query 14

course mth 271

IzͮY͂assignment #014

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014. `query 14

Applied Calculus I

11-05-2007

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22:04:13

**** 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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RESPONSE -->

1st take the deriv... which im not sure about, but is it y'= x^(-1/2)

so y'(1)= 1

y'(4)= .5

delta h/ delta t= (.5-1)/ (4-0) = -.125

confidence assessment: 3

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22:05:08

STUDENT 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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RESPONSE -->

im not getting that answer... what is the deriv of 1/sq root x?

self critique assessment: 1

The derivative of x^n is n x^(n-1). Here n = -.5 so the derivative is -.5 x^(-.5 - 1) = -.5 x^(-1.5), also written as 1/2 x^(-3/2), or 1 / ( 2 x^(3/2)).

However the question asked for the average rate of change over the interval, not an instantaneous rate of change.

When x = 1, x^-.5 = 1; when x = 4, x^-.5 = 1/2. So the change in y is (1/2 - 1) = -1/2 and the change in x is 4 - 1 = 3. The average rate of change is therefore

ave rate = change in y / change in x = (-1/2) / 3 = -1/2 * 1/3 = -1/6.

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22:05:34

**** How does the average compare to the instantaneous rates at the endpoints of the interval?

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RESPONSE -->

not sure b/c i dont have the correct deriv

confidence assessment: 0

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22:10:32

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 these two 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 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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RESPONSE -->

yup

self critique assessment: 3

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22:14:49

Query 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/dt and what is its meaning?

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RESPONSE -->

i get h'= 330v^(.5)-33 not sure about that though

so at v=2, h'= 433.69

and at v=5 h'= 704.902

confidence assessment: 1

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22:18:04

ERRONEOUS STUDENT SOLUTION:

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

dH/dt 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 or the derivative, which 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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RESPONSE -->

im not following ur math on the deriv

self critique assessment: 0

See also my previous notes related to the derivative of the square root.

sqrt(v) = v^.5 and the derivative is .5 v^(.5 - 1) = .5 v^(-.5) or 1/2 v^-(1/2) or 1 / (2 v^.5) or 1 / (2 sqrt(v)).

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22:19:30

Query 2.3.20 C = 100(9+3`sqrt(x)); marginal cost **** What is the marginal cost for producing x units?

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RESPONSE -->

dc/dx= 150x^-.5 ????

confidence assessment: 1

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22:20:10

STUDENT SOLUTION: To get the marginal cost for producing x units, I think you take the first derivative of the cost function. If this is true, the marginal cost for 100(9+3sqrt(x)) is 100 * 3 * .5 x^-.5 = 150 x^-.5.

The marginal cost is the rate at which cost changes with respect to the number of units produced.

For this problem x is the number produced and C = 100 ( 9 + 3 sqrt(x) ).

Marginal cost is therefore dC/dx = 100 * 3 / (2 sqrt(x)) = 150 / sqrt(x). **

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RESPONSE -->

yup

self critique assessment: 3

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See my notes and let me know if you have questions.