical direction, it will end up at (-4, -4 + 3) = (-4, -1).

NOTE: We can express this sequence of transformations in a single step as

(-3 + (-1), 4 * -1 + 3) = (-4, -1).

Self-critique:

Instead of multiplying I just moved it to right 4 times. Now I understand why I got the last 2 points wrong.

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Self-critique rating: 3

Question `q002: Starting with the point P = (0, 0):

Sketch the point you get if you shift this point -1 units in the horizontal direction. What are the coordinates of your point?

Sketch the point you get if you shift the original point 3 units in the vertical direction. What are the coordinates of your point?

Sketch the point you get if you move the original point 4 times as far from the x axis. What are the coordinates of your point?

If you move the original point 4 times as far from the x axis, then shift the resulting point -1 units in the horizontal direction, and finally shift the point 3 units in the vertical direction, what are the coordinates of the final point?

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Your Solution:

(-1, 0); (0,3); (0,0); (-1, 3)

confidence rating #$&*:

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Given Solution:

Shifting the point -1 units in the horizontal direction we end up at the point (0 + (-1), 0) = (-1, 0).

Shifting the point 3 units in the vertical direction we end up at the point (0, 0 + ( 3)) = (0, 3).

The point (0, 0) is 0 units from the x axis.

• If the point is moves 4 times further from the x axis, the y coordinate will be 4 * 0 = 0.

• The x coordinate will not change.

• So the coordinates of the new point will be (0, 0).

If you then shift the resulting point -1 units in the horizontal direction, it will end up at (-1, 0).

If you shift this new point 3 units in the vertical direction, it will end up at (-1, 3)

Self-critique: ok

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Self-critique rating: 3

Question `q003: Plot the points (0, 0), (-1, 1) and (1, 1) on a set of coordinate axes.

Now plot the points you get if you move each of these points 4 times further from the x axis, and put a small circle around each point. What are the coordinates of your points?

Plot the points that result if you shift each of your three circled points -1 units in the x direction. Put a small 'x' through each point. What are the coordinates of your points?

Plot the points that result if you shift each of your three new points (the ones with the x's) 3 units in the y direction. Put a small '+' through each point. What are the coordinates of your points?

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Your Solution:

If I move each of the 4 times further from the x axis we will get (-4, 1); (0,0); & (4,1)

If I move each -1 units in the x direction I get (-5,1); (-1, 0); & (3,1).

If I move each of these 3 units in the y direction I get (-5, 4); (-1, 3); & (3,4).

confidence rating #$&*:

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Given Solution:

Moving each point 4 times further from the x axis:

The point (0, 0) is 0 units from the x axis. Multiplying this distance by 4 still gives you 0. So the point (0, 0) will remain where it is.

The points (-1, 1) and (1, 1) are both 1 unit above the x axis. Multiplying this distance by 4 gives us 4 * 1 = 4. The x coordinates will not change, so our new points are (-1, 4) and (1, 4).

At this stage our three points are

• (-1, 4)

• (0, 0)

• (1, 4)

Horizontally shifting each point -1 units, our x coordinates all change by -1. We therefore obtain the points

(-1 + (-1), 4) = (-2, 4),

(0 + -1, 0) = (-1, 0) and

((1 + (-1), 4) = ( 0, 4), so our points are now

• (-2, 4)

• (-1, 0)

• ( 0, 4)

Vertically shifting each point 3 units, our y coordinates all change by 3. We therefore obtain the points

(-2, 4 + 3) = (-2, 7)

(-1, 0 + 3) = (-1, 3) and

( 0, 4 + 3) = ( 0, 7)

Self-critique:

I do not understand how I got the first points wrong??? Why do the coordinates not change??? I multiplied -1 * 4 = -4 & 1* 4 = 4 therefore I got the points (-4, 1) & (4, 1). But that makes all of the rest wrong as well.

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Self-critique rating: 1

@&

(-1, 1) is 1 unit above the x axis.

(-4, 1) is 1 unit above the x axis.

If you multiply only the x coordinate of a point, you move it somewhere along a horizontal line, but you don't move it further from the x axis.

What you did was to move the point 4 times further from the y axis.

*@

Question `q004: On the coordinate axes you used in the preceding, sketch the parabola corresponding to the three basic points (0, 0), (-1, 1) and (1, 1).

Then sketch the parabola corresponding to your three circled basic points.

Then sketch the parabola corresponding to three basic points you indicated with 'x's'.

Finally sketch the parabola corresponding to the three basic points you indicated with '+'s'.

Describe how each parabola is related to the one before it.

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Your Solution:

They all have the same shape which is in the vertical direction. The first parabola is narrower than both of the others. The 2nd parabola of the circled point is one point to the left being -1 units mostly to the left. The 3rd parabola of the + sign points is similar to the second one but 3 units vertical of the 2nd parabola.

confidence rating #$&*:

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Given Solution: Your 'circled-points' parabola will be narrower than the original parabola through (-1, 1), (0, 0) and (1, 1). In fact, each point on the 'circled-points' parabola will lie 4 times further from the x axis than the point on the original parabola.

