course mth 271 ??W????????{?assignment #019
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14:36:40 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 --> simplify -4{(t+2)^ -2)} f(x)=z^ -2 f'(x)= -2z^ -3 g(x)=t+2 g'(x)=1 chain rule: f'(x)= -4{(1)(-2(t+2)^ -3)} f'(x)= -8(t+2)^ -3 2nd deriv: f(x)=z^ -3 f'(x)=-3z^ -4 g(x)=t+2 g'(x)=1 f''(x)= -8{(1)(-3(t+2)^ -4)}= 24(t+2)^ -4 confidence assessment: 3
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14:36:55 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 --> yup self critique assessment: 3
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14:39:45 2.6.30 f'''' if f'''=2`sqrt(x-1)
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RESPONSE --> f'''(x)= 2[sqroot(x+1)] f(x)=sqroot(z) f'(x)=.5z^ -.5 g(x)=x-1 g'(x)=1 f''''(x)=2[(1)(.5(x-1)^ -.5)]= (x-1)^ -.5= 1/(sqrt(x-1)) confidence assessment: 3
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14:39:55 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 --> yup self critique assessment: 3
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14:53:15 2.6.42 brick from 1250 ft
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RESPONSE --> a) position f(x) the book says that on earth a= -32 accel is the deriv of vel so v'= -32t+c vel is the deriv of the pos function s so v=s' s'=-32t+c which is of the form 2at+b therefore, s=-16t^2+ct+k where k is a constant we know that at t=0, s=1250ft solve s(0) for k k=1250 so s(t)= -16t^2+ct+1250 b/c the ball was dropped from rest, the initial velocity is zero v(0)=0 so c=0 and this gives s(t)=-16t^2+1250 b) time to hit sidewalk a=0 -16t^2+1250=0 t=8.8s (the neg answer is n/a) c) speed on impact -32(8.8)= -280 ft/s confidence assessment: 1
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14:54:57 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 wasnt sure on my answer, and i dont feel very confident about how i got there self critique assessment: 1
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