#$&* course mth 174 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In a linear action we can say that t=d/(vf+vi)^2. In this case t= time d=distance vf=velocity final vi=velocity initial. I am not sure how to show this solution is true??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: ** Using s for the distance fallen we can translate Galileo's statement as follows: t = s / [ (vf + v0)/2 ]. An object accelerating for time t, starting with initial velocity v0 at t = 0, has velocity function v = v0 + a * t and position function s = .5 a t^2 + v0 t + s_0 where s_0 is the position at clock time t = 0. Both of these results are easily obtained, for the given initial condition that initial velocity is zero, by a straightforward integration of the acceleration function (to get the velocity function) then the velocity function (to get the position function). For given displacement s we can solve the position equation for t. The equation is rearranged to the form of a standard quadratic: .5 a t^2 + v0 t – s = 0, with solutions t = (-v0 +- sqrt(v0^2 + 2 a s) ) / a ). Substituting this into the velocity function we obtain the final velocity: Final velocity = v0 + a t = v0 + a * (-v0 +- sqrt(v0^2 + 2 a s) / a) = +- sqrt(v0^2 + 2 a s) . The average of the initial and final velocities is therefore ave. of init and final vel = (initial vel + final vel) / 2 = (v0 +- sqrt(v0^2 + 2 a s)) / 2. Traveling at this velocity for time t = (-v0 +- sqrt(v0^2 + 2 a s) ) / a ) the displacement will be Displacement = average velocity * time of travel = ( v0 +- sqrt(v0^2 + 2 a s)) (-v0 +- sqrt(v0^2 + 2 a s) ) / a ) / 2 The numerator is the product of the sum and the difference of two quantities and simplifies to 2 a s. The denominator is 2 a so the expression simplifies to just s. This confirms that the distance traveled is the same as the distance that would be traveled at the average of the initial and final velocities. Alternative solution: For uniform acceleration the velocity function can be expressed as v0 + a t. Integrating this function between clock times t1 and t2 we obtain the displacement s: s = 1/2 a (t2^2 – t1^2) + v0 ( t2 – t1). Dividing the integral (i.e., the displacement) by the length of the interval we get the average value of the function—i.e., the average velocity. The length of the interval is t2 – t1 so displacement / interval = Integral / interval = (1/2 a (t2^2 – t1^2) + v0 ( t2 – t1)) / (t2 – t1) = 1/2 a ( t2 + t1) + v0. The initial value of the velocity on the interval is v0 + a t1, the final velocity v0 + a t2, so the average of initial and final velocities is Ave of init and final velocity = (v0 + a t1 + v0 + a t2) / 2 = v0 + 1/2 a ( t1 + t2). This is identical to the expression obtained previously. Thus the average of the initial and final velocities is equal to the average velocity on the interval, and the result is proven. Still another solution: If an object is dropped from rest and falls for time t it will reach velocity vf = a t. So the average of its initial and final velocities will be Ave of init and final vel =(vf + v0) / 2 = (32 t + 0) / 2 = 16 t. The distance fallen is known to be s = 16 t^2. But this is exactly the distance the object would travel in time t if its average velocity was 16 t. Thus the two quantities are identical. A numerical example for a given s: When s = 100, then, t is the positive solution to 100 = 16 t^2. This solution is t = 2.5. The velocity function is v = a t. We have to find v when s = 100. When s = 100, we have t = 2.5, as just seen. So v = a t = 32 * 2.5 = 80, representing 80 ft/sec. Now we know that s = 100, vf = 80 and v0=0. Substituting into t = s / [ (vf + v0)/2 ] we get t = 100 / [ (80 + 0) / 2 ] = 100 / 40 = 2.5. This agrees with the t we got using s = .5 a t^2. ** Another argument: The average value of any linear function over an interval is equal to the average of its initial and final values over the interval. In a nutshell, because the v vs. t graph is linear, the average velocity is equal to the average of the initial and final velocities. Since time of fall = displacement / average velocity, it follows that time of fall = displacement / (ave of initial and final vel). This latter expression is just the time that would be required to fall at a constant velocity which is equal to the average of initial and final velocities. More rigorously: The graph is linear, so the area beneath the graph is the area of a triangle. The base of the triangle is the time of fall, and its altitude is the final velocity. By a simple construction we know that the area of the triangle is equal to the area of a rectangle whose length is equal to the base of the triangle, and whose width is equal to half the altitude of the triangle. It follows that the area of the triangle is equal to half the final velocity multiplied by the time