Your 'x'-points parabola will have the same shape as your 'circled-points' parabola, but will lie to the right or left of that parabola, having been shifted -1 units in the horizontal direction.

Your '+'-points parabola will have the same shape as the 'x-points' parabola (and the 'circled-point' parabola), but will lie above or below that parabola, having been shifted 3 units in the vertical direction.

Self-critique: ok

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Self-critique rating: 3

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Question: `q005.

Begin to solve the following system of simultaneous linear equations by first eliminating the variable which is easiest to eliminate. Eliminate the variable from the first and second equations, then from the first and third equations to obtain two equations in the remaining two variables:

2a + 3b + c = 128

60a + 5b + c = 90

200a + 10 b + c = 0.

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Your solution:

I chose the first & 2nd equations to start with which gave me 58a + 2b = -38 & 198 a + 7b = -128

confidence rating #$&*: 3

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Given Solution:

The variable c is most easily eliminated. We accomplish this if we subtract the first equation from the second, and the first equation from the third, replacing the second and third equations with respective results.

Subtracting the first equation from the second, are left-hand side will be the difference of the left-hand sides, which is

• 2d eqn - 1st eqn left-hand side: (60a + 5b + c )- (2a + 3b + c ) = 58 a + 2 b.

The right-hand side will be the difference 90 - 128 = -38, so the second equation will become

• new' 2d equation: 58 a + 2 b = -38.

The 'new' third equation by a similar calculation will be

• 'new' third equation: 198 a + 7 b = -128.

You might well have obtained this system, or one equivalent to it, using a slightly different sequence of calculations. (As one example you might have subtracted the second from the first, and the third from the second).

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q006. Solve the two equations

58 a + 2 b = -38

198 a + 7 b = -128

which can be obtained from the system in the preceding problem, by eliminating the easiest variable.

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Your solution:

By eliminating the easiest variable I chose to eliminate b. When this happened I was left with 396a =-256 & -406a = 266. Leaving me with -10a = 10. Therefore a = -1.

confidence rating #$&*: 3

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Given Solution:

Neither variable is as easy to eliminate as in the last problem, but the coefficients of b are significantly smaller than those of a. So here we choose eliminate b.It would also have been OK to choose to eliminate a.

To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the multiplications:

-7 * ( 58 a + 2 b) = -7 * -38

2 * ( 198 a + 7 b ) = 2 * (-128)

Doing the arithmetic we obtain

-406 a - 14 b = 266

396 a + 14 b = -256.

Adding the two equations we obtain

-10 a = 10,

so we have

a = -1.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q007. Having obtained a = -1, use either of the equations

58 a + 2 b = -38

198 a + 7 b = -128

to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation.

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Your solution:

I chose the first equation 58 (-1) + 2 b = -38. Solving this would give us -58 + 2b = -38. Then adding 58 we are left with 2b = 20. Divide 20 by 2 and get b = 10.

confidence rating #$&*: 3

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Given Solution:

You might have completed this step in your solution to the preceding problem.

Substituting a = -1 into the first equation we have

58 * -1 + 2 b = -38, so

2 b = 20 and

b = 10.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q008. Having obtained a = -1 and b = 10, determine the value of c by substituting these values for a and b into any of the 3 equations in the original system

2a + 3b + c = 128

60a + 5b + c = 90

200a + 10 b + c = 0.

Verify your result by substituting a = -1, b = 10 and the value you obtained for c into another of the original equations.

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Your solution:

I chose to substitute these values into the first equation giving 2 (-1) + 3 (10) + c = 128. When we multiply both values we get -2 + 30 + c = 128. Adding 30 & -2 together we get 28 + c = 128. Then subtract 28 from 128 and we get that c = 100.

confidence rating #$&*: 3

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Given Solution:

Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which we easily solve to get c = 100.

Substituting these values into the second equation, in order to check our solution, we obtain

60 * -1 + 5 * 10 + 100 = 90, or

-60 + 50 + 100 = 90, or

90 = 90.

We could also substitute the values into the third equation, and will again obtain an identity. This would completely validate our solution.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q009. The graph you sketched in a previous assignment contained the given points (1, -2), (3, 5) and (7, 8).

We are going to use simultaneous equations to obtain the equation of that parabola.

• A graph has a parabolic shape if its the equation of the graph is quadratic.

• The equation of a graph is quadratic if it has the form y = a x^2 + b x + c.

• y = a x^2 + b x + c is said to be a quadratic function of x.

To find the precise quadratic function that fits our points, we need only determine the values of a, b and c.

• As we will discover, if we know the coordinates of three points on the graph of a quadratic function, we can use simultaneous equations to find the values of a, b and c.

The first step is to obtain an equation using the first known point.

• What equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c?

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Your solution:

The equation we get if we substitute x & y values into the form is -2 = a * 1^2 + b * 1 + c, or a + b + c = -2

confidence rating #$&*: 3

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Given Solution:

We substitute y = -2 and x = 1 to obtain the equation

-2 = a * 1^2 + b * 1 + c, or

a + b + c = -2.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q010. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as in the preceding question, then what two equations do we get if we substitute the x and y values corresponding to the point (3, 5), then the point (7, 8) into the form y = a x^2 + b x + c? (each point will give us one equation)

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Your solution:

If we substitute these values in we will get 5 = a*3^2 + b * 3 + c or 9a + 3b + c = 5

And 8 = a*7^2 + b*7 + c or 49a + 7b + c = 8

confidence rating #$&*: 3

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Given Solution:

Using the second point we substitute y = 5 and x = 3 to obtain the equation

5 = a * 3^2 + b * 3 + c, or

9 a + 3 b + c = 5.

Using the third point we substitute y = 8 and x = 7 to obtain the equation

8 = a * 7^2 + b * 7 + c, or

49 a + 7 b + c = 8.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q011. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then we obtain three equations with unknowns a, b and c. You have already done this.

Write down the system of equations we got when we substituted the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c.

Solve the system to find the values of a, b and c.

• What is the solution of this system?

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Your solution:

The equations we have obtained are:

A + b + c = -2

9a + 3b + c = 5

49a + 7b + c = 8

To solve for this system of equation we must first add the first 2 equations together and eliminate the easiest variable which will be c. When we do this we must multiply the first equation by -1 in order to get them to eliminate. We are left with 9a + 3b + c = 5 & -a-b-c = 2. When we solve this we get 8a + 2b =7

We must then take the first and third equations again eliminating c and get 49 a + 7b + c = 8 & -a-b-c = 2. When we solve this we get 48a + 6b = 10.

Adding these equations together we get -24a - 6b = -21 & 48 + 6b = 10. When we add these together we get 24 a = -11. Then divide -11 by 24 and get a = -0.458333.

Then we must substitute a in for one of these equations we have obtained which gives us 8 * (-0.458333) + 2b = 7. We multiply and get -3.6664 + 2b = 7. Add that to 7 and get 2b = 10.6664. Then divide by 2 and get b = 5.3332.

To get c we must go back to the beginning equations and put in values a & b and solve for c. I choose the first equation giving -0.458333 + 5.3332 + c = -2. We then add those together and get 4.875 + c = -2. We then subtract 4.875 from -2 and get c = -6.875.

confidence rating #$&*: 3

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Given Solution:

The system consists of the three equations obtained in the last problem:

a + b + c = -2

9 a + 3 b + c = 5

49 a + 7 b + c = 8.

This system is solved in the same manner as in the preceding exercise. However in this case the solutions don't come out to be whole numbers.

The solution of this system, in decimal form, is approximately

a = - 0.45833,

b = 5.33333 and

c = - 6.875.

If you obtained a different solution, you should show your solution. Start by indicating the system of two equations you obtained when you eliminated c, then indicate what multiple of each equation you put together to eliminate either a or b.

ADDITIONAL DETAILS ON SOLUTION OF SYSTEM

You should have enough practice by now to be able to solve the system; however signs can trip us all up, and I've decided to append the following:

The second equation minus the first gives us 8a + 2 b = 7.

• To avoid a common error in subtracting these questions, note that the right-hand sides of these equations are 5 and -2, and that 5 - (-2) = 5 + 2 = 7. It is very common for students (and the rest of us as well) to get a little careless and calculate the right-hand side as 5 - 2 = 3.

The third equation minus the first gives us 48 a + 6 b = 10 (again the right-hand side can trip us up; 8 - (-2) = 10. I often see the incorrect calculation 8 - 2 = 6).

Now we solve these two equations, 8 a + 2 b = 7 and 48 a + 6 b = 10:

• If you subtract 3 times the first from the second you will get 24 a = -11, so that a = -.45833.

• Substituting this into 8 a + 2 b = 7 and solving for b you get b = 5.33333.

• Substituting these values of a and b into any of the three original equations you get c = -6.875.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q012. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c, in order to obtain aspecific quadratic function.

• What is your function?

• What y values do you get when you substitute x = 1, 3, 5 and 7 into this function?

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Your solution:

When we substitute those values just obtained into the equation we get y = (-0.45833) x^2 + 5.3332x - 6.875.

When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2.

When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5.

When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333.

When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8.

confidence rating #$&*: 3

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Given Solution:

Substituting the values of a, b and c into the given form we obtain the equation

y = - 0.45833 x^2 + 5.33333 x - 6.875.

• When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2.

• When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5.

• When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333.

• When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8.

Thus the y values we obtain for our x values yield the points (1, -2), (3, 5) and (7, 8). These are the points we used to obtain the formula. We also get the additional point (5, 8.33333).

