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.

 

 

 

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?

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

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

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

 

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

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

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. **

 

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

 

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?

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

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. **

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

 

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

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

Given Solution:

`aSTUDENT 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). **

 

STUDENT COMMENT:

 

I got the end of the problem 150 / 'sqrtx,
But what I don't understand is where the 900 went and why it is not included in the answer.

INSTRUCTOR RESPONSE:

 

900 is a constant. It doesn't change. So its rate of change with respect to any variable is zero. This is the conceptual answer, which is very important.

Rules are also important. In terms of rules, the derivative of a constant is zero.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating: