open qa 19

#$&*

course Mth 271

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:

-4/(t+2) ^ 2 = -4 * (t+2) ^ -2

Y’ = 8 * (t+2) ^ -3

Y’’ = -24 * (t+2)^ -4 = -24/(t+2)^4

confidence rating #$&*:

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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. **

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Self-critique (if necessary):

###Can this be done the way I did it??? My sign seems to be off though… not sure where the mistake occurred. Actually I think there might be a mistake in a given solution when it comes to sign:

Ok This is from the given solution:

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

I think there might be a mistake in sign : -4 * -2 gives 8 and not -8.

@& The error occurred in the first derivative, which should have been positive; there were two negative factors in the expression, but I apparently ignored one of them. Your solution is correct. The given solution should read

`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. ***@

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Self-critique Rating:

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

F’’’ = 2 * (x-1)^1/2

F’’’’ = 2 * ½ (x-1) ^ -1/2 * (1) = 1/sqrt(x-1)

confidence rating #$&*:

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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). **

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Self-critique (if necessary):

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Self-critique Rating:

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

h = -16 t^2 + Vot + ho

h = -16t^2 + Vot + 1250

Initial velocity Vo = 0

h = --16t^2 + 1250

brick will hit the sidewalk when h = 0

16t^2 + 1250 = 0

16t^2 = -1250

t^2 = -1250/-16

t= sqrt(78.125)

t = 8.84

It takes 8.84 seconds approximately for the brick to hit the sidewalk.

Velocity function is:

h’’ = -32 t

h’’ = -32 * 8.84 = -282.8

confidence rating #$&*:

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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):

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Self-critique rating:

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

h = -16 t^2 + Vot + ho

h = -16t^2 + Vot + 1250

Initial velocity Vo = 0

h = --16t^2 + 1250

brick will hit the sidewalk when h = 0

16t^2 + 1250 = 0

16t^2 = -1250

t^2 = -1250/-16

t= sqrt(78.125)

t = 8.84

It takes 8.84 seconds approximately for the brick to hit the sidewalk.

Velocity function is:

h’’ = -32 t

h’’ = -32 * 8.84 = -282.8

confidence rating #$&*:

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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):

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Self-critique rating:

#*&!

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

h = -16 t^2 + Vot + ho

h = -16t^2 + Vot + 1250

Initial velocity Vo = 0

h = --16t^2 + 1250

brick will hit the sidewalk when h = 0

16t^2 + 1250 = 0

16t^2 = -1250

t^2 = -1250/-16

t= sqrt(78.125)

t = 8.84

It takes 8.84 seconds approximately for the brick to hit the sidewalk.

Velocity function is:

h’’ = -32 t

h’’ = -32 * 8.84 = -282.8

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):

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Self-critique rating:

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open qa 19

&#Very good responses. Let me know if you have questions. &#