NOTE THAT ADDITIONAL QUESTIONS RELATED TO THIS EXERCISE CONTINUE IN q_a_ ASSIGNMENT FOR ASSIGNMENT 3.

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Self-critique (if necessary): ok

Exercises:

Here are some data for the temperature of a hot potato vs. time:

Time (minutes) Temperature (Celsius)

0 95

10 75

20 60

30 49

40 41

50 35

60 30

70 26

Graph these data below, using an appropriate scale:

Pick three representative points and circle them.

Write the equations that result from the assumption that the appropriate mathematical model is a quadratic function y = a t^2 + b t + c.

The equations are 95 = 0a + 0b + c

75 = 100a + 10b + c

60 = 400a + 20b + c

Eliminate c from your equations to obtain two equations in a and b.

When we eliminate c from the equations we get the equations 25 = -400a -20b

& 30 = 600a + 20 b.

Solve for a and b.

When we solve for a & b we get a = -0.025 & b = -0.75

Write the resulting model for temperature vs. time.

The resulting model for temp. vs. time :

y = -0.025x^2 - 0.75x + 70

Make a table for this function:

Time (minutes) Model Function's Prediction of Temperature

0 70

10 62.25

20 54.5

30 46.75

40 41

50 31.25

60 113.5

70 15.75

Sketch a smooth curve representing this function on your graph.

Expand your table to include the original temperatures and the deviations of the model function for each time:

Time (minutes) Temperature (Celsius) Prediction of Model Deviation of Observed Temperature from Model

0 95 70 25

10 75 62.25 12.75

20 60 54.5 5.5

30 49 46.75 2.25

40 41 41 0

50 35 31.25 3.75

60 30 23.5 6.5

70 26 15.75 10.25

Find the average of the deviations.

The average deviation when you add up all the deviations and divide by 8 gives you 8.25.

Comment on how well the function model fits the data. (Note: the model might or might not do a good job of fitting the data. Some types of data can be fit very well by quadratic functions, while some cannot).

The function mode fits closely to the actual model. Being only 8.25 off isn’t too much, but it is more than we would like.

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Self-critique Rating: 3

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Question: `qAssignment 2

For the temperature vs. clock time model, what were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?

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Your solution:

The temperature vs. time model for the 1st, 3rd, & 5th data points are:

(0, 95), (20, 60); & (40, 41).

confidence rating #$&*: 3

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Given Solution:

** Continue to the next question **

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Self-critique (if necessary): ok

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Self-critique Rating: 3

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Question: `qAccording to your graph what would be the temperatures at clock times 7, 19 and 31?

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Your solution:

confidence rating #$&*:

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Given Solution:

** Continue to the next question **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat three points did you use as a basis for your quadratic model (express as ordered pairs)?

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Your solution:

I used (0, 95), (10, 75), & (20, 60).

confidence rating #$&*: 3

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Given Solution:

** A good choice of points `spreads' the points out rather than using three adjacent points. For example choosing the t = 10, 20, 30 points would not be a good idea here since the resulting model will fit those points perfectly but by the time we get to t = 60 the fit will probably not be good. Using for example t = 10, 30 and 60 would spread the three points out more and the solution would be more likely to fit the data. The solution to this problem by a former student will be outlined in the remaining nswers'.

STUDENT SOLUTION (this student probably used a version different from the one you used; this solution is given here for comparison of the steps, you should not expect that the numbers given here will be the same as the numbers you obtained when you solved the problem.)

For my quadratic model, I used the three points

(10, 75)

(20, 60)

(60, 30). **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat is the first equation you got when you substituted into the form of a quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The first equation I got was for (0, 95) was 95 = 0a + 0b + c or 95 = c

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat is the second equation you got when you substituted into the form of a quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The second equation I got was for (10,75) was 75 = 100a + 10b + c.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat is the third equation you got when you substituted into the form of a quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The third equation was for (20, 60) was 60 = 400a + 20b + c.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **

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Self-critique (if necessary):ok

------------------------------------------------

Self-critique Rating:3

*********************************************

Question: `qWhat multiple of which equation did you first add to what multiple of which other equation to eliminate c, and what is the first equation you got when you eliminated c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I used -1 as a multiple to eliminate equations 2 & 3 from each other. This gave me 75 = 100a + 10 b + c & -60 = -400a - 20b - c. Solving for this we get15 = -300a - 10b.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: First, I subtracted the second equation from the third equation in order to eliminate c.

By doing this, I obtained my first new equation

3200a + 40b = -30. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating:3

*********************************************

Question: `qTo get the second equation what multiple of which equation did you add to what multiple of which other quation, and what is the resulting equation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I again used -1 as a multiple to eliminate equations 1 & 3 giving 95 = 0a + 0b + c & -60a = -400a - 20b -c. Solving this equation give 25 = -400a - 20b

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: This time, I subtracted the first equation from the third equation in order to again eliminate c.

