#$&* course Mth 174 Question: Problem 4 section 6.3.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: s’= -32t+100, so we can then integrate and get s= -16t^2+100t+C, if we plug in our given values, we get 50= -16(0)^2 + 100(0) + C, which means C=50. Our solution is s= -16t^2 +100t + 50. confidence rating #$&* Very ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: Problem 6 section 6.3 problem 6.3.17 was 6.3.6 (previously 6.3 #16) water balloon from 30 ft, v(t) = -32t+40
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: s= -16t^2 +40t + C, and C=30ft. So, we can then say 0= -16t^2 +40t + 30. After a bit of calculator work, we know t= -.604 or 3.1. Time cannot be negative, so t=3.1 is our correct time. If we plug time back into our equation, we get v(3.1)= -32(3.1)+40 = -59.2 ft/s. confidence rating #$&* Very ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: How fast is the water balloon moving when it strikes the person's head?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The balloon has to fall 24 feet, so -16t^2 + 40t + 24 = 0. We can then find t, which is 3. V(3)= -32(3) + 40, so v(3) = -56 ft/s. confidence rating #$&* Very ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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12:13:49
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: What is the average velocity of the balloon between the two given clock times?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, we need to find the position at 1.5 seconds, which is the anti-derivative of the velocity function, so s= -16(1.5)^2 + 40(1.5) + 30 which gives us s(1.5) = 54. vAve = ‘ds/’dt, so (6ft. - 54 ) / (3sec – 1.5 sec). So, vAve = -32ft/s. confidence rating #$&* Very ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: What function describes the position of the balloon as a function of time? How can this function be used to answer the various questions posed in this problem?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: s= -16t^2 + 32t + C. Since we know s(0)=30, we know C=30. This let us find t= 3.1, for an object dropped off a 30ft building to hit the ground. We can use the same logic for any distance fallen for any object (like to our 6ft tall person’s head). We can also use the same equations if we’re given any time interval.
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Given Solution: ** On this problem you are given s(0) = 30. So we have 30 = -16 * 0^2 + 40 * 0 + c or 30 = c. Thus c = 30 and the solution satisfying the initial condition is s(t) =- 16 t^2 + 40 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 + 40 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 + 40 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): I think my answer is right, but I’m not sure if it suffices or not.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Well, the derivative “undoes” the integral, so we get ln1 – lnx which is equivalent to –ln x. confidence rating #$&* Very ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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12:23:36
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: Why do we use something besides x for the integrand?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The variable for the limit of integration can’t be the same variable being integrated because they stand for two different things. confidence rating #$&* Very ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We will plug our upper limit into t, then multiply that by the derivative of the upper limit. E^-((x^3))^2 * 3x^2. If we follow the same method to subtract our lower limit, we get 0, so I ignored it. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If we were finding (int(e^-(t^2),t, 0,x) the answer would just be e^-(x^2) by the Second Fundamental Theorem. However the upper limit on the integral is x^3. • This makes the expression int(e^(-t^2),t,0,x^3) a composite of f(z) = (int(e^-(t^2),t, 0,z) and g(x) = x^3. • Be sure you understand how the composite f(g(x)) = f(x^3) would be the original expression int(e^(-t^2),t,0,x^3). Now we apply the chain rule: g'(x) = 3x^2, and by the Second Fundamental Theorem f'(z) = e^-(z^2). Thus the derivative is g'(x) f'(g(x))= 3x^2 * e^-( (x^3)^2 ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: Find the derivative with respect to x of the integral of e^(t^2) between the limits cos(x) and 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: d/dx(int.(e^(t^2)),x,cosx,3 = (e^9 * 0) – (e^(cos^2 x)) * -sinx). With simplification, we get sinx*e(cos^2 x). confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we were finding (int(e^(t^2),t, x, 3) the answer would just be -e^(x^2) by the Second Fundamental Theorem (along with the reversal of integration limits and therefore sign). However the lower limit on the integral is cos(x). This makes the expression int(e^(t^2),t,cos(x), 3) a composite of f(z) = (int(e^(t^2),t, z, 3) and g(x) = cos(x). Be sure you see that the composite f(g(x)) = f(cos(x)) would be the original expression int(e^(t^2),t,cos(x),3). g'(x) = -sin(x), and by the Second Fundamental Theorem f'(z) = -e^(z^2) (again the negative is because of the reversal of integration limits). The derivative is therefore • g'(x) f'(g(x))= -sin(x) * (-e^( (cos(x))^2 ) = sin(x) e^(cos^2(x)). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: Section 6.3 Problem 5 6.3.11 was 6.3.5 (was 6.3 #10) dy/dx = 2x+1 What is the general solution to this differential equation?
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11:18:57 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 + x +c confidence rating #$&* Very ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: student answer: x^2 / 2+x Instructor response: ** Good. This is an antiderivative of the given function. So is x^2 + x + c for any constant number c, because the derivative of a constant is zero. The general solution is therefore the function y(x) = x^2 + x + c . **
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11:18:58
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: What is the solution satisfying the given initial condition (part c)?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t see a 6.3 #5, but I know you mean a y(#)= # type problem. If we know a value to set our equation equal to, and a value for X, we can solve for C and get a specific solution. confidence rating #$&* Relatively…based on not knowing the problem until I read the answer. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If for example you are given the initial condition y(0) = 12, then since you know that y(x) = x^2 / 2 + x + c, you have y(0) = 0^2 / 2 + 0 + c = 12. Thus c = 12 and your particular solution is y(x) = x^2 + x + 12. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: What three solutions did you graph, and what does your graph of the three solutions look like?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Well, since x^2 + x is a parabola opening upward, adding “c” would just make the same parabola with the vertex moved accordingly. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Very ** To graph the three solutions you could choose three different values of c. The graph of x^2 + x is a parabola; you can find its zeros and its vertex using the quadratic formula. The graph of x^2+ x + c just lies c units higher at every point than the graph of x^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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