interval. Since the initial velocity is 0, the average of initial and final velocities is half the final velocity. So the area of the triangle is the product of the time of fall and the average of initial and final velocities area beneath graph = time of fall * ave of init and final vel The area beneath the graph is equal to the integral of the velocity function over the time interval, which we know is equal to the displacement. So we have displacement = time of fall * ave of init and final vel, so that time of fall = displacement / ave of init and final vel. This leads to the same conclusion as above. The same argument, expressed more symbolically: A graph of v vs. t graph is linear. Over any time interval t = t1 to t = t1 + `dt, displacement is represented by the area of the corresponding trapezoid, which is d = (v1 + v2) / 2 * `dt. Solving for `dt we again obtain `dt = d / [ (v1+v2)/2 ]. **** query problem 7.1.18 integral of ` sqrt(cos(3t) ) * sin(3t)
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23:20:37 ** TYPICAL INCORRECT SOLUTION: (-2/3) (cos3t)^(3/2) INSTRUCTOR COMMENT: The derivative of cos(3t) is -3 sin(3t), wo the derivative of -2/3 (cos(3t))^(-3/2) is 3 sqrt(cos(3t) sin(3t). This solution is not consistent with the Chain Rule. To perform the integral use substitution. Very often the first substitution you want to try involves the inner function of a composite, and that is the case here. Cos(3t) is the inner function of the composite sqrt(cos(3t)), so we try using this as our substitution: w = cos (3t) dw = -3 sin (3t) so that sin(3t) = -dw / 3. Thus our expression becomes w^(1/2) * (-dw / 3). The integral of -1/3 w^(3/2) with respect to w is -1/3 * (2/3) w^(3/2), treating w^(1/2) as a power function. This simplifies to -2/9 w^(3/2) or -2/9 * (cos(3t))^(3/2). The general antiderivative is -2/9 * (cos(3t))^(3/2) + c, where c is an arbitrary constant.** DER 22:16:28
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I get the basic concept in writing out the verbalized equation but I am having some difficulty determining how to prove the equation??????
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I started by breaking this problem into individual components, u=x^3+1 du=3x^2 and x^2=2x du this can then be applied to solve for the equation X^2s^(x3+1)+c confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Substituting for the inner function of the composite: u = x^3 + 1 yields du = 3 x^2 dx, so x^2 dx = 1/3 du. This makes the integrand 1/3 e^u du. An antiderivative is then antiderivative = 1/3 e^u, or 1/3 e^(x^3+1) leading to the general indefinite integral indefinite integral = 1/3 e^(x^3+1) + c If we take the derivative of this function we do indeed obtain our original integrand. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I do not understand why the first term is 1/3 can you explain this portion of the problem to me????? ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: Section 7.1 Problem 9 7.1.9 (previously 7.1.35) Find an antiderivative of (t+1)^2 / t^2
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I started by distributing the ^2 over the quantities in perensetheses this resulted in t^2+2t^-1+t^-2 I then factored out t^2 from the equation to result in the new term of 1+ 2t^-1+t^-2 I then solved for the antiderivative to be t+2ln(t)-t^-1+c. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: expand (t+1)^2 to get t^2+2t+1. divide by t^2 to get 1 + 2t^-1 + t^-2. Each term of this function is a power function. Integrate term-by-term and add the integration constant to get the general antiderivative general antiderivative = t + 2ln(t) - t^-1 +C &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: Section 7.1 Problem 13 7.1.13 (previously 7.1.60). int(1/(t+7)^2, t, 1, 3) **** What did you get for the definite integral?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am okay with the initial part of the problem and I was fine breaking the problem into components and applying the variations of t to the problem but after I solve for the two u values I am lost on what to do. Can you explain the problem from the u=8&10 point to the end ?????????????? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 0
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Given Solution: This situation involves a composite of the power function 1 / z^2 and the linear function t + 7. The latter is the ‘inner’ function of the composite. We therefore substitute u = t+7 to get du = dt; limits t = 1 and t = 3 become u = 8 and u = 10. So the integral becomes integral( u^-2, u from 8 to 10). The integrand is the power function u^-2, or 1 / u^2. Its simplest antiderivative can be expressed as -u^-1 or -1/u. So the definite integral is change in antiderivative = -1/10 - (-1/8) = 1/8 - 1/10 = 1/40. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):