I obtained my second new equation:

3500a + 50b = -45**

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhich variable did you eliminate from these two equations, and what was the value of the variable for which you solved these equations?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

From these 2 equations I eliminated b. I used the multiple of -2. This gave me 25 = -400a -20b & -30 = 600a + 20b. Solving for this gave -5 = 200a. Then divide -5 by 200 and get a = -0.025.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: In order to solve for a and b, I decided to eliminate b because of its smaller value. In order to do this, I multiplied the first new equation by -5

-5 ( 3200a + 40b = -30)

and multiplied the second new equation by 4

4 ( 3500a + 50b = -45)

making the values of -200 b and 200 b cancel one another out. The resulting equation is -2000 a = -310. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat equation did you get when you substituted this value into one of the 2-variable equations, and what did you get for the other variable?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

When I substituted a into the first equation 25 = -400 * -0.025 - 20b. I multiplied giving 25 = 10 - 20b. Then subtracted 10 from 25 giving 15 = 20b. Then divide 15 by 20 and get b = -0.75.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: After eliminating b, I solved a to equal .015

a = .015

I then substituted this value into the equation

3200 (.015) + 40b = -30

and solved to find that b = -1.95. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat is the value of c obtained from substituting into one of the original equations?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I get c from putting it back into the first original equation and get the equation 75 = 100 * -0.025 + 10 * -0.75 + c. Multiplying and simplifying we get 75 = -2.5 + 7.5 +c. When we add these we get 75 = 5 + c. Then subtract 5 and get c = 70

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat is the resulting quadratic model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The resulting quadratic model I get is y = 0.025 x ^2 - 0.75x + 70.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was

y = (.015) x^2 - (1.95)x + 93. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat did your quadratic model give you for the first, second and third clock times on your table, and what were your deviations for these clock times?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

My quadratic model gave me (0, 70), (20, 54.5) & (40, 41).

The deviations for these times are 25, 5.5, & 0

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: This model y = (.015) x^2 - (1.95)x + 93 evaluated for clock times 0, 10 and 20 gave me these numbers:

First prediction: 93

Deviation: 2

Then, since I used the next two ordered pairs to make the model, I got back

}the exact numbers with no deviation. So. the next two were

Fourth prediction: 48

Deviation: 1

Fifth prediction: 39

Deviation: 2. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat was your average deviation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

My average deviation was 8.25

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: My average deviation was .6 **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qIs there a pattern to your deviations?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

There is no real patter to my devaitions

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: There was no obvious pattern to my deviations.

INSTRUCTOR NOTE: Common patterns include deviations that start positive, go negative in the middle then end up positive again at the end, and deviations that do the opposite, going from negative to positive to negative. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qHave you studied the steps in the modeling process as presented in Overview, the Flow Model, Summaries of the Modeling Process, and do you completely understand the process?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Yes

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qHave you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Yes I have and will

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: Yes, sir, I have memorized the steps of the modeling process at this point. I also printed out an outline of the steps in order to refresh my memory often, so that I will remember them forever!!!

INSTRUCTOR COMMENT: OK, I'm convinced. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qQuery Completion of Model first problem: Completion of model from your data.Give your data in the form of depth vs. clock time ordered pairs.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

My data is:

Time (minutes) Temperature (Celsius) Prediction of Model Deviation of Observed Temperature from Model

0 95 70 25

10 75 62.25 12.75

20 60 54.5 5.5

30 49 46.75 2.25

40 41 41 0

50 35 31.25 3.75

60 30 23.5 6.5

70 26 15.75 10.25

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION: Here are my data which are from the simulated data provided on the website under randomized problems.

(5.3, 63.7)

(10.6. 54.8)

(15.9, 46)

(21.2, 37.7)

(26.5, 32)

(31.8, 26.6). **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat three points on your graph did you use as a basis for your model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(0, 70), (10, 62.5), & (20, 54.5)

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: As the basis for my graph, I used

( 5.3, 63.7)

(15.9, 46)

(26.5, 32)**

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qGive the first of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The first equation I got was for (0, 70) 70 = a * 0^2 + b* 0 + c

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qGive the second of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The second equation I got was for (10,62.5) was 100a + 10b +c = 62.5

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qGive the third of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The third equation was for (20, 54.5) is 54.5 = 400a + 20b + c

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qGive the first of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I used -1 as a multiple to eliminate equations 1 & 2 from each other. This gave me -70 = -0a - 0b - c & 62.5 = 100a + 10b + c. Solving this gives -7.5 = 100 a + 10b.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qGive the second of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I again used -1 as a multiple to eliminate equations 2 & 3 giving 62.5 = 100a + 10 b + c & -54.5 = -400a - 20b - c. Solving this gives 8 = 300a - 10b.

