course Mth 174 Question: Section 6.3 #14, ds / dt = -32 t + 100, s = 50 when t = 0 . Find the function s(t).......!!!!!!!!...................................
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Given Solution: * s' = 100 - 32t. Integrating with respect to t we obtain * s= 100t - 16t^2 + C. Since s = 50 when t = 0 we have * 50 = 100(0) - 16(0)^2 + C, which we easily solve to obtain * 50 =C. this into the expression for s(t) we have * s(t) = 100(t) - 16t^2 + 50
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: Section 6.3 #16 water balloon from 30 ft, v(t) = -32t+40
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********************************************* Your solution: A. s = -16t^2 + 40t + 30 = -8t^2+20t+15 =quadratic formula gives 3.12=t so v(3.12) = -60 Confidence Assessment:
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Given Solution: Since v(t) = s ' (t), it follows that the antiderivative of the v(t) function is the s(t) function so we haves * s(t) = -16 t^2 + 40 t + c. Since the building is 30 ft high we know that s(0) = 30. Following the same method used in the preceding problem we get * s(t) = - 16 t^2 + 40 t + 30. The water balloon strikes the ground when s(t) = 0. This occurs when -16 t^2 + 40 t + 30 = 0. Dividing by 2 we have -8 t^2 + 20 t + 15 = 0. The quadratic formula gives us t = [ -20 +- `sqrt( 400 + 480) ] / ( 2 * -8) or t = 1.25 +- sqrt(880) / 16 or t = 1.25 +- 29.7 / 16, approx. or t = 1.25 +- 1.87 or t = 3.12 or -.62. The solution t = 3.12 gives us v(t) = v(3.12) = -32 * 3.12 + 40 = -60, approx. The interpretation of this result is that when it strikes the ground the balloon is moving downward at 60 feet / second.**
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: How fast is the water balloon moving when it strikes the person's head?
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********************************************* Your solution: Quadratic formula gives us 3 for 24ft. So, v = -32(3) + 40 = -56ft/sec Confidence Assessment: 3
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Given Solution:
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12:13:49
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: What is the average velocity of the balloon between the two given clock times?
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********************************************* Your solution: difference in time = 1.5 sec. s(3) = -16(3^2) + 40(3) + 30 = 6 s(1.5) = -16(1.5^2) + 40(1.5) + 30 = 54 Avg. velocity = (6ft – 54ft)-(1.5sec) = -33ft/sec Confidence Assessment: 3
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Given Solution: Average Velocity=-32 m/s average velocity = change in position / change in clock time = (s(3) - s(1.5) ) / (3 sec - 1.5 sec) = (6 ft - 54 ft) / (1.5 sec) = -32 ft / sec. Alternatively since the velocity function is linear, the average velocity is the average of the velocities at the two given clock times: * vAve = (v(1.5) + v(3) ) / 2 = (-56 ft / sec + (-8 ft / sec) ) / 2 = -32 ft / sec. This method of averaging only works because the velocity function is linear.
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12:15:31
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: What function describes the velocity of the balloon as a function of time?
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********************************************* Your solution: v(t) = -32t + 40 Confidence Assessment: 3
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Given Solution: ** On this problem you are given s(0) = 30. So we have 30 = -16 * 0^2 + 100 * 0 + c or 30 = c. Thus c = 30 and the solution satisfying the initial condition is s(t) =- 16 t^2 + 100 t + 30. To find the clock time when the object strikes the ground, note that at the instant of striking the ground s(t) = 0. So we solve - 16 t^2 + 100 t + 30 = 0, obtaining solutions t = -.60 sec and t = 3.10 sec. The latter solution corresponds to our 'real-world' solution, in which the object strikes the ground after being released. The first solution, in which t is negative, corresponds to a projectile which 'peaks' at height 30 ft at clock time t = 0, and which was at ground level .60 seconds before reaching this peak. To find when the height is 6 ft, we solve - 16 t^2 + 100 t + 30 = 6, obtaining t = -.5 sec and t = 3.0 sec. We accept the t = 3.0 sec solution. At t = 3.0 sec and t = 3.10 sec the velocities are respectively v(3.104) = -32 * 3.10 + 40 = -59.2 and v(3.0) = -32 * 3 + 40 = -56, indicating velocities of -59.2 ft/s and -56 ft/s at ground level and at the 6 ft height, respectively. From the fact that it takes .104 sec to travel the last 6 ft we conclude that the average velocity during this interval is -6 ft / (.104 sec) = -57.7 ft / sec. This is how we find average velocity. That is, ave velocity is displacement / time interval, vAve = `ds / `dt. Since velocity is a linear function of clock time, a graph of v vs. t will be linear and the average value of v over an interval will therefore occur at the midpoint clock time, and will be equal to the average of the initial and final velocities over that interval. In this case the average of the initial and final velocities over the interval during which altitude decreased from 6 ft to 0 is vAve = (vf + v0) / 2 = (-59.2 + (-56) ) / 2 ft / sec = -57.6 ft / sec. This agrees with the -57.7 ft / sec average velocity, which was calculated on the basis of the .104 sec interval, which was rounded in the third significant figure. Had the quadratic equation been solved exactly and the exact value ( sqrt(55) + 5 - 3) / 2 of the time interval been used, and the exact corresponding initial and final velocities, the agreement would have been exact.
