course Mth 174 ê䧷Ÿ–Ä’ŸÝù“ÓèåüáÇ‹nãˆéöº¬äÇEˆassignment #002
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16:46:34 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> OK
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16:47:27 Query Section 6.3 #8, ds / dt = -32 t + 100, s = 50 when t = 0
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RESPONSE -->
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16:57:09 What is the solution satisfying the given initial condition?
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RESPONSE --> ds/dt = -32t +100 is the velocity equation and the derivative of the distance equation, therefore: S(t) = antiderivative of (-32t+100) dt S(t) = -16t^2 + 100t + C Given S(0) = 50, subsitute: S(0) = 50 = -16(0^2) + 100(0) + C C = 50 S(t) = -16t^2 + 100t + 50
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16:58:21 What is the general solution to the differential equation?
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RESPONSE --> The general solution is: S(t) = -16t^2 + 100t + C
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16:58:54 Query Section 6.3 #14 water balloon from 30 ft, v(t) = -32t+40
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RESPONSE -->
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17:15:41 How fast is the water balloon moving when it strikes the ground?
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RESPONSE --> The distance function is the antiderivative of the velocity function, so: S(t) = antiderivative of (-32t + 40) dt S(t) = -16t^2 + 40t + C feet Given S(0) = 30 feet: S(0) = 30 = -16(0^2) + 40(0) + C C = 30, thus: S(t) = -16t^2 + 40t + 30 feet Since S = 0 when the balloon strikes the ground: 0 = -16t^2 + 40t + 30 This equation has solutions of t = -.6 sec (approx.) and t = 3.1 sec (approx.) Since only positive time is possible, we us t= 3.1 sec. Substituting into the velocity function: v(3.1) = -32(3.1) + 40 = -59.2 feet/sec (i.e., 59.2 feet/second downward)
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17:20:12 How fast is the water balloon moving when it strikes the 6 ft person's head?
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RESPONSE --> Using the same methodology as the last response, with S=6 feet: 6 = -16t^2 +40t + 30 This equation has the solutions: t = -0.5 sec and t = 3 sec. Again only the positive t value is valid. Substituting into the velocity function: v(3) = (-32*3) + 40 = -56 feet/second (56 feet/second downward)
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18:28:57 What is the average velocity of the balloon between the two given clock times?
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RESPONSE --> I calculated the average velocity two ways: Solving the velocity function for t = 1.5 sec and t = 3 sec resulted in velocities of -8 ft/sec and -56 ft/sec, respectively. Since the velocity function is linear, the average velocity can be found by averaging the beginning and ending velocities for the interval, thus: v(average) = [-8+(-56)]/2 = -32 feet/sec I also found the acceleration function (the antiderivative of the velocity function), which is: a = -32 feet/sec (no constant is needed because of the next step). I then used the average value of f rule"" v(average) = [1/(b-a)]* antiderivative of a dt from a to b v(average) = [1/(3 - 1.5)] * antiderivative of -32 dt from 1.5 to 3 v(average) = (1/1.5) * (-32t) evaluated from 1.5 to 3 v(average) = (1/1.5)*[-96-(-48)] = -32 ft/sec (same result)
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18:30:12 What function describes the velocity of the balloon as a function of time?
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RESPONSE --> The velocity function is -32t + 40 feet/sec.
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18:31:05 Query Section 6.4 #19 (#18 3d edition) derivative of (int(ln(t)), t, x, 1)
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RESPONSE -->
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18:37:40 What is the desired derivative?
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RESPONSE --> Since the integral is from x to 1, the derivative is subtacted from zero, thus the desired derivative is - ln x
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19:02:08 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 --> If we look at the proof of the Second Fundamental theorem, it assumes that h is positive. However, if h is negative (as it would be if the variable is the lower limit), then the area represented by F(x+h) is less than the area represented by F(x). We then get the negative of F(x+h) - F(x). Working through the proof, we end up with -F'(x). Therefore, if the variable is the lower limit, the derivative is multiplied by negative 1.
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19:13:30 Why do we use something besides x for the integrand?
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RESPONSE --> We use a dummy variable in place of x in the integrand to distinguish it from the x used to identify a particular value of the variable in the integrand.
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19:13:50 Query Section 6.4 #26 (3d edition #25) derivative of (int(e^-(t^2),t, 0,x^3)
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RESPONSE -->
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19:16:53 What is the desired derivative?
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RESPONSE --> The desired derivative is found by substituting sin t for the variable x and using the chain rule, thus: cos(sin t)^2 * cos t = cos (sin^2 t)(cos t)
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19:28:21 How did you apply the Chain Rule to this problem?
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RESPONSE --> The chain rule is applied by substituting sin t for x, producing cos (sin t)^2 for the derivative of the outside function, then multiplying by the derivative of the inside function (sin t), which is cos t, thus: cos(sin^2 t)(cos t) I had difficulty with this problem, since I had only used the chain rule to find the derivatives of elementary functions, rather than antiderivatives. I initially thought that this problem was solved the same way as #15, with straight substitution. I'm still not sure I totally understand this use of the chain rule.
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19:30:12 Why was the Chain Rule necessary?
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RESPONSE --> The chain rule was necessary because we have a composite function: cosine of a sine.
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