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qExplain how you solved for one of the variables.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

From these 2 equations I eliminated b. I did not have to use a multiple to solve it because they were already ready to eliminate. When doing this I got 0.5 = 400a. Then divide 0.5 by 400 we get that a = 0.00125.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated b by multiplying the first equation by 21.2, which was the b value in the second equation. Then, I multiplied the seond equation by -10.6, which was the b value of the first equation, only I made it negative so they would cancel out. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating:3

*********************************************

Question: `qWhat values did you get for a and b?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

a = 0.00125 as I stated in the question above

When I substituted a into the first equation -7.5 = 100a + 10 b. I had the equation -7.5 = 100 * 0.00125 + 10b. Multiply that and get -7.5 = .125 + 10b. Subtract .125 from -7.5 and get -7.625 = 10b. Then divide by 10 and get b = -0.7625.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating:3

*********************************************

Question: `qWhat did you then get for c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I get c from putting it back into the first original equation and get the equation 62. 5 = 100 * 0.00125 + 10 + -0.765 + c. Multiply both and get 62.5 = 0.125 - 7.625 + c. Then subtract these and get 62.5 = -7.5 + c. Add 7.5 to 62.5 and get c = 70.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: c = 73.4 **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat is your function model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The resulting function model I get is y = 0.00125 x ^2 - 0.765x + 70.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat is your depth prediction for the given clock time (give clock time also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

My depth prediction for the given clock time of 40 sec would be 41cm.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qWhat clock time corresponds to the given depth (give depth also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The clock time that corresponds to the given depth of 70cm is 0 sec.

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** INSTRUCTOR COMMENT: The exercise should have specified a depth.

The specifics will depend on your model and the requested depth. For your model y = (.0165)x^2 + (-2)x + 73.4, if we wanted to find the clock time associated with depth 68 we would note that depth is y, so we would let y be 68 and solve the resulting equation:

68 = .01t^2 - 1.6t + 126

using the quadratic formula. There are two solutions, x = 55.5 and x = 104.5, approximately. **

STUDENT QUESTION

I have done what I could with the completion of flow model page 7 directions when I hit the sqrt

button on calculator so sqrt -.652 I would get (0,.807465169527) I do not remember ever doing a problem with 0,.8…. so I

hope I used the correct numbers to solve the rest of quad equation using quad formula

INSTRUCTOR RESPONSE:

Short answer:

The square root of a negative isn't a real number, so there is no solution to the equation for your given depth. Your calculator indicated a complex-number solution.

Longer answer:

The square root of a negative number is an imaginary number; the result you got is a point in the complex-number plane, on the imaginary axis. You might not understand what that means, but the point is that there is no real-number solution. You can't square a real number and get a negative, so the square root of a negative isn't a real number.

What this means is that the equation has no real-number solution. There is no clock time t for which the depth takes the y value you used in the equation.

In terms of the graph, note that the graph of the quadratic function is a parabola, which opens upward. So there are y values that lie completely below the parabola. If you try to solve the quadratic for one of these y values, you won't get a solution.

This sort of thing can certainly happen with a mathematical model. When it does, the answer is simply that there is no such solution.

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Self-critique (if necessary):

I did not realize that I needed to solve with the quadratic equation. I assumed I could look at my data table and receive the information. Now I understand.

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qCompletion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I am using the table and the table is as follows:

Percent of Assignments Reviewed Grade Average

0 1

10 1.790569

20 2.118034

30 2.369306

40 2.581139

50 2.767767

60 2.936492

70 3.09165

80 3.236068

90 3.371708

100 3.5

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION: Grade vs. percent of assignments reviewed

(0, 1)

(10, 1.790569)

(20, 2.118034)

(30, 2.369306)

(40, 2.581139)

(50, 2.767767)

(60, 2.936492)

(70, 3.09165)

(80, 3.236068)

(90, 3.371708)

(100, 3.5). **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating:3

*********************************************

Question: `qWhat three points on your graph did you use as a basis for your model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I used the points (0, 1) ; (10, 1.790569); & (20, 2.118034)

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED:

(20, 2.118034)

(50, 2.767767)

(100, 3.5)**

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qGive the first of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

My first equation is 1 = 0^2a + 0b + c or 1 = c

confidence rating #$&*:1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating:1

*********************************************

Question: `qGive the second of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

My second equation is 1.790569 = 100a + 10b + c

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qGive the third of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2.118034 = 400a + 20b + c

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **

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Self-critique (if necessary):ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `qGive the first of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The first equation I got when I eliminated c was -0.327465 = -300a - 10b.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **

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Self-critique (if necessary): ok

------------------------------------------------

Self-critique Rating: 3

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Question: `qGive the second of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The second equation I get is -0.790569 = -100a - 10b

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go

9600a + 80b = 1.381966 **

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Self-critique (if necessary): ok

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Self-critique Rating: 3

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Question: `qExplain how you solved for one of the variables.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I solved for a by eliminating b and using the multiple of +1. Giving me the equations -0.327465 = -300a - 10b & 0.790569 = 100a + 10b. We eliminate the 10b and are left with -0.327465 = -300a & 0. 790569 = 100a. we subtract and get 0.463104 = -200a. divide by 200 and get 0.00231552.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated the variable b. In order to do this, I multiplied the first new equation by 80 and the second new equation by -50. **

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Self-critique (if necessary): ok

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Self-critique Rating:3

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Question: `qWhat values did you get for a and b?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A = 0.00231552 as shown above.

