course MTH 174 2:55 p.m. 6/11/2010 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htmYour solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In terms of time we get t= s/((vmax+v0)/2), where t= time, s= position, vmax is maximum velocity and v0 is initial velocity. If acceleration is uniform we have v(t)= a*t + v0. Looking at a graph of this function we see it is linearly increasing and that the distance traveled is the area under the line. So the distance traveled in any interval t1 to t2, where t2 can be represented as t1+dt, is found by finding the trapezoidal area of A= d = ((v(t1)+v(t2))/2) * dt. Solving for dt gives dt= d/((v(t1)+v(t2))/2). Which was the original expression we formulated since dt = time interval, v(t1) = v0 and v(t2)= vmax and d= s. 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, assuming that s = 0 at 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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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int(sqrt(cos3t)*sin(3t). We use u= cos 3t and du= -3*sin 3t dt. Now we have Int(-1/3 sqrt(u)*du) Evaluating gives: -2/9 (cos 3t)^3/2 + c. ** 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.** ......!!!!!!!!................................... Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: 7.1.6 (previously 7.1.21) antiderivative of x^2 e^(x^3+1)
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using u= x^3 + 1, then du= 3x^2 dx. Substituting we get: Int( 1/3 e^u du) = 1/3 e^u+ c= 1/3 e^(x^3+1)+ c confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: Substituting for the inner function of the composite: • u = x^3 + 1 yields • du = 3 x^2 dx, so • x^2 dx = 1/3 du. This makes the integrand 1/3 e^u du. An antiderivative is then • antiderivative = 1/3 e^u, or 1/3 e^(x^3+1) leading to the general indefinite integral • indefinite integral = 1/3 e^(x^3+1) + c If we take the derivative of this function we do indeed obtain our original integrand. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* 3 ********************************************* Question: 7.1.9 (previously 7.1.35) Find an antiderivative of (t+1)^2 / t^2
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My Solution: This is equivalent to (t+1)*(t+2)*(t^-2)= (t^2+2t+1)t^-2= 1+2t^-1+ t^-2. Integration then gives an antiderivitive to t+2 ln |x| + t^-1 + c 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 #$&* ********************************************* Question: 7.1.13 (previously 7.1.60). int(1/(t+7)^2, t, 1, 3) **** What did you get for the definite integral?
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Your solution: Int((t+7)^-2, t, 1, 3). First: -(t+7)^-1, evaluating over the interval gives: -0.1- -0.125 = 0.025. Confidence Assessment: 3 Given Solution: This situation involves a composite of the power function 1 / z^2 and the linear function t + 7. The latter is the ‘inner’ function of the composite. We therefore substitute u = t+7 to get du = dt; limits t = 1 and t = 3 become u = 8 and u = 10. So the integral becomes integral( u^-2, u from 8 to 10). The integrand is the power function u^-2, or 1 / u^2. Its simplest antiderivative can be expressed as -u^-1 or -1/u. So the definite integral is • change in antiderivative = -1/10 - (-1/8) = 1/8 - 1/10 = 1/40. Self-critique (if necessary): Didn't mess with substitution as integral was fairly straightforward. Self-critique rating #$&* 3 ********************************************* Question: 7.1.16 (previously 7.1.86). World population P(t) = 5.3 e^(0.014 t). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For 1990: p(0)= 5.3 billion and 2000: p(10)= 6.1 billion. To find the average population over an interval we use the mean value theorem. 1/(10-0) Int(p(t), t, 0, 10) First we get: 1/10*(379*e^(0.014t)) then evaluate over the interval to get 5.695 billion. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: What were the populations in 1990 and 2000? What is the average population between during the 1990's and how did you find it?
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In 1990, t = 0 so the population is P(0) = 5.3 billion In 2000, t = 0 so the population is P(10) = 6.1 billion To find the average population we integrate the population function over the 10-year interval, then divide by the 10 years: An antiderivative of 5.3 e^(0.014 t) is found by letting u equal the ‘inner function’ of the composite: u = .014 t so du = .014 dt and dt = du / .014. Thus the expression 5.3 e^(0.014 t) dt becomes 5.3 e^u * du / .014. An antiderivative of e^u with respect to u is just e^u, so the antiderivative of the function is • 5.3 e^u / .014 = 380 e^u, approximately. In terms of t this is • 380 e^(.014 t). At t = 0 this antiderivative would have value about 380. At t = 0 your antiderivative would have value about 435. • The change in the value of the antiderivative is therefore about 435 – 380 = 55. • The change in the antiderivative is the definite integral of the function. The average value of the function is equal to its definite integral divided by the time interval: • average value = 56.89 / 10 = 5.689. This is quite close to, but a little less than the average of the initial and final populations. It is less because the exponential function is concave upward, and the population curve therefore 'dips' below the straight line connecting the initial and final points. However the curvature on this interval is not pronounced, and the function never 'dips' far below the straight line, so the average is close to the straight-line average you gave in your initial solution. INCORRECT IF NOT UNREASONABLE ANSWER: The average population seems as if it would be the average of the two, therefore 6.1 + 5.3/ 2 = 5.7 billion INSTRUCTOR RESPONSE: Review the definition of the average value of a function: • The average value of a function over an interval is equal to its integral over that interval, divided by the length of the interval. Now if the function is linear, then the average of its initial and final values is equal to its average value (see the above given solution to the Galileo problem). However: • If the function is not linear, it is very unlikely that the average of its initial and final values is equal to its average value • If the function has nonzero positive or negative concavity on the entire interval it will never have an average value equal to the average of its initial and final values. The exponential function is not linear, and is in fact concave upward on this interval, so averaging initial and final values on the interval won't work here. You have to integrate the function and divide by the 10-year interval of integration.
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COMMON INTEGRATION ERRORS: integral( x^2 * e^(x^3+1) ) = integral( x^2) * integral (e^(x^3 + 1) ) = x^3 / 3 * e^(x^3 + 1) INSTRUCTOR RESPONSE: You have essentially assumed that the integral of a product function is equal to the product of the integrals. The integral is of the form • integral (f * g) with f(x) = x^2 and g(x) = e^(x^3 + 1). However integral (f * g) is not the same as integral (f) * integral(g): The derivative of our antiderivative function must equal the original integrand. That is, the derivative of integral(f * g) must be f * g. However the derivative of integral (f) * integral(g) is expressed by the product rule as (integral (f) * integral(g)) ' = ( integral (f) ) ' * integral(g) + integral (f) * (integral(g)) ' = f * integral(g) + g * integral(f). This is much different than the original integrand f * g, so integral ( f * g) is not generally the same as integral(f) * integral(g). Without going into the details, we'll also assert that • integral(f / g) is not the same as integral(f) / integral(g), which you can if you wish verify for yourself by taking the derivative of integral(f) / integral(g), using the quotient rule. What you get is nothing like the original integrand f / g. So for example integral( (t+1)^2 / t^2 ) is not the same as integral ( t + 1) / integral (t^2). Note also that, while the integral of e^x with respect to x is e^x + c, the integral of (e^(x^3 + 1) ) with respect to x is NOT the same as e^(x^3 + 1) + c: • The derivative of e^(x^3 + 1) is 3 x^2 * e^(x^3 + 1), not e^(x^3 + 1). So e^(x^3 + 1) is not an antiderivative of e^(x^3 + 1). "