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course Mth 173

2/11 9:00 PM

008. Approximate depth graph from the rate function

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Question: `q001. Note that there are 5 questions in thie assignment.

Sketch a graph of the function y ' = .1 t - 6 for t = 0 to t = 100. Describe your graph.

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Your solution:

this looks like a quadratic derivative, so the equation would be

y= .05t^2-6t+c. I will leave c as zero with no specification to what desired constant it should be.

for x =0 y = 0

when x=10 y=-55

y'=-6 for x=0

y'=1-6 or -5 for x=10 -5.5 average rate, which confirms the two points on the graph.

y= -100 at x= 20

y'=-4 at x=10 and x=20 the average is -4.5 in 10 seconds that is -45, added to -55 that is -100, which checks out.

y=-135 at x=30

y'=-3 from x=20 to x=30 the average is -3.5 and times 10 seconds that is -35. added to the -100 this checks out.

y= - 160 at x = 40

y'=-2 from x=30 to x=40 this average would be -2.5 times 10 is -25, added to -135, it is -160, checks out.

y= - 175 at x= 50

y'= -1 from x= 40 to x=50 this average would be -1.5, times 10 -15, added to -160 this is -175, checks out.

????had a thought, is this in a way constructing the tiny trapezoids, finding the slope of each one?

y= -180 at x= 60

y'=0 from x=50 to 60 the average would be -.5, times 10 is -5, added to -175 this is -180, checks out

???is it a plausable assumption to say at y' = 0 this will be the lowest point on a graph, following a decreasing path, or following an increasing path, the highest point, and that if you can tell a line on a graph has both decreasing points and increasing points, that y'=0 will either be its minimum or maximum, at least for parabolas, and lines of that sort (I do recall doing a trick question for a math teacher where he showed me only part of a graph that looked like it remained constant rate of zero, and then he said something to the effect ""but you didnt ask me what the graph looks like if we 'zoom' it out more"" and then the graph changed drastically after a bit.) I just rethought this, will this only work on a quadratic equation? because every graph I am thinking is a parabola of sorts--or a basic line that bottoms off.

y= -175 at x=70

y'=1 from x= 60 to 70 the average would be .5, times 10 5, added to -180 is -175, so check.

???would it be plausable to get a few more data points, to see the trend, and understand a parabola and that the values will be the same, just increasing instead of decreasing, so a mirror image on x=60 ?

y= -160 at x= 80

y'= 2 from x=70 to 80 the average would be 1.5, times 10 15, added to -175 is -160, checks out.

y=-135 at x=90

y'=3 from x= 80 to 90 the average would be about 2.5, times 10 25, added to -160 is -135, checks out

y=-100 at x=100

y'=4 from x=90 to x=100 the average would be about 3.5, times 10 35, added to -135 is -100, checks out.

this graph decreasing at a decreasing rate to the point (60,-180), then it starts to increase at an increasing rate. it is a parabola from the looks of it. and using a quadratic equation this should be correct.

confidence rating #$&*:2

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Given Solution:

`aThe graph of this function has an intecept on the y' axis at (0,-6) and an intercept on the x axis at (60,0). The graph is a straight line with slope .1. At t = 100 the graph point is (100,4).

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Self-critique (if necessary):

Oops, I did not think you meant to take the function as is. the function would give a tangent of the quadratic i created, would it not.

???I am thinking of my geometry class I had a while back, tangent is one touching, and is a derivative, there is a secant, where two is touching, is this when we take the averages? and what about cosine, how would it relate?

@&

Right. You did a good job, but of answering a question much more complicated than the question being asked.

The tangent line is the line that touches the graph at just the one point.

A secant line is the line between two points of the graph.

The use of the terms 'tangent' and 'secant' in this context are not related to the tangent and secant functions you use in trigonometry, so there is no connection with the cosine.

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Self-critique Rating:

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Question: `q002. This problem is a continuation of the preceding, in which y ' = .1 t - 6.

Now sketch the graph of a function y which starts at the point (0, 100), and whose slope at every t is equal to the value of y ' at t.

Thus, for example, at t = 0 we have y ' = .1 * 0 - 6 = -6, so our graph of y will start off a the t = 0 point (0,100) with a slope of -6, and the graph begins by decreasing rather rapidly.

But the slope won't remain at -6. By the time we get to t = 10 the slope will be y ' = .1 * 10 - 6 = -5, and the graph is decreasing less rapidly than before.

Then by the time we get to t = 20 the slope will be y ' = . 1 * 20 - 6 = -4, etc..

If you sketch a graph of y vs. t with these calculated slopes at the corresponding t values, what does the graph look like? Will it be increasing or decreasing, and will it be doing so at an increasing, decreasing or constant rate? Is the answer to this question different for different parts of the graph? If so over what intervals of the graph do the different answers apply?

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Your solution:

this would be the previous values i got, which creates a parabola, and is decreasing at a decreasing rate, and after x=60, starts to increase at an increasing rate.

???could we split these into two different functions, the split being at x=60? that isn't an inverse if you could, is it?

@&

An inverse function is where the x and the y values interchange.

If you interchanged the x and y values for a quadratic function you wouldn't get a function, since each y value (except that for the vertex) occurs for two different x values. That's no problem for a function, but when you inverse it you get two y values corresponding to a single x value, and that is a problem.

Now if you split the function as you say, there would be only one x value for each y value, so that each half would have an inverse function.

*@

confidence rating #$&*: 3

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Given Solution:

`aThe graph will have slopes -6 at t = 0, -5 at t = 10, -4 at t = 9, etc.. At least for awhile, the slope will remain negative so the graph will be decreasing. The negative slopes will however become less and less steep. So the graph be decreasing at a decreasing rate.