For b I substituted it into the first equation I got giving me -0.327465 = -300 * 0.00231552 - 10b. we then multiply and add that to -0.327465 giving us 10b = .367091. we then divide by 10 and get b = 0.0367091

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED:

a = -.0000876638

b = .01727 **

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Self-critique (if necessary):ok

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Self-critique Rating: 3

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Question: `qWhat did you then get for c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

To get c I went back to the very first equation and plugged in a & b in for it and got 1.790569 = 100 * 0.00231552 + 10 * 0.0367091 + c. We follow through with addition and subtraction getting c = 1.191926.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: c = 1.773. **

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Self-critique (if necessary): ok

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Self-critique Rating: 3

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Question: `qWhat is your function model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

My functional model is thus reduced to

Y = 0.00231552 x^2 + 0.0367091x + 1.191926

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** y = -.0000876638 x^2 + (.01727)x + 1.773 **

STUDENT QUESTION

Hello! I am working on the Modeling Project #1 still and I am having such issues with the data sets for the rest of the

worksheet. I keep reading and I see that your doing grade average versus percentage of assignments, but I am confused on what

it is asking or what method I am supposed to be using. I got the first question, solving for a, b, and c and I am familiar

with the quadratic forumla, I am just missing something on how to start these next two problems.

Could you give me a boost to what to do?

INSTRUCTOR RESPONSE

What that boils down to can be summarized by a table.

For example, consider the following

x y

2 20

5 50

12 130

From this table and the form y = a x^2 + b x + c you get the equations

20 = 4 a + 2 b + c

50 = 25 a + 5 b + c

130 = 144 a + 12 b + c

which you can solve by elimination, as you did with the first question.

Now you are given data for grade ave. vs. percent of review.

• You could make a table of y vs. x, with y the grade average and x the percent of review.

• You could replace the heading 'x' in the first column with the identifier 'percent of review' and the 'y' in the second column with 'grade ave', so your table would represent percent of review vs. grade average.

Your table would have several additional rows (one additional row for every 'data point').

• You are instructed to choose three data points, and to base your model on those three points.

• You could for example make a 'shortened table' with just the three points you choose, very similar to the table given above (but with different numbers).

• To get a quadratic model you would again use the form y = a x^2 + b x + c to get three equations, one for each point.

• Solving the equations for a, b and c and plugging those values back into the form y = a x^2 + b x + c gives you your model.

Let me know if this doesn't help.

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Self-critique (if necessary): ok

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Self-critique Rating: 3

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Question: `qWhat is your percent-of-review prediction for the given range of grades (give grade range also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

For the percent of review prediction of 0 I would get an average of 1.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** The precise solution depends on the model desired average.

For example if the model is y = -.00028 x^2 + .06 x + .5 (your model will probably be different from this) and the grade average desired is 3.3 we would find the percent of review x corresponding to grade average y = 3.3 then we have

3.3 = -.00028 x^2 + .06 x + .5.

This equation is easily solved using the quadratic formula, remembering to put the equation into the required form a x^2 + b x + c = 0.

We get two solutions, x = 69 and x = 146. Our solutions are therefore 69% grade review, which is realistically within the 0 - 100% range, and 146%, which we might reject as being outside the range of possibility.

To get a range you would solve two equations, on each for the percent of review for the lower and higher ends of the range.

In many models the attempt to solve for a 4.0 average results in an expression which includes the square root of a negative number; this indicates that there is no real solution and that a 4.0 is not possible. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat grade average corresponds to the given percent of review (give grade average also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** Here you plug in your percent of review for the variable. For example if your model is y = -.00028 x^2 + .06 x + .5 and the percent of review is 75, you plug in 75 for x and evaluate the result. The result gives you the grade average corresponding to the percent of review according to the model. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qHow well does your model fit the data (support your answer)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

It fits pretty closely

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** You should have evaluated your function at each given percent of review-i.e., at 0, 10, 20, 30, . 100 to get the predicted grade average for each. Comparing your results with the given grade averages shows whether your model fits the data. **

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Self-critique (if necessary): ok

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Self-critique Rating:3

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Question: `qillumination vs. distance

Give your data in the form of illumination vs. distance ordered pairs.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Percent of Assignments Reviewed Grade Average Predicted Values

0 1 1.1919

10 1.790569 1.7906

20 2.118034 2.852

30 2.369306 4.3772

40 2.581139 6.3651

50 2.767767 8.8162

60 2.936492 11.73

70 3.09165 15.108

80 3.236068 18.948

90 3.371708 23.251

100 3.5 28.019

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION: (1, 935.1395)

