course MTH 174 X}ʿ`«ѽMassignment #002
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16:03:05 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
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16:08:41 Query Section 6.3 #8, ds / dt = -32 t + 100, s = 50 when t = 0
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RESPONSE --> This seems to be #7 in the second edition text. ds/dt = -32t + 100 where s=50 when t=0 s = -32/2 * t^2 + 100t + C s = -16t^2 + 100t + C 50 = -16 * (0)^2 + 100 * (0) + C C = 50 s = -16 t^2 + 100t + 50
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16:09:11 What is the solution satisfying the given initial condition?
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RESPONSE --> The solution satisfying the initial condition is C = 50
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16:09:53 What is the general solution to the differential equation?
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RESPONSE --> The general solution is in the form of an equation. s = -16t^2 + 100t + 50
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16:26:54 Query Section 6.3 #14 water balloon from 30 ft, v(t) = -32t+40
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RESPONSE --> This seems to be #16 in the second edition. a) v(t) = -32t + 40 s(t) = -16t^2 + 40t + C When the balloon is 30 ft. off the ground C = 30. The general equation is then: s(t) = -16t^2 + 40 +30 b) The average velocity between 1.5 and 3 seconds is 1/(3 - 1.5) * integral from 1.5 to 3 of (-32t + 40) dt 2/3 * [-16 (3)^2 + 40 (3) - [-16 (1.5)^2 + 40 (1.5)]] 2/3 * [ -24 - 24] = -32 ft/s c) Balloon to 6ft tall person on the ground. s(t) = 6 6 = -16t^2 + 40t + 30 0 = -16t^2 + 40t + 24 0 = 8(-2t^2 + 5t + 3) 0/8 = (-2t - 1)(t - 3) t = -1/2 or t = 3 t = -1/2 does not make sense in this context, so 3 is the answer. Substituting back into the equation shows the height at t=3 -16(3)^2 + 40(3) + 30 = 6
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16:44:29 How fast is the water balloon moving when it strikes the ground?
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RESPONSE --> Solving 0 = -16t^2 + 40t + 30 for t using the quadratic formula: a = -16, b = 40, c = 30 [-40 +- sqrt( 40^2 - 4 * -16 * 30)]/(2 * -16) t= 3.10s or t = -.604. s Discarding negative time, t = 3.10s velocity at 3.10 seconds is v(3.10) = -32(3.10) + 40 = -59.20 ft/s
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16:47:45 How fast is the water balloon moving when it strikes the 6 ft person's head?
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RESPONSE --> I forgot this portion previously. Here is the complete solution to part c) s(t) = 6 6 = -16t^2 + 40t + 30 0 = -16t^2 + 40t + 24 0 = 8(-2t^2 + 5t + 3) 0/8 = (-2t - 1)(t - 3) t = -1/2 or t = 3 t = -1/2 does not make sense in this context, so 3 is the answer. Substituting back into the equation shows the height at t=3 -16(3)^2 + 40(3) + 30 = 6 Substitutingn back into the velocity function: v(3) = -32(3) + 40 = -56 ft/s
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16:48:11 What is the average velocity of the balloon between the two given clock times?
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RESPONSE --> b) The average velocity between 1.5 and 3 seconds is 1/(3 - 1.5) * integral from 1.5 to 3 of (-32t + 40) dt 2/3 * [-16 (3)^2 + 40 (3) - [-16 (1.5)^2 + 40 (1.5)]] 2/3 * [ -24 - 24] = -32 ft/s
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16:49:45 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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17:04:24 Query Section 6.4 #19 (#18 3d edition) derivative of (int(ln(t)), t, x, 1)
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RESPONSE --> This seems to be #12 in the second edition. Instead of going left to right, we are going right to left. ln x is undefined at 0, so making x > 0 is required. Knowing this, I believe that the proper derivative for this would be ln at -ln x with the limitation that x > 0
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17:04:44 What is the desired derivative?
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RESPONSE --> - ln x where x > 0
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17:05:28 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 reversed the sign on the area to show movement from right to left. In this case, x needs to be greater than 0.
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17:08:06 Why do we use something besides x for the integrand?
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RESPONSE --> We use something besides x for the integrand because we are taking the derivative with respect to x. We are not looking at the variable of the function in the integrand, but the variable in the boundaries of the integral.
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17:26:10 Query Section 6.4 #26 (3d edition #25) derivative of (int(e^-(t^2),t, 0,x^3)
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RESPONSE --> This seems to be #18 in the second edition. d/dx integral from 0 to x^3 of e^(-t^2) dt The derivative will be e^-(x^3)^2 * 3x^2
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17:26:28 What is the desired derivative?
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RESPONSE --> e^-(x^3)^2 * 3x^2
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17:28:04 How did you apply the Chain Rule to this problem?
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RESPONSE --> The chain rule must be used inside e^-(x^3)^2 because we are taking the derivative with respect to x.
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17:28:38 Why was the Chain Rule necessary?
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RESPONSE --> Because we are taking the derivative with respect to x.
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