#$&* course Mth 174 Question: Section 7.6 Problem 3
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21:44:07 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We will be using 1/3 for all 3 new calculations since 10/30=1/3. We will also be using 2.346-4=-1.654 for all of our calculations since that is the “error”. (1/3)*-1.654= -.551, so our approximation for Left/Right would be 3.45. For Trap/Mid, we will us (1/3)^2 * -1.654, which equals -.184, so our approximation is 3.816. Finally, for Simpsons, we will us (1/3)^4 * -1.654 = -.02, so our approximation is 3.98. confidence rating #$&*: Very ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: LEFT and RIGHT approach the exact value in proportion to the number of steps used. MID and TRAP approach the exact value in proportion to the square of the number of steps used. SIMP approachs the exact value in proportion to the fourth power of the number of steps used. Using these principles we can work out this problem as follows: ** The original 10-step estimate is 2.346, which differs from the actual value 4.000 by -1.654. If the original estimate was done by LEFT then the error is inversely proportional to the number of steps and the n = 30 error is (10/30) * -1.654 = -.551, approximately. So the estimate for n = 30 would be -.551 + 4.000 = 3.449. If the original estimate was done by TRAP then the error is inversely proportional to the square of the number of steps and the n = 30 error is (10/30)^2 * -1.654 = -.184, approximately. So the estimate for n = 30 would be -.184 + 4.000 = 3.816. If the original estimate was done by SIMP then the error is inversely proportional to the fourth power of the number of steps and the n = 30 error is (10/30)^4 * -1.654 = -.020, approximately. So the estimate for n = 30 would be -.02 + 4.000 = 3.98. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: Section 7.6 Problem 5 problem 7.6.5 (previously problem 7.6.10) If TRAP(10) = 12.676 and TRAP(30) = 10.420, estimate the actual value of the integral. **** What is your estimate of the actual value and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (1/3)^2 is 1/9, and 10.42-12.676= -2.256. These numbers represent, respectively, is the error of TRAP(30) versus TRAP(10) and the difference between our trapezoid’s values. So, trap(30) is 8/9 the error of trap(10), which we then invert to get (9/8), and multiply by our difference, -2.256, and get a value of -2.538. We will add this to our TRAP(10) value, and get 10.138. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You need to use the inverse square proportionality of the error with the number of steps. Trap(30) is approximately (10/30)^2 = 1/9 of TRAP(10). So the difference 10.420 - 12.676 = -2.256 between TRAP(10) and TRAP(30) is approximately 8/9 of the error of TRAP(10). It follows that the error of TRAP(10) is 9/8 * -2.256 = -2.538. Our best estimate of the integral is therefore -2.538 + 12.676 = 10.138. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: Section 7.7 Problem 2 problem 7.7.2 (previously 7.7.12) integrate 1 / (u^2-16) from 0 to 4 if convergent YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: int(1/(u-4)(u+4)),0,4 1/8< 1/u+4 < ¼. So, 1/8int 1/u-4,0,4 < original integral < 1/4int 1/u-4,0,4 Int1/u-4 = ln|u-4| which = ln4 as u->0 since its in abs. value Lim as x->4 of the int 1/u-4,0,x = lim x->4 ln(x-4) – ln4 As x->4, the ln(x-4) approaches –infinite, so our integral diverges. confidence rating #$&*: Very ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 1 / (u^2-16) = 1 / [(u+4)(u-4)] . Since for 0 < x < 4 we have 1/8 < 1 / (u+4) < 1/4, the integrand is at most 1/4 times 1/(u-4) and at least 1/8 of this quantity, so the original integral is at most 1/4 as great as the integral of 1 / (u-4) and at least 1/8 as great. That is, 1/8 int(1 / (u-4), u, 0, 4) < int(1 / (u^2-4), u, 0, 4) < 1/4 int(1 / (u-4), u, 0, 4). Thus if the integral of 1 / (u-4) converges or diverges, the original integral does the same. An antiderivative of 1 / (u-4) is ln | u-4 |, which is just ln(4) at the limit u=0 of the integral but which is undefined at the limit u = 4. We must therefore take the limit of the integral of 1/(u-4) from u=0 to u=x, as x -> 4. The integral of 1 / (u-4) from 0 to x is equal to ln(x-4) - ln(4). As x -> 4, ln(x - 4) approaches -infinity, Thus the integral diverges. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: Section 7.7 Problem 7 problem 7.7.7 (previously 7.7.30) rate of infection r = 1000 t e^(-.5t) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we will need the 2nd derivative test for this, I just derived at the start. f(t)=1000t e^(-.5t), f’(t)=1000e^(-.5t) – 500t e^(-.5t), f’’(t)= -1000e^(-.5t) +250t e^(-.5t). f’(t)/e^(-.5t)=0 will give us our relative maximum, which is also our fastest rate of sickness. F’(t)/e^(-.5t)=0 = 1000-500t=0, so t=2 To describe the graph, we will set f’’=0 which is a true statement when t=4, this makes t=4 a Point of Inflection, so our graph is c.down from –inf, to 4 and c.up from 4 to inf. Finally, to find our total sick, we will integrate 1000int t*e^(-.5t) dt. This is a uv substitution where u=t and v=-2e^(-.5t) Our integral will solve to 1000[-2te^(-.5t) – 4e^(-.5t) Putting this on the interval from 0 to x, we will subtract f(x)-f(0) to find out total sick. 1000[-2te^(-.5x) -4e^(-.5x)] -1000[-2(0)e^(-.5(0)) – 4e^(-.5(0))] This solves to 1000e^(-.5x)[-2t-4] + 4000…..as x->inf, 1000e^(-.5x)[-2t-4] approaches 0, so we are left with 4000 as our maximum number sick. confidence rating #$&*: I sure hope its right……. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** First graph the function using standard graphing techniques from first-semester calculus: This graph increases at first as you move to the right from t = 0. However e^(-.5 t) eventually approaches zero much faster than t increases so the graph has an asymptote at the positive t axis. So it increases for small positive t but eventually returns almost to the t axis, and it can't be strictly increasing. Its concavity changes from downward (negative) for small positive t to upward for larger t; the point at which the concavity changes is important. We use the standard technique from first-semester calculus to find the point at which this function maximizes. The first derivative is dr/dt = 1000 e^(-.5 t) - 500 t e^(-.5 t). Setting this derivative equal to 0 we get • 1000 e^(-.5 t) - 500 t e^(-.5 t) = 0; dividing through by e^-.5 t we get the equation 1000 - 500 t = 0, which is easily solved to obtain t = 2. A first-or second-derivative test confirms that the t = 2 graph point is a relative maximum. Concavity is determined by the second derivative r'' = e^(-.5 t) [ -1000 + 250 t ], which is 0 when t = 4. This is a point of inflection because the second derivative changes from negative to positive at this point. So the function is concave downward on the interval (-infinity, 4) and concave upward on (4, infinity). The first derivative has a critical point where the second derivative is zero. This occurs at x = 4, which was identified in the preceding paragraph as the point of inflection for the original function. Since the second derivative goes from negative to positive, this point is a minimum of the first derivative. The first derivative is a decreasing function from t = 0 to t = 4 (2d derivative is negative) and is then an increasing function with asymptote y = 0, the x axis, which it approaches through negative values. Its maximum value for t >= 0 is therefore at t = 0. ** People are getting sick the fastest when the rate of infection is highest. This occurs at the relative maximum of the rate function, which was found above to occur at t = 2. Thus people are getting sick the fastest 2 days after the epidemic begins. To find how many people get sick during a time interval, you integrate the rate function over that interval. In this case the interval doesn't end; so you need to integrate the rate function r = 1000 t e^(-.5t) from t = 0 until forever, i.e., from t = 0 to t = infinity. An antiderivative of the function is F(t) = 1000 int ( t e^(-.5 t)) = 1000 [ -2 t e^(-.5t) - int ( e^(-.5 t) ) ] = 1000 [ -2 t e^(-.5 t) - 4 e^(-.5 t) ]. Integrating from 0 to x gives F(x) - F(0) = 1000 [ -2 t e^(-.5 x) - 4 e^(-.5 x) ] - 1000 [ -2 * 0 e^(-.5 *0 ) - 4 e^(-.5 * 0 ) ] = 1000 e^-(.5 x) [ -2 t - 4 ] - (-4000). As x -> infinity, e^-(.5 x) [ -2 t - 4 ] -> 0 since the exponential will go to 0 very much faster than (-2 x - 4) will approach -infinity. This leaves only the -(-4000) = 4000. ** The calculator is fine for checking yourself, but you need to use the techniques of calculus to determine inflection points, maxima, minima etc.. The careful use the calculator to enhance rather than replace mathematical understanding. I get a lot of students in these courses who are now at 4-year institutions and who have taken courses based on the graphing calculator, or even TI-92, and many of them tend to have a very difficult time in courses that don't permit them, and in courses were mathematical understanding is required. ** ** You have to use the techniques of calculus to determine these behaviors. Plugging values in won't show you the exact location of intercepts, maxima, minima, etc.. ** STUDENT QUESTION I didn’t know where to go with the antiderivative but I think I understand your conclusion on that as well. INSTRUCTOR RESPONSE The infection is the rate-of-change function, so the antiderivative is the change-in-amount function. Specificly we have the rate of change of the number of people who are or have been sick, with respect to clock time. The 'amount' is the number of people, so the antiderivative function is the change in the number of people (i.e., in the number who have been or are sick). The definite integral between two clock times therefore tells you how many people are or have been sick between those clock times. If we integrate from some clock time from the initial instant to infinity, we get the total number of people who will get sick. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating "