course Mth 174 ݔ᪦^Ɉɘassignment #002
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15:52:41 Query Section 6.3 #8, ds / dt = -32 t + 100, s = 50 when t = 0
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RESPONSE --> ds/dt = -32t + 100 ds/dt = -16t^2 + 100 50 = 16t^2 + 100 -50 = 16t^2 -50/16 = t^2 3.125 = t^2 t = squart of 3.125 = 1.767 seconds
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15:52:50 What is the solution satisfying the given initial condition?
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RESPONSE --> ds/dt = -32t + 100 ds/dt = -16t^2 + 100 50 = 16t^2 + 100 -50 = 16t^2 -50/16 = t^2 3.125 = t^2 t = squart of 3.125 = 1.767 seconds
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15:53:23 What is the general solution to the differential equation?
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RESPONSE --> ds/dt = -32t + c
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16:04:55 How fast is the water balloon moving when it strikes the ground?
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RESPONSE --> The building is 30 ft tall. v(t) = -32t +40 s = -16t^2 + 40 30 = -16t^2 + 40 -10 = -16t^2 -10/-16 = t^2 .625 = t^2 squart of .625 = .79 sec = t v= -32 (.79) +40 v = -25.28 + 40 v= 14.72 (I wasn't sure if I was supposed to add 40 to this or not.)
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16:10:29 How fast is the water balloon moving when it strikes the 6 ft person's head?
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RESPONSE --> 30 - 6 = 24 ft the ball drops 24 = -16t^2 +40 -16 = -16t^2 1 = t^2 t= 1 sec v = -32 (1) = 40 v = 8 I'm not sure I solved this right because the time at 24ft should be less than the time at 30 ft, and it's not. I don't understand what I have done wrong here.
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16:19:34 What is the average velocity of the balloon between the two given clock times?
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RESPONSE --> Average velocity = change in position / change in time s1 = 30 s2 = 24 t1= .79 t2 = 1.0 v = 6 / .21 In the given equation v (t) = -32t + 40, what does 40 represent? I thought c always represented the position at t=0, but the building is only 30 ft tall so I am confussed.
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16:20:37 What function describes the velocity of the balloon as a function of time?
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RESPONSE --> v (t) = -32t + 40
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16:36:49 Query Section 6.4 #19 (#18 3d edition) derivative of (int(ln(t)), t, x, 1)
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RESPONSE --> I know that the derivative of (int( ln(t))dt = -lnx but I don't know how to arrive at this answer. I know that the derivative of lnx= 1/x. Can you use substitution here?
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16:37:08 What is the desired derivative?
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RESPONSE --> -lnx
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16:41:26 The Second Fundamental Theorem applies to an integral whose upper limit is the variable with respect to which we take the derivative. How did you deal with the fact that the variable is the lower limit?
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RESPONSE --> I do not understand how to treat the problem differently if the variable is the lower limit.
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16:44:22 Why do we use something besides x for the integrand?
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RESPONSE --> You can you something besides x if it is a fixed value.
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16:53:21 What is the desired derivative?
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RESPONSE --> F(3) = (int (e^-t2)dt = 1.23 I thought that when finding the integral, you are finding the antiderivitive.
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16:55:05 How did you apply the Chain Rule to this problem?
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RESPONSE -->
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17:02:28 Why was the Chain Rule necessary?
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RESPONSE --> It was necessary because there were to parts of the problem that needed to be integrated.
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