#$&* course MTH 173 8/7/11 022. `query 22
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Given Solution: `aThe derivative is y ' = e^x - 10, which is zero when e^x - 10 = 0, or e^x = 10. This occurs at x = ln(10), or approximately x = 2.30258. The second derivative is just y '' = e^x, which is always positive. The graph is therefore always concave up. Thus the 0 of the derivative at x = ln(10) implies a minimum of the function at (ln (10), 10 - 10 ln(10)), or approximately (2.30258, -13.0258). Since e^x -> 0 for large negative x, e^x - 10 x will be very close to -10 x as x becomes negative, so for negative x the graph is asymptotic to the line y = - 10 x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhere is the first derivative positive, where is it negative and where is it zero, and how does the graph show this behavior? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&& ******************* y' = e^x - 10 negative up until its zero at approx. x =2.3, then positive to infinity. y = e^x -10x tells us this because it is decreasing (neg. slope) until the x=2.3 point, then increasing onward (positive slope). &&&&&&&&&&&&&&&&&&& ******************* confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The derivative e^x - 10 isn't always positive. For example when x = 0, e^x - 10 = 1 - 10 = -9. The value of the derivative e^x - 10 at a given value of x is equal to the slope of the graph of the original function, at that value of x. When this derivative is negative, as it is for x < 2.3 or so, the slope of the graph of e^x - 10 x will be negative so the graph will be decreasing. At the point where e^x - 10 becomes 0, indicating 0 slope for the graph of e^x - 10x, that graph reaches its minimum. For x > 2.3 (approx) the graph of e^x - 10 x will be increasing. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhere is the second derivative positive, where is in negative where is its zero, and how does the graph show this behavior? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&& ******************* The second derivative is always positive because the slope of the graph is always increasing. &&&&&&&&&&&&&&&&&&& ******************* confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The second derivative, e^x, is always positive. So the derivative is always increasing and the graph of the original function is always concave upward. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery (problem has apparently been edited out of the 5th edition; it's an excellent problem as students using the 5th edition should attempt it) problem 4.1.29 (3d edition 4.1.26) Given the function y = f(x) = a x e^(bx) What are the values of a and b such that f(1/3) = 1 and there is a local maximum at x = 1/3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&& ******************* assuming a and b are constants, f'(x) = a((1)(e^(bx) + b(x)(e^(bx)) = ae^(bx)(1 + bx) and f(1/3) = 1, so 1 = a(1/3)e^(b*1/3) 3/(e^(b*1/3)) = a and 0 = ae^(b(1/3))(1 + b(1/3)) 0 = (3/(e^(b*1/3)))(e^(b(1/3))(1 + b(1/3))) 0 = 3(1 + (1/3)b) 0 = 3 + b -3 = b then 3/(e^(-3*1/3)) = a 3/e^-1 = a 3*e = a 8.15 =~ a &&&&&&&&&&&&&&&&&&& ******************* confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** At a local maximum the derivative is zero so we have y'(1/3)=0. y ' (x) = a e^(bx) + a b x e^(bx) so if x = 1/3 we have ae^(1/3 b)+a b * 1/3 * e^(1/3b)=0. Factoring out ae^(1/3 * b) we get 1+1/3 b=0 which we easily solve for b to obtain b = -3. So now the function is y = a x e^(-3 x). We also know that f(1/3) = 1 so a * 1/3 e^(-3 * 1/3) = 1 or just a / 3 * e^-1 = 1, which is the same as a / ( 3 * e) = 1. We easily solve for a, obtaining a = 3 * e. So the function is now y = a x e^(-3x) = 3 * e * x e^(-3x). We can combine the e and the e^(-3x) to get y = 3 x e^(-3x+1). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 4.2.23 (3d edition 4.2.18) family x - k `sqrt(x), k >= 0. Explain how you showed that the local minimum of any such function is 1/4 of the way between its x-intercepts. