course Mth 174 wʅxľOaassignment #002
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18:11:05 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The second fundamental theorem looks to be quite useful, though a little confusing at times. The construction of antiderivatives between acceleration, velocity, and distance is interesting as well, and shows another good use of the ds/dt notation rather than just s'. I have to be really careful about general and specific answers as I forget the C quite often.
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18:15:16 Query Section 6.3 #8, ds / dt = -32 t + 100, s = 50 when t = 0
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RESPONSE --> First thing to do is the find the general equation then solve for C. General antiderivative: s = -16t^2 + 100t + C. s(0) = 50 = -16 (0)^2 + 100 (0) + C C = 50 Specific antiderivative: -16t^2 + 100t + 50
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18:15:25 What is the solution satisfying the given initial condition?
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RESPONSE --> -16t^2 + 100t + 50
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18:15:31 What is the general solution to the differential equation?
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RESPONSE --> -16t^2 + 100t + C
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21:24:49 Query Section 6.3 #14 water balloon from 30 ft, v(t) = -32t+40
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RESPONSE --> a) To find the distance formula, I need to find the antiderivative of the velocity function v = -32t + 40. General: -16t^2 + 40t + C Specific: -16t^2 + 40t + 30 b) Average velocity is found by adding the two endpoints together and dividing by two. I found the average velocity to be: -32 ft/sec c) Now I need to find the speed when d=6. First I find the time, which I found to be t=3. The velocity at t=3 is -56 ft/sec.
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21:25:39 How fast is the water balloon moving when it strikes the ground?
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RESPONSE --> When it strikes the ground: t=3.104 v=-59.328
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21:25:51 How fast is the water balloon moving when it strikes the 6 ft person's head?
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RESPONSE --> -56 ft/sec
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21:26:01 What is the average velocity of the balloon between the two given clock times?
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RESPONSE --> -32 ft/sec
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21:26:28 What function describes the velocity of the balloon as a function of time?
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RESPONSE --> Given function v = -32t + 40
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21:47:41 Query Section 6.4 #19 (#18 3d edition) derivative of (int(ln(t)), t, x, 1)
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RESPONSE --> I must confess I am rather puzzled by this one. The answer in the book is: -ln(t) The only thing I can guess is that the way this integral is set up the slope is a negative value. By moving from a positive number (which the ln function is only defined for) to 0, this makes the slope negative and therefore makes the derivative negative.
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21:47:52 What is the desired derivative?
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RESPONSE --> I guess -ln(t). ** In the following we'll use the format [int(f(t), t, c, x] to stand for 'the integral of f(t) with respect to t with lower limit c and upper limit x'. {}{}The 2d Fundamental Theorem says that d/dx [ int(f(t)), t, c, x ] = f(x). When applying this Theorem you don't integrate anything. {}{}The integral we are given has limits x (lower) and 1 (upper), and is therefore equal to -int(ln(t), t, 1, x). This expression is in the form of the Fundamental Theorem, with c = 1, and its derivative with respect to x is therefore - ln(x). {}{}Note that this Theorem is simply saying that the derivative of an antiderivative is equal to the original function, just like the derivative of an antiderivative of a rate-of-depth-change function is the same rate function we start with. **
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21:50:31 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 --> By reversing the sign of the derivative is the method that the book used, I think.
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21:59:47 Why do we use something besides x for the integrand?
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RESPONSE --> This is a mnemonic device to remind us that while x might be a variable, the x in the interval is a different number from the variable used in the integrand. To put it another way, x is an integration limit, while the variable in the integrand is a point in the area to be integrated. I think.
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22:29:02 Query Section 6.4 #26 (3d edition #25) derivative of (int(e^-(t^2),t, 0,x^3)
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RESPONSE --> 6.4.35 4th edition. My answer is e^-(x^6), but the book's answer is 3x^2e^-(x^6). I know I am missing something, but I am not sure what it is.
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22:29:14 What is the desired derivative?
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RESPONSE --> See prior answer. I don't really know.
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22:55:50 How did you apply the Chain Rule to this problem?
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RESPONSE --> Wait, I just realized the 3x^2 on the 3x^2e^-(x^6) came from x^3. However, I can't seem to find a good path to get from one end to the other. In other words, what happens to the -t^2? ** If we were finding (int(e^-(t^2),t, 0,x) the answer would just be e^-(x^2) by the Second Fundamental Theorem. However the upper limit on the integral is x^3. This makes the expression int(e^(-t^2),t,0,x^3) a composite of f(z) = (int(e^-(t^2),t, 0,z) and g(x) = x^3. Be sure you see that the composite f(g(x)) = f(x^3) would be the original expression int(e^(-t^2),t,0,x^3). g'(x) = 3x^2, and by the Second Fundamental Theorem f'(z) = e^-(z^2). Thus the derivative is therefore g'(x) f'(g(x))= 3x^2 * e^-( (x^3)^2 ). This one takes a bit of thinking.**
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22:56:05 Why was the Chain Rule necessary?
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RESPONSE --> Again, because I don't know the answer, I can't answer this.
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