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.

 

 

019. `query 19

 

 

Question: `q 1d   7th edition 2.6.12 2d der of -4/(t+2)^2

 

What is the second derivative of your function and how did you get it?

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

Given Solution:

`a You need to use the Constant Rule. [ -4(t+2)^-2 ] ' = -4 [ (t+2)^-2 ] '

 

By the chain rule, with g(z) = z^-2 and h(t) = t + 1 this gives us -4 [ h '(t) * g ' ( h(t) ] =

 

-4 [ (t+2) ' * -2(t+2)^-3 ] = 8 ( t+2)^-3.

 

So g ' (t) = 8 ( t+2)^-3.

 

Using the same procedure on g ' (t) we obtain

 

g '' (t) = -24 ( t + 2)^-4. **

 

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: `q 3d 7th edition 2.6.28    2.6.28 f'''' if f'''=2`sqrt(x-1)

 

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

Given Solution:

`a The fourth derivative f '''' is equal to the derivative of the third derivative. So we have

 

f '''' = (f ''') ' = [ 2 sqrt(x-1) ] ' = 2 [ sqrt(x-1) ] '.

 

Using the Chain Rule (noting that our function is sqrt(z) with z = x - 1, and that sqrt(z) ' = 1 / (2 sqrt(z) ) we get

 

2 [ (x-1)' * 1/(2 sqrt(x-1) ) ] = 2 [ 1 * 1/(2 sqrt(x-1) ) ] = 1 / sqrt(x-1). **

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: `q 5    7th edition 2.6.43 (was 2.6.40) brick from 1250 ft

 

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

Given Solution:

`a The detailed analysis is as follows:

 

The equation comes from a = -32; since a = v' we have v' = -32 so that v = -32 t + c. Then s' = v so s' = -32 t + c and s = -16 t^2 + c t + k. c and k are constants.

 

If s(t) = -16 t^2 + c t + k and at t = 0 we have s = 1250, then

 

s(0) = -16 * 0^2 + c * 0 + k so k = 1250 and s(t) = -16 t^2 + c t + 1250.

 

If the ball is dropped from rest then the initial velocity is v(0) = 0 so

 

v(0) = -32 t + c = 0 and -32 * 0 + c = 0 so c = 0.

 

So s(t) = -16 t^2 + c t + k becomes s(t) = -16 t^2 + 1250.

 

To find how long it takes to hit the sidewalk:

 

Position function, which gives altitude, is y = -16 t^2 + 1250.

 

When the brick hits the sidewalk its altitude is zero.

 

So -16 t^2 + 1250 = 0, and t = + - `sqrt(1250 / 16) = + - 8.8, approx.

 

The negative value makes no sense, so t = 8.8 seconds.

 

To find how fast the brick was moving when it hit the sidewalk:

 

velocity = -32 t so when t = 8.8 we have velocity = -32 * 8.8 = -280 approx.

 

That is, when t = 8.8 sec, v = -280 ft/sec. **

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating: