Assignment 19 Query

course Mth 271

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17:01:01

2.6.12 2d der of -4/(t+2)^2

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

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RESPONSE -->

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

= -4(t+2)^-2

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

g''(t) = -24(t+2)-4

confidence assessment: 3

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17:01:05

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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RESPONSE -->

self critique assessment: 3

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17:01:15

2.6.30 f'''' if f'''=2`sqrt(x-1)

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RESPONSE -->

f'''=2`sqrt(x-1)

= 2 (x-1)^1/2

f'''' = 1(x-1)^-1/2 = 1/sqrt(x-1)

confidence assessment: 3

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17:01:18

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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RESPONSE -->

self critique assessment: 3

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17:01:30

2.6.42 brick from 1250 ft

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RESPONSE -->

s(t) = -16 t^2 + c t + 1250

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

position function s(t) = -16t^2 + 1250

velocity function is -32t

acceleration function is -32

s(t) = -16t^2 + 1250 = 0

t = 8.8 sec.

v(8.8) = 281.6 ft/sec.

confidence assessment: 3

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17:01:32

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. **

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RESPONSE -->

self critique assessment: 3

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This looks very good. Let me know if you have any questions. &#