#$&* course Mth 173 7/20 4:06 pm 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): Im not sure why you calculated the y coordinate the way you did? (ln(10), 10 - 10 ln(10)) ???? can you explain this to me? I think I understand the asymptote part ------------------------------------------------ Self-critique Rating: 1
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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, so it is always concave up for all x values 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: To have a local max the derivative must equal 0 So f(1/3) = 0 y ' (x) = a e^(bx) + a b x e^(bx) y (1/3) = ae^(1/3 b)+a b * 1/3 * e^(1/3b)=0 1+1/3 b=0 (after factoring out the variables (ae^(1/3 b) ) b = -3. The function is now y = axe^( -3 x) we know that f(1/3)=1 so: a * 1/3 e^(-3 * 1/3) = 1 or: a / 3 * e^-1 = 1 a / ( 3 * e) = 1 a = 3 * e. So the function now becomes: y = a x e^(-3x) y=3 * e * x e^(-3x) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). ** STUDENT QUESTION I understand the derivative rules used, but the simplification is a mystery to me. INSTRUCTOR RESPONSE You are adept with algebra, and given the equations a e^(1/3 b) + a b * 1/3 * e^(1/3b) = 0 and a * 1/3 e^(-3 * 1/3) = 1. I believe you could easily solve them without looking at the given solution. I could try to explain the given solution further, but I'm afraid in this case that would further complicate the issue. So rather than trying to follow the given solution, try the following: Write down, on paper, the equation a e^(1/3 b) + a b * 1/3 * e^(1/3b) = 0 and solve it for b. If you didn't get b = -3, then answer these questions: What do you get when you divide both sides by a? What do you get when you then divide both sides by e^(1/3 b)? What is the solution to the resulting equation? Then take another look at the given solution. If you don't understand it after working it out yourself, submit an outline of what you did and I'll be happy to comment. Then do the same with the equation a * 1/3 e^(-3 * 1/3) = 1 Solve the equation for a. Note that e^(-3 * 1/3) = e^(-1) = 1 / e. STUDENT QUESTION I am still having trouble with problems like this that I am given a formula and a value at certain points. Where can I go in the notes to try to relearn this concept????? INSTRUCTOR RESPONSE It's a question of problem solving, and as I often state it takes practice. In this situation there are two unknown quantities a and b. To find two unknown quantities we generally need two independent equations. (Think back. This is a principle that was probably covered in Algebra I, Algebra II and Precalculus, though most students don't really learn it until their first calculus course.) Here f(x) = a x e^(bx). We're given that f(1/3) = 1, which means that a * 1/3 * e^(b * 1/3) = 1. Simplified, 1/3 * a e^(b/3) = 1. This is one equation. A local maximum occurs when f ' (x) = 0. This means that a e^(b x) + a x * b e^(b x) = 0. Factoring out a e^(bx) we have a e^(b x) (1 + b x), which we set equal to 0, giving us a second equation. The two simultaneous equations are then solved, as shown in the given solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: To find the local min I used the first derivative test: The derivative of y = x - k `sqrt(x) is y ' = 1 - k / (2 sqrt(x) ) we set this equal to 0: 0 = 1 - k / (2 sqrt(x) ) 0= 2sqrt(x) - k Sqrt(x) = k/2 x= k^2 / 4 this is a local minimum for the function The zeros of the function occur when 1 - k / (2 sqrt(x) ) is equal to zero, x - k `sqrt(x) = 0. x = 0 or x = k^2. The x coordinate of the minimum, x = k^2 / 4, therefore is 1/4 of the way between the zeros x = 0 and x = k^2 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 intercepts occur when x-k sqrt(x) = 0 solving for x when get: sqrt(x) - k = 0 sqrt(x) = k x = k^2 so the x intercepts is at (k^2, 0) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** This problem requires the use of derivatives and other calculus-based tools (which do not include the graphing calculator). You have to find intercepts, critical points, concavity, etc.. We first find the zeros: x-intercept occurs when x - k `sqrt(x) = 0, which we solve by by dividing both sides by `sqrt(x) to get `sqrt(x) - k = 0, which we solve to get x = k^2. So the x intercept is at (k^2, 0). ** STUDENT QUESTION Can I get a more detailed steps for solving y=x-ksqrt(x)=0 because I am missing some simple part to doing it somewere. INSTRUCTOR RESPONSE Sure. Start with x - k sqrt(x) = 0. Divide both sides the sqrt(x): x / sqrt(x) - k sqrt(x) / sqrt(x) = 0 / sqrt(x). Now x / sqrt(x) = sqrt(x). Clearly k sqrt(x) / sqrt(x) = k, and 0 / sqrt(x) = 0. So our equation x / sqrt(x) - k sqrt(x) / sqrt(x) = 0 / sqrt(x) becomes sqrt(x) - k = 0. Add k to both sides to get sqrt(x) = k. Then square both sides to get x = k^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhere is(are) the critical point(s) of x - k `sqrt(x) and how did you find it(them)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We find the critical points by setting the derivative of the function equal to 0 f(x) = x - k `sqrt(x) f ' (x) = 1 - k / (2 `sqrt(x)). 1 - k / (2 `sqrt(x)) = 0. 2 `sqrt(x) - k = 0 x = k^2 / 4. The second derivative will be positive for this point if k>0 so this is a minimum confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** STUDENT QUESTION I am given y=x-ksqrt(x) and from this I can solve that y`=1-k/2sqrt(x) I then set this up to the expression 1-k/2sqrt(x)=0. This is were my problem is for some reason I am having trouble solving this for x due to the sqrt x I know I know how to do this but I cant figure it out. Is there anyway you could give me a little explanation of how to do this and I feel that I will be able to jog my memory enough to work this then???? INSTRUCTOR RESPONSE sqrt(x) * sqrt(x) = x. It follows that sqrt(x) = x^(1/2), since x^(1/2) * x^(1/2) = x. I think you know this, but I misread your question. A general rule in solving equations is, if denominators occur, multiply both sides by a common denominator (and be careful about the possibility that your expression is zero). If you multiply both sides of the equation by sqrt(x) you get sqrt(x) * 1 - sqrt(x) * k / (2 sqrt(x)) = sqrt(x) * 0, which gives you sqrt(x) - k / 2 = 0. Adding k/2 to both sides you get sqrt(x) = k / 2. Squaring both sides gives you x = k^2 / 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: U = b( a^2/x^2 - a/x) U = b(-2a^2 / x^3 + a / x^2) -2a^2 / x^3 + a / x^2 = 0 -2a^2 + a x = 0 x = 2 a^2 / a = 2a this point is a minimum the function is at its minimum at (2a, -b/4). Im not sure how to find the asymptotes confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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, while (provided a is positive) -a / x will approach - infinity to the right of the y axis, and +infinity to the left. However, near x=0 we see that x^2 is much smaller than x so its reciprocal will be much larger, with the result that the a^2 / x^2 term will dominate the -a/x term and the vertical asymptote will therefore be the positive y axis, and it will be so on both sides of that 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): Most of this explanation was very hard for me to understand I dont understand asymptotes very well ------------------------------------------------ Self-critique Rating: 0
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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): The fact about the hovering makes a lot of sense after reading your solution, I was a little stumped on that ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!