#$&* course MTH 271 04/077 pm 019. `query 19
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Given Solution: `a You need to use the Constant Rule. [ -4(t+2)^-2 ] ' = -4 [ (t+2)^-2 ] ' By the chain rule, with g(z) = z^-2 and h(t) = t + 1 this gives us -4 [ h '(t) * g ' ( h(t) ] = -4 [ (t+2) ' * -2(t+2)^-3 ] = -8 ( t+2)^-3. So g ' (t) = -8 ( t+2)^-3. Using the same procedure on g ' (t) we obtain g '' (t) = 24 ( t + 2)^-4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 3d 7th edition 2.6.28 2.6.28 f'''' if f'''=2`sqrt(x-1) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f’’’’ is the derivative of f’’’: 2 * (x - 1)^1/2 → 2 * (1/2) * ( x - 1)^-1/2 → 1/ 2(sqrt(x - 1)) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 5 7th edition 2.6.43 (was 2.6.40) brick from 1250 ft YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Assuming that a = -32, from a previous problem, Acceleration = s’’(t) = V’ = -32 Velocity = s’(t) = -32t + b Position = s(t) = -16t^2 + bt + c - When t = 0, s(0) = -16(0)^2 +b(0) + c → When t = 0, position is 1250, so c = 1250 Also, when t = 0, s’(0) = -32(0) + b → b = 0 So the position function is s(t) = -16t^2 + 1250 When the ball hits the sidewalk, acceleration hits 0 so: 0 = -16t^2 + 1250 → -16t^2 = -1250 → t^2 = 1250/16 → t = +- sqrt(1250/16) → t = +- 8.84 so when the ball hits the sidewalk, at acceleration = 0, time = 8.84 seconds. v(t) = -32t + b → v(8.84) = -32(8.84) → -282.88 ft/sec confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!