qu 14

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course mth 271

12/11 2

014. `query 14*********************************************

Question: `q **** Query 2.3.8 ave rate compared with inst rates at endpts on [1,4] for x^-.5 **** What is the average rate of change over the interval and how did you get it?

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Your solution:

for the average rate of change i inserted both points into the function and put the difference of those functions over the difference of the points.

(4^-.5) - (1^-.5) / (4-1)

(.5) - (1) / 3 = -.5 / 3 = -1/6

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Given Solution:

`aSTUDENT SOLUTION: The average rate of change over the interval is -1/6.

I got this answer by taking the difference of the numbers obtained when you plug both 1 and 4 into the function and then dividing that difference by the difference in 1 and 4. f(b)-f(a)/b-a

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Question: `q **** How does the average compare to the instantaneous rates at the endpoints of the interval?

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Your solution:

To figure the instantaneous rates at the points 1,4, we use the derivative of the function (x^-.5) = -.5 x^-1.5

Place each point into the derivative function

-.5* 1^-1.5= -.5

-.5* 4^-1.5= -.0625

The average of those two is (-.5 - -.0625) / 2= -.281

The rate at each point as well as the average, none equal that of the average rate of change.

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Given Solution:

`a The average rate of change is change in y / change in x.

For x = 1 we have y = 1^-.5 = 1.

For x = 4 we have y = 4^-.5 = 1 / (4^.5) = 1 / 2.

So `dx = 1/2 - 1 = -1/2 and `dy = 4 - 1 = 3.

`dy / `dx = (.5-1) / (4-1) = -.5 / 3 = -1/6 = -.166... .

To find rates of change at endpoints we have to use the instantaneous rate of change:

The instantaneous rate of change is given by the derivative function y ' = (x^-.5) ' = -.5 x^-1.5.

The endpoints are x=1 and x=4. There is a rate of change at each endpoint.

The rate of change at x = 1 is y ' = -.5 * 1^-1.5 = -.5.

The rate of change at x = 4 is y ' = -.5 * 4^-1.5 = -.0625.

The average of the two endpoint rates is (-.5 -.0625) / 2 = -.281 approx, which is not equal to the average rate -.166... .

Your graph should show the curve for y = x^-.5 decreasing at a decreasing rate from (1, 1) to (4, .5). The slope at (1, 1) is -.5, the slope at (4, .5) is -.0625. and the average slope is -.166... . The average slope is greater than the left-hand slope and less than the right-hand slope.

That is, the graph shows how the average slope between (1,1) and (4,.5), represented by the straight line segment between those points, lies between the steeper negative slope at x=1 and the less steep slope at x = 4.

If your graph does not clearly show all of these characteristics you should redraw the graph so that it does. **

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Question: `qQuery 2.3.14 H = 33(10`sqrt(v) - v + 10.45): wind chill; find dH/dv, interpret; rod when v=2 and when v=5

What is dH/dv and what is its meaning?

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Your solution:

Needing the derivative for instantaneous rates, the derivative is 165 v^-.5 - 33.

insert 2 and 5 each into the function

when v is 2, the function is 83.7

when v is 5, the function is 40.8.

this is the relation of heat loss to widn velocity.

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Given Solution:

`aERRONEOUS STUDENT SOLUTION:

dH/dv is equal to [33(10 sqrt v+x - v+x + 10.45) + (33(10 sqrt v - v + 10/45))] / [(v+x)-x].

dH/dv represents the average heat loss from a person's body between two difference wind speeds; v+x and v

INSTRUCTOR COMMENT:

You give the difference quotient, which in the limit will equal the rate of change, i.e., the derivative.

The derivative of h with respect to v is

dH / dv = 33 * 10 * .5 * v^-.5 + 33 * -1 = 165 v^-.5 - 33.

When v = 2, dH / dv is about 85 and when v = 5, dH / dv is about 40. Check my mental approximations to be sure I'm right (plug 2 and 5 into dH/dv = 165 v^-.5 - 33).

H is the heat loss and v is the wind velocity.

On a graph of H vs. v, the rise measures the change in heat loss and the run measures the change in wind velocity. So the slope measures change in heat loss / change in wind velocity, which is the change in heat loss per unit change in wind velocity. We call this the rate of change of heat loss with respect to wind velocity.

dH / dv therefore measures the instantaneous rate of change of heat loss with respect to wind velocity. **

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