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&& ******************* 0 = x - ksqrt(x) x(1 - k(x)^-.5) = 0 x = 0 and 1 - k(x)^-.5 = 0 -k(x)^-.5 = -1 (x)^-.5 = 1/k ((x)^-.5)^-2 = (1/k)^-2 x = k^2 so x = 0, k^2 y' = 1 - .5k(x)^-.5 0 = 1 - .5k(x)^-.5 -1 = - .5k(x)^-.5 2 = k(x^-.5) 2/k = (x^-.5) (2/k)^-2 = (x^-.5)^-2 (1/4)k^2 = x y'' = .25k(x)^-1.5 y''(.25k^2) = .25k(.25k^2)^-1.5 = positive &&&&&&&&&&&&&&&&&&& ******************* confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a To find the local minimum we use the first-derivative test. The first derivative of y = x - k `sqrt(x) is y ' = 1 - k / (2 sqrt(x) ). y ' = 0 when 1 - k / (2 sqrt(x) ) = 0. We solve this equation: 1 - k / (2 sqrt(x) ) = 0. Solving for x we first multiply through by the common denominator, obtaining 2 sqrt(x) - k = 0 so that sqrt(x) = k / 2 and x = k^2 / 4 Checking to be sure this is a maximum we take the second derivative, obtaining y '' = k / (4 x^(3/2) ). which is positive for x = k^2 / 4, showing that the critical point found previously yields a minimum. The zeros of the function occur when the function 1 - k / (2 sqrt(x) ) is equal to zero, giving us the equation x - k `sqrt(x) = 0. This is easily solved for x. We get solutions x = 0 or x = k^2. The x coordinate of the minimum, x = k^2 / 4, therefore does lie 1/4 of the way between the zeros x = 0 and x = k^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat are the x-intercepts of f(x) = x - k `sqrt(x) and how did you find them? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&& ******************* x = 0, k^2 since x - k sqrt(x) = 0 x(1 - k(x)^-.5) = 0 x = 0 and 1 - k(x)^-.5 = 0 -k(x)^-.5 = -1 (x)^-.5 = 1/k ((x)^-.5)^-2 = (1/k)^-2 x = k^2 &&&&&&&&&&&&&&&&&&& ******************* confidence rating #$&*:OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** This problem requires the use of derivatives and other calculus-based tools (which do not include the graphing calculator). You have to find intercepts, critical points, concavity, etc.. We first find the zeros: x-intercept occurs when x - k `sqrt(x) = 0, which we solve by by dividing both sides by `sqrt(x) to get `sqrt(x) - k = 0, which we solve to get x = k^2. So the x intercept is at (k^2, 0). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhere is(are) the critical point(s) of x - k `sqrt(x) and how did you find it(them)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&& ******************* y' = 1 - .5k(x)^-.5 0 = 1 - .5k(x)^-.5 -1 = - .5k(x)^-.5 2 = k(x^-.5) 2/k = (x^-.5) (2/k)^-2 = (x^-.5)^-2 (1/4)k^2 = x y'' = .25k(x)^-1.5 y''(.25k^2) = .25k(.25k^2)^-1.5 = positive, so minimum &&&&&&&&&&&&&&&&&&& ******************* confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We now find the critical point, where f ' (x) = 0: If f(x) = x - k `sqrt(x) then f ' (x) = 1 - k / (2 `sqrt(x)). f'(x) = 0 when 1 - k / (2 `sqrt(x)) = 0. Multiplying both sides by the denominator we get 2 `sqrt(x) - k = 0 so that x = k^2 / 4. The critical point occurs at x = k^2 / 4. If k > 0 the second derivative is easily found to be positive at this point, so at this point we have a minimum. The derivative is a large negative number near the origin, so the graph starts out steeply downward from the origin, levels off at x = k^2 / 4 where it reaches its minimum, then rises to meet the x axis at (k^2, 0). We note that the minumum occurs 4 times closer to the origin than the x intercept. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery (this problem has been omitted from the 5th edition; however 5th edition students should solve it) problem 4.2.37 (3d edition 4.2.31) U = b( a^2/x^2 - a/x). What are the intercepts and asymptotes of this function? At what points does the function have local maxima and minima? Describe the graph of the function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&& ******************* vert. asymptotes where x = 0 since you can't have a 0 in the denominator. horizontal asymp. at y = 0 (if you add them degree of highest numerator is one, highest denominator is 2) x-int.: 0 = b(a^2 - ax)/x^2 0 = b(a^2 - ax) 0 = ba(a - x) 0 = a - x -a = x y-int.: none, since x can't be 0 U' = b(-2a^2/x^3 + a/x^2) 0 = b(-2a^2/x^3 + a/x^2) 0 = (-2a^2/x^3 + a/x^2) 