It's obvious that the slopes will eventually reach 0, and since y' is the slope of the y vs. t graph it's clear that this will happen when y' = .1 t - 6 becomes 0. Setting .1 t - 6 = 0 we get t = 60. Note, and think about about the fact, that this coincides with the x-intercept of the graph of y' vs. t. At this point the slope will be 0 and the graph will have leveled off at least for an instant.

Past t = 60 the values of y' will be positive, so the slope of the y vs. t graph will be positive and graph of y vs. t will therefore be increasing. The values of y' will also be increasing, so that the slopes of the y vs. t graph will be increasing, and we can say that the graph will be increasing at an increasing rate.

STUDENT QUESTION

The graph will decrease at a decreasing rate however the slope is decreasing at a constant

rate? I believe this is where I get confused because if the slope is constant how does it cause the graph to decrease at a

decreasing rate?

INSTRUCTOR RESPONSE

Be careful to distinguish between the two graphs.

We have a graph of y ' vs. t, and we have a graph of y vs. t. They are two separate graphs with different, though related, characteristics.

A graph of y ' vs. t is a straight line.

y ' gives us the slope of the y vs. t graph (a little more precisely, the value of y ' evaluated some value of t is the slope of the y vs. t graph at that same value of t).

Whatever is true about the y ' vs. t graph, it true of the slope of the y vs. t graph.

y ' is changing (as you say it is increasing slightly as we move to the right). So the slope of the y vs. t graph is increasing. The y vs. t graph will therefore not be a straight line.

At the current point, the value of y ' is in fact negative, so the slope of the y vs. t graph is negative. That is, the y vs. t graph is decreasing.

As you have pointed out the y ' vs. t graph is 'slightly increasing', in your words. So the slope of the y vs. t graph is increasing.

The y vs. t graph is therefore decreasing, but with an increasing slope. The slope of the y vs. t graph is negative, but is increasing (however slowly) toward 0. Its rate of decrease is therefore decreasing, and we can make the following equivalent statements:

The graph is decreasing at a decreasing rate.

The graph is decreasing but its slope is increasing.

The slope is negative but increasing.

The graph is concave up.

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `q003. The graph of y vs. t corresponding to the given rate function y ' = .1 t - 6 has slope -6 at the point (0,100). This slope immediately begins changing, and becomes -5 by the time t = 10. However, let us assume that the slope doesn't change until we get to the t = 10 point. This assumption isn't completely accurate, but we're going to see how it works out.

If the slope remained -6 for t = 0 to t = 10, then starting at (0, 100) what point would we reach when t = 10?

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Your solution:

ten seconds past, times amount of decrease, -6, -60, so 40

confidence rating #$&*:

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Given Solution:

`aThe slope of the graph is the ratio slope = rise / run. If the slope remains at -6 from t = 0 to t = 10, then the difference between 10 is the run. Thus the run is 10 and the slope is -6, so the rise is

rise = slope * run = -6 * 10 = -60.

The y coordinate of the graph therefore changes by -60, from y = 100 to y = 100 + (-60) = 40. The corresponding point is (10, 40).

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Self-critique (if necessary): OK

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Self-critique Rating:

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Question: `q004. We see that we reach the point (10, 40) by assuming a slope of -6 from t = 0 to t = 10. We have seen that at t = 10 the slope will be y ' = .1 * 10 - 6 = -5. If we maintain this slope for the interval t = 10 to t = 20, what will be the coordinates of the t = 20 point?

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Your solution:

that would be 50 subtracted from the current, so -10.

confidence rating #$&*:3

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Given Solution:

`aThe run from t = 10 to t = 20 is 10. With a slope of -5 this implies a rise of

rise = slope * run = -5 * 10 = -50.

Starting from point (10,40), a rise of -50 and a run of 10 takes the graph to the point (20, -10).

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q005. Continue this process up to the t = 70 point, using a 10-unit t interval for each approximation. Describe the graph that results.

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Your solution: (70,-110) I did this in a simple way using units on the graph, each move right of one unit, decreased the amount of units i would move down by one, so just counted them out and found the ordered pair.

confidence rating #$&*: 3

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Given Solution:

`aThe slope at t = 20 is y ' = .1 * 20 - 6 = -4. From t = 20 to t = 30 the run is 10, so the rise is rise = slope * run = -4 * 10 = -40. Starting from (20,-10) a rise of -40 and a run of 10 takes us to (30, -50).

The slope at t = 30 is y ' = .1 * 30 - 6 = -3. From t = 30 to t = 40 the run is 10, so the rise is rise = slope * run = -3 * 10 = -30. Starting from (30,-50) a rise of -30 and a run of 10 takes us to (40, -80).

The slope at t = 40 is y ' = .1 * 40 - 6 = -2. From t = 40 to t = 50 the run is 10, so the rise is rise = slope * run = -2 * 10 = -20. Starting from (40,-80) a rise of -20 and a run of 10 takes us to (50, -100).

The slope at t = 50 is y ' = .1 * 50 - 6 = -1. From t = 50 to t = 60 the run is 10, so the rise is rise = slope * run = -1 * 10 = -10. Starting from (50,-100) a rise of -10 and a run of 10 takes us to (60, -110).

The slope at t = 60 is y ' = .1 * 60 - 6 = -0. From t = 60 to t = 70 the run is 10, so the rise is rise = slope * run = -0 * 10 = 0. Starting from (60,-110) a rise of and a run of 10 takes us to (70, -110).

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Self-critique (if necessary):OK

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Self-critique Rating:OK"

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#*&!

&#Your work looks good. See my notes. Let me know if you have any questions. &#