(2, 264..4411)

(3, 105.1209)

(4, 61.01488)

(5, 43.06238)

(6, 25.91537)

(7, 19.92772)

(8, 16.27232)

(9, 11.28082)

(10, 9.484465)**

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat three points on your graph did you use as a basis for your model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED:

(2, 264.4411)

(4, 61.01488)

(8, 16.27232) **

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Self-critique (if necessary): ok

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Self-critique Rating: 3

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Question: `qGive ththe first of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The first equation is 1.19 = 0a + 0b + c

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**

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Self-critique (if necessary): ok

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Self-critique Rating:3

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Question: `qGive the second of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The second equation is 1.79 = 100a +10b + c

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**

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Self-critique (if necessary): ok

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Self-critique Rating:3

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Question: `qGive the third of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The third equation is 2.85 = 400a + 20b + c

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**

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Self-critique (if necessary): ok

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Self-critique Rating:3

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Question: `qGive the first of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The first equation I got was 0.6 = 100a + 10b.

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**

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Self-critique (if necessary):ok

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Self-critique Rating:3

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Question: `qGive the second of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The second equation was 1.66 = 400a + 20b.

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**

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Self-critique (if necessary): ok

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Self-critique Rating: 3

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Question: `qExplain how you solved for one of the variables.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I solved for a by eliminating b and getting .46 = -200a. Divide by 200 and get a = 0.0023.

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: I solved for a by eliminating the variable b. I multiplied the first new equation by 4 and the second new equation by -6 **

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Self-critique (if necessary):ok

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Self-critique Rating:3

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Question: `qWhat values did you get for a and b?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A = 0.0023 as stated above

B = 0.037

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **

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Self-critique (if necessary): ok

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Self-critique Rating:3

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Question: `qWhat did you then get for c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

C = 1.19

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: c = 588.5691**

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Self-critique (if necessary):ok

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Self-critique Rating:3

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Question: `qWhat is your function model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

My functional model is y = 0.0023x^2 + 0.037x + 1.19

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **

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Self-critique (if necessary):ok

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Self-critique Rating: 3

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Question: `qWhat is your illumination prediction for the given distance (give distance also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

For 1 AU from the sun I predict 936(W/m^2) away.

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: The given distance was 1.6 Earth distances from the sun. My illumination prediction was 319.61 w/m^2, obtained by evaluating my function model for x = 1.6. **

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Self-critique (if necessary):ok

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Self-critique Rating:3

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Question: `qWhat distances correspond to the given illumination range (give illumination range also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

For 2 AU I predict about 265 (W/m^2) away.

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** The precise solution depends on the model and the range of averages.

For example if the model is y =9.4 r^2 - 139 r + 500 and the illumination range is 25 to 100 we would find the distance r corresponding to illumination y = 25, then the distance r corresponding to illumination y = 100, by solving the equations

25=9.4 r^2 - 139 r + 500

and

100 =9.4 r^2 - 139 r + 500

Both of these equations are easily solved using the quadratic formula, remembering to put both into the required form a r^2 + b r + c = 0. Both give two solutions, only one solution of each having and correspondence at all with the data.

The solutions which correspond to the data are

r = 3.9 when y = 100 and r = 5.4 when y = 25.

So when the distance x has range 3.9 - 5.4 the illumination range is 25 to 100.

Note that a quadratic model does not fit this data well. Sometimes data is quadratic in nature, sometimes it is not. We will see as the course goes on how some situations are accurately modeled by quadratic functions, while others are more accurately modeled by exponential or power functions. **

"

Self-critique (if necessary):

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Self-critique rating:

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Question: `qWhat distances correspond to the given illumination range (give illumination range also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

For 2 AU I predict about 265 (W/m^2) away.

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** The precise solution depends on the model and the range of averages.

For example if the model is y =9.4 r^2 - 139 r + 500 and the illumination range is 25 to 100 we would find the distance r corresponding to illumination y = 25, then the distance r corresponding to illumination y = 100, by solving the equations

25=9.4 r^2 - 139 r + 500

and

100 =9.4 r^2 - 139 r + 500

Both of these equations are easily solved using the quadratic formula, remembering to put both into the required form a r^2 + b r + c = 0. Both give two solutions, only one solution of each having and correspondence at all with the data.

The solutions which correspond to the data are

r = 3.9 when y = 100 and r = 5.4 when y = 25.

So when the distance x has range 3.9 - 5.4 the illumination range is 25 to 100.

Note that a quadratic model does not fit this data well. Sometimes data is quadratic in nature, sometimes it is not. We will see as the course goes on how some situations are accurately modeled by quadratic functions, while others are more accurately modeled by exponential or power functions. **

"

Self-critique (if necessary):

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Self-critique rating:

#*&!