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12:16:06
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 3 I answered by stating the velocity function with respect to time since the question only asks for the function. Is the rest of the given solution required as part of the correct response? Self-critique Rating: OK ********************************************* Question: Section 6.4 #12 derivative of (int(ln(t)), t, x, 1). {}{}What is the derivative of this function?
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********************************************* Your solution: derivative of (Int(ln(t)), t, x, 1) = -ln(x) Confidence Assessment:
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Given Solution:
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12:23:36
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: Why do we use something besides x for the integrand?
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********************************************* Your solution: In order to differentiate between the antiderivative and the derivative. Confidence Assessment:
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Given Solution: ** In the statement of the Theorem the derivative is taken with respect to the variable upper limit of the interval of integration, which is different from the variable of integration. The upper limit and the variable of integration are two different variables, and hence require two different names. **
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12:24:24
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: Query Section 6.4 #18 derivative of (int(e^-(t^2),t, 0,x^3){}{}What is the desired derivative?
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********************************************* Your solution: This problem is not part of the assigned problem set. Confidence Assessment:
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Given Solution: ** 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 understand how the composite f(g(x)) = f(x^3) would be the original expression int(e^(-t^2),t,0,x^3). Now we apply the chain rule: g'(x) = 3x^2, and by the Second Fundamental Theorem f'(z) = e^-(z^2). Thus the derivative is g'(x) f'(g(x))= 3x^2 * e^-( (x^3)^2 ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This problem was not included in the assigned set of problems. Self-critique Rating: ********************************************* Question: Find the derivative with respect to x of the integral of e^(t^2) between the limits t and cos(x). ********************************************* Your solution: Derivative of the int.(e^(t^2) x, t, cosx) = Sinx*e^(cos^2(x)) Confidence Assessment: 3
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Given Solution: If we were finding (int(e^(t^2),t, x, 0) the answer would just be -e^(x^2) by the Second Fundamental Theorem (along with the reversal of integration limits and therefore sign). However the lower limit on the integral is cos(x). This makes the expression int(e^(t^2),t,cos(x), 3) a composite of f(z) = (int(e^(t^2),t, z, 3) and g(x) = cos(x). Be sure you see that the composite f(g(x)) = f(cos(x)) would be the original expression int(e^(t^2),t,cos(x),0). g'(x) = -sin(x), and by the Second Fundamental Theorem f'(z) = -e^(z^2) (again the negative is because of the reversal of integration limits). The derivative is therefore * g'(x) f'(g(x))= -sin(x) * (-e^( (cos(x))^2 ) = sin(x) e^(cos^2(x)). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: Query Section 6.3 #10 dy/dx = 2x+1
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11:18:57 Int.(2x+1) = (x^2)/2 + x + C student answer: (x^2)/2+x Instructor response: ** Good. This is an antiderivative of the given function. So is x^2 / 2 + x + c for any constant number c, because the derivative of a constant is zero. The general solution is therefore the function y(x) = x^2 / 2 + x + c . **
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11:18:58
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: What is the solution satisfying the given initial condition (part c)?
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for y(1) = 5, we get: 1/2 + 1 =C so C = 7/2 ** If for example you are given the initial condition y(0) = 12, then since you know that y(x) = x^2 / 2 + x + c, you have y(0) = 0^2 / 2 + 0 + c = 12. Thus c = 12 and your particular solution is y(x) = x^2 / 2 + x + 12. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: What three solutions did you graph, and what does your graph of the three solutions look like?
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********************************************* Your solution: (x^2/2)+x+0 (x^2/2)+x+1 (x^2/2)+x-1 The graphs were merely parabolas that varied by on either up or down at every point. Confidence Assessment: 3
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Given Solution: ** To graph the three solutions you could choose three different values of c. The graph of x^2 / 2 + x is a parabola; you can find its zeros and its vertex using the quadratic formula. The graph os x^2 / 2 + x + c just lies c units higher at every point than the graph of x^2 / 2 + x. So you get a 'stack' of parabolas. Be sure you work through the details and see the graphs. **
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11:20:51
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK