course Mth 271 ¢Ö¥À¡w¯z†â²y±Ûw£ÔæäYéassignment #019
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17:15:46 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 g(t) = -4(t+2)^-2 I use the general power rule to get g'(t) = 8(t+2)^-3 I use the general power rule again to get g' '(t) = -24(t+2)^-4 g' '(t) = -24 / (t+2)^4 confidence assessment: 3
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17:17:31 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 assessment: 3
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17:21:50 2.6.30 f'''' if f'''=2`sqrt(x-1)
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RESPONSE --> f'''(x) = 2`sqrt(x-1) f'''(x) = 2(x-1)^.5 Now, I use the general power rule. f^4(x) = (x-1)^-.5 f^4(x) = 1 / `sqrt(x-1) confidence assessment: 3
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17:22:43 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 assessment: 3
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17:52:18 2.6.42 brick from 1250 ft
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RESPONSE --> a) I had to find the position function of a brick falling from 1250 ft. I used the position function for a falling object on earth s = -16t^2 + v0 t + h0, where t = time in seconds, v0 is the initial veloctiy, and h0 is the initial height. My function is s = -16t^2 + 1250. The initial velocity is zero since the brick had just begun to fall. b) Next, I needed the velocity and acceleration functions. To get the velocity function, I took the derivative of the postition function. s = -16t^2 + 1250 ds/dt = -32t To get the acceleration function, I took the derivative of the velocity funciton. ds/dt = -32t (d^2 s)/(d^2 t) = -32 c) To find how long it took the brick to hit the sidewalk, I went back to the original position function and set it equal to zero -16t^2 + 1250 = 0 16t^2 = 1250 t^2 = 78.125 t = +-`sqrt(78.125) = 8.8 approx I used the positive value because we can't use a negative time answer. So it took approximately 8.8 seconds for the brick to hit the sidewalk. d) Finally, I needed to know how fast the brick was falling when it hit the sidewalk. I substituted the time when the brick's position equaled zero into the velocity function. ds/dt = -32t (ds/dt)(8.8) = -32 * 8.8 = -282 approx. The brick was falling at approximately 282 ft/sec when it hit the sidewalk. confidence assessment: 2
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17:54:07 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 assessment: 3
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