Assignment 27 

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course mth 158

7/17 12

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

027. `* 27

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Question: * 3.6.2 / 10. P = (x, y) on y = x^2 - 8.

Give your expression for the distance d from P to (0, -1)

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Your solution:

D= sqrt( (x-0)^2 - (-1 -(x^2-8)^2))

D= sqrt(x^2 + (8- x^2)^2

Sqrt( x^2 +64 -16x^2 + x^4)

Sqrt(x^4 - 15x^2 + 64)

@&

The quantity that need to be squared is -1 - (x^2 - 8), not just x^2 - 8.

-1 - (x^2 - 8) = -x^2 + 7, and the square of this quantity is x^2 - 14 x + 49.

*@

confidence rating #$&*: 3

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Given Solution:

* * ** P = (x, y) is of the form (x, x^2 - 8).

So the distance from P to (0, -1) is

sqrt( (0 - x)^2 + (-1 - (x^2-8))^2) =

sqrt(x^2 + (-7-x^2)^2) =

sqrt( x^2 + 49 - 14 x^2 + x^4) =

sqrt( x^4 - 13 x^2 + 49). **

What are the values of d for x=0 and x = -1?

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Your solution:

If x=0 then

Sqrt(0^4 - 13(0)^2 +49)

sqrt(0-0+49)

Sqrt(49)

d=7

If x= -1 then

Sqrt( -1^4 -13(-1)^2 +49)

Sqrt( 1-13 +49)

Sqrt(37)

D= 6.08

confidence rating #$&*: 3

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Given Solution:

* * If x = 0 we have

sqrt( x^4 - 13 x^2 + 49) = sqrt(0^4 - 13 * 0 + 49) = sqrt(49) = 7.

If x = -1 we have

sqrt( x^4 - 13 x^2 + 49) = sqrt((-1)^4 - 13 * (-1)^2 + 49) = sqrt( 63).

sqrt(64) = 8, so sqrt(63) is a little less than 8 (turns out that sqrt(63) is about 7.94).

Note that these results are the distances from the x = 0 and x = 1 points of the graph of y = x^2 - 8 to the point (0, -1). You should have a sketch of the function and you should verify that these distances make sense. **

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Self-critique (if necessary):

On the first problem:

So the distance from P to (0, -1) is

sqrt( (0 - x)^2 + (-1 - (x^2-8))^2) =

sqrt(x^2 + (-7-x^2)^2) =

sqrt( x^2 + 49 - 14 x^2 + x^4) =

sqrt( x^4 - 13 x^2 + 49). **

I don’t understand where the -7 came from in the second step. I put that it was 8 which threw off my entire problem.

In the second problem:

sqrt( x^4 - 13 x^2 + 49) = sqrt((-1)^4 - 13 * (-1)^2 + 49) = sqrt( 63).

sqrt(64) = 8, so sqrt(63) is a little less than 8 (turns out that sqrt(63) is about 7.94).

How did you get the sqrt(64) instead of 37? My answer was 6.08 instead of 7.94 because I took the square root of 37. Where did I go wrong?

@&

I was careless with the sign of -13 ( -1)^2.

sqrt(37) is correct.

*@

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Self-critique Rating: 2

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Question: * 3.6.9 / 18 (was and remains 3.6.18). Circle inscribed in square.

What is the expression for area A as a function of the radius r of the circle?

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Your solution:

The circle has a radius of 2, making its diameter 4. the length of each side of the rectangle is equal to 2r. We find the area of the rectangle by:

2r^2=4r^2

The area of the circle is found by the formula pi*r^2 so

The area of the rectangle not covered by the circle is

4r^2-pi*r^2 = (4- pi)r^2

Perimeter is = to 4*2r= 8r.

confidence rating #$&*: 3

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Given Solution:

* * ** A circle inscribed in a square touches the square at the midpoint of each of the square's edges; the circle is inside the square and its center coincides with the center of the square. A diameter of the circle is equal in length to the side of the square.

If the circle has radius r then the square has sides of length 2 r and its area is (2r)^2 = 4 r^2.

The area of the circle is pi r^2.

So the area of the square which is not covered by the circle is 4 r^2 - pi r^2 = (4 - pi) r^2. **

What is the expression for perimeter p as a function of the radius r of the circle?

The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. **

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Self-critique (if necessary):

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Self-critique Rating: 3

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Question: * 3.6.19 / 27 (was 3.6.30). one car 2 miles south of intersection at 30 mph, other 3 miles east at 40 mph

Give your expression for the distance d between the cars as a function of time.

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Your solution:

I have no idea how to approach this problem. I drew a graph with the x axis representing east and west and the y axis represent north and south. I also plotted the points of both cars on the x and y axis in the appropriate direction. I’m just not sure how to find the distance between the two points when I don’t know the actual time it will take for those cars to reach the origin (intersection).

confidence rating #$&*: 1

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Given Solution:

* * ** Assume a coordinate system with the y axis pointing north and south, the x axis east and west.

At time t= 0 the position of one car is 2 miles south of the origin, and its distance from the origin is increasing at 30 mph. So at time t it will have traveled distance 30 t from the 2-mile position. Its position will therefore be at distance 2 + 30 t from the origin along the negative y axis.

The position of the other is found by similar reasoning to be 3 + 40 t east of the origin, putting it at distance 3 + 40 t along the positive x axis.

At clock time t, then, a straight line from the position of the first car to that of the second will therefore form a right triangle with the x and y axes. The legs of the triangle will be 2 + 30 t and 3 + 40 t.

The distance between the cars is the hypotenuse of this triangle, so

distance = sqrt( (2 + 30 t)^2 + (3 + 40t)^2 ) = sqrt( 4 + 120 t + 900 t^2 + 9 + 240 t + 1600 t^2) = sqrt( 2500 t^2 + 360 t + 13). **

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Self-critique (if necessary):

Ohhhhh so I didn’t actually have to know the time it took. I could have just used t to represent the time. This makes a ton of sense now that I have read through your solution.

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Self-critique Rating: 3

&#Good responses. See my notes and let me know if you have questions. &#