course Mth 271 ???????????assignment #019
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21:30:34 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 Take the derivative of the first and get g'(t)= -8(t+2)^-3 Take the derivative of the second and get g''(t)= 24(t+2)^-4 confidence assessment: 2
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21:31:21 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 --> This is the answer that I got when I solved for the derivative functions. self critique assessment: 2
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21:33:49 2.6.30 f'''' if f'''=2`sqrt(x-1)
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RESPONSE --> You first solve for f'(x) and get 1/2sqrtx-1 Then you solve for f''(x) = 2(1/2sqrtx-1). The 2s cancel out and you are left with f'''(x)=1/sqrtx-1 confidence assessment: 2
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21:34:20 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 --> I got the same answer on this one also. self critique assessment: 2
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21:36:51 2.6.42 brick from 1250 ft
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RESPONSE --> a. The position function of the brick is -16t^2+t+1250 b. The velocity function would be -32t + t The acceleration would be -32 I'm not sure how to solve for c and d. I can't find an example to go by for these. confidence assessment: 2
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21:40:56 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 --> I have made a note of how you solved for how long it takes the brick to hit the sidewalk and how fast the brick is traveling when it hits the sidewalk. I understand better now that I have read your critique. self critique assessment: 2
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