course MTH 271 11/01,1630 assignment #019019. `query 19
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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?
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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 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. **
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3
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16:01:39 2.6.30 f'''' if f'''=2`sqrt(x-1)
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RESPONSE --> f''(x)=2 (x-1)^(1/2) f'''(x)=(x - 1)^(-1/2) f'''(x)= 1 / (x - 1)^(1/2) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). **
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3
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16:02:09 2.6.42 brick from 1250 ft
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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 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. **
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3
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