0 = -2a^2/x + a -a = -2a^2/x -ax = -2a^2 x = 2a U(2a) = -b/4, so (2a, -b/4) &&&&&&&&&&&&&&&&&&& ******************* confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We use the standard techniques to analyze the graphs: To find the intercepts we set x = 0 (giving us the 'vertical' intercepts), and we set the function equal to zero (to get the 'horizontal' intercepts). To find vertical asymptotes we check near points where the denominator is zero. Alternatively we can look for points where the derivative approaches infinity. To find horizontal asymptotes we consider what happens when x has a very large magnitude. To find maxima and minima we find critical points and a first- or second-derivative test. If the domain is bounded we also consider endpoints of the domain. The derivative of U = b( a^2/x^2 - a/x) is dU/dx = b(-2a^2 / x^3 + a / x^2), which is 0 when - 2a^2 / x^3 + a / x^2 = 0. This is easily solved: we get -2a^2 + a x = 0 so x = 2 a^2 / a = 2a. A first-or second-derivative test tells us that this point is a minimum. The function therefore reaches its minimum at (2a, -b/4). The parameter a determines the x coordinate of the minimum, and the parameter b independently determines the value of the minimum. Adjusting the constant a changes the x-coordinate of the minimum point, moving it further out along the x axis as a increases, without affecting the minimum. Adjusting b raises or lowers the minimum: larger positive b lowers the minimum, which for positive b is negative; and an opposite effect occurs for negative b. Therefore by adjusting two parameters a and b we can fit this curve to a wide variety of conditions. As x -> +-infinity, both denominators in the U function get very large, so that U -> 0. Thus the graph will have the x axis as an asymptote at both ends. As x -> 0, a^2 / x^2 will approach -infinity for positive a, while -a / x will approach + infinity. However, since near x=0 we see that x^2 is much smaller than x so its reciprocal will be much larger, the a^2 / x^2 term will dominate and the vertical asymptote will therefore be the positive y axis. As x gets large, x^2 will increasingly dominate x so the -a/x term will become larger than the a^2 / x^2 term, and if a is positive the graph must become negative and stay that way. Note that a graph of a smooth continuous function that becomes and stays negative while eventually approaching 0 is bound to have a minimum. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 4.2.36 was 4.3.31 (3d edition 4.3.29) f(v) power of flying bird vs. v; concave up, slightly decreasing for small v; a(v) energy per meter. Why do you think the graph has the shape it does? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&& ******************* In general, you would be using energy at a greater rate at a greater velocity due to air flow resistance, but when the velocity gets extremely low, you would also need greater flow of energy to stay aloft. &&&&&&&&&&&&&&&&&&& ******************* confidence rating #$&*:OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** the graph actually doesn't give energy vs. velocity -- the authors messed up when they said that -- it gives the rate of energy usage vs. velocity. They say this in the problem, but the graph is mislabeled. The graph says that for high velocities the rate of energy usage, in Joules / second, increases with increasing velocity. That makes sense because the bird will be fighting air resistance for a greater distance per second, which will require more energy usage. To make matters worse for the bird, as velocity increases the resistance is not only fought a greater distance every second but the resistance itself increases. So the increase in energy usage for high velocities isn't too hard to understand. However the graph also shows that for very low velocities energy is used at a greater rate than for slightly higher velocities. This is because low velocities imply hovering, or near-hovering, which requires more energy than the gliding action the bird achieves at somewhat higher velocities. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!