QUERY 19

course MTH 271

11/01,1630

assignment #019019. `query 19

Applied Calculus I

11-01-2009

......!!!!!!!!...................................

15:59:21

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

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

......!!!!!!!!...................................

RESPONSE -->

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

(0)(t+2)^-2 + (-4)(-2)(t + 2)^-3

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

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

= 24 / (t + 2)^4

confidence rating: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.................................................

......!!!!!!!!...................................

15:59:36

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. **

......!!!!!!!!...................................

RESPONSE -->

------------------------------------------------

Self-critique Rating:ent: 3

.................................................

......!!!!!!!!...................................

16:01:39

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

......!!!!!!!!...................................

RESPONSE -->

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

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

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

confidence rating: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.................................................

......!!!!!!!!...................................

16:01:48

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). **

......!!!!!!!!...................................

RESPONSE -->

------------------------------------------------

Self-critique Rating:ent: 3

.................................................

......!!!!!!!!...................................

16:02:09

2.6.42 brick from 1250 ft

......!!!!!!!!...................................

RESPONSE -->

Downward acceleration: -32 ft/s^2

V(initial)=0 ft/s

a=v'=-32 ft/s^2

v(t)= -32t + c

v(0)=0= -32*0 + c

0=c

s'= v

s'= -32t + c

s= -16t^2 + ct + k =1250

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

k=1250

0=-16t^2 + 1250

-1250=-16t^2

78=t^2

t=8.9 seconds

v(8.9)=-32 ft/s^2 * 8.9s + 0

v(8.9)=285 ft/s

v(final)=285 ft/s

confidence rating: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.................................................

......!!!!!!!!...................................

16:02:17

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. **

......!!!!!!!!...................................

RESPONSE -->

------------------------------------------------

Self-critique Rating:ent: 3

.................................................

"

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