Excruciating Practice Final, Babson 21A, Fall 13.

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1 Excruciating Practice Final Excruciating Practice Final, Babson 2A, Fall 3. This is an *optional* practice final. It has been designed with intention to be more difficult that what should be encountered on the actual final. However, I have not seen the actual final, and have had no hand in designing it. Therefore, I will try to cover all bases in this practice final to adequately prepare you. As with the second practice midterm, I am including a solutions set at the end of this document, should you get stuck along the way or wish to check your answers. Note that, although I tried to work slowly through the solutions, I may be prone to error, or may have gone about calculating my answers differently that the way you have; feel free to me if you think I made a mistake (jrsclark@math.ucdavis.edu). I strongly encourage you in during your finals week to meet with your fellow peers and develop study groups within which to interact and work on your personal difficulties. You are in college now: you will become very accustomed to study groups (just like you will become very accustomed to office hours) over the next four years, so begin taking advantage now. As always, good luck, and may the force be with you. Part One: Limits. Calculate each of the following limits using any methods at hand. Do not skip steps. IM- PORTANT NOTE: (e) is VERY long (at least the way I proceeded to do it). Check the solutions for it for more info. (a) lim x (x 7 + )(x 8 6x 2 + x) 7x 2 + 9x 7 4x 5 ) (d) lim x x( x 2 + x) (b) lim x 0 cos 2 (x) sin(x) (e) lim x 0 x x x x 2 8 (c) lim x 9 x x (f) lim x x Part Two: Derivatives. Calculate the derivatives (f (x) or dy ) of each of the following functions. If you are given dx a function in terms of x alone, your answer should have only x s in it. Otherwise, it can contain both x s and y s (a) f(x) = ln(ln(sec x)) (b) f(x) = x5 + sin( x ) e x2 (c) f(x) = 3 csc(2x + ) 7 (x 7 5x) 2 (d) xy = x y (x 7 5x) 4 e 2x+7 (e) sin(x + 3y) = cos(3x y) (f) f(x) = (x + x 2 + x 3 ) 4 x 5 x 3

2 Excruciating Practice Final 2 Part Three: Some theory. (a) If a function y = f(x) is differentiable at a point a, write down the formal definition of f (a). Use the formal definition to calculate the derivative of y = e x at x = a (you may assume l Hôpital s rule). (b) State the intermediate value theorem (IVT) in full. Use it to prove that the polynomial p(x) = x 5 + x has a root in the interval [0, ]. (c) State the mean value theorem (MVT) in full. Use it to solve the following conceptual problem: Suppose that y = f(x) is continuous and differentiable on [6, 5], and that f(6) = 2 and f (x) 0. What is the largest possible value for f(5)? (d) State the formal definition of what lim x x0 f(x) = L is. Prove that lim x (ax + b) = a + b using this definition, where a and b are some real numbers. Part Four: Applications. (a) History lesson: Once upon a time, the Babylonians needed ways of estimating the square roots of numbers. They developed a seemingly miraculous formula for doing it: let S be the number you wanted to find the square root of, and let x 0 be your guess of S (so if I wanted to know the square root of 5, I might take x 0 = 2). Then they could obtain a much better guess y according to the formula y = ) (x 0 + Sx0. 2 When S = 5 and x 0 = 2, this formula gives me y = 2.25, which is very close to the real answer Problem: I m not sure how the Babylonians figured it out the above (called the Babylonian method), but I want you to derive the above formula, given some value S I want you to square root, and an initial guess x 0. (b) History lesson: Back in the golden age of algebra, all mathematicians were interested in was solving for roots of polynomials y = p(x); i.e.: finding when p(x) = 0. When p(x) = ax+b, this is easy, since the answer is just x = b. When you have a quadratic a ax 2 +bx+c, it was more challenging, but luckily for us, there is a quadratic formula. In fact, although they are terrible to write down, there are equations for cubic polynomials ax 3 + bx 2 + cx + d and for quartic polynomials ax 4 + bx 3 + cx 2 + dx + e. However, as HARD as they tried, no one could figure out a formula for a quintic (degree 5) polynomial. But eventually in the early 9 th -century, a mathematician named Évariste Galois was able to prove that such an equation was impossible, and the methods he used dawned the revolutionary study of group theory and to some extend, modern algebra. But we still need to know roots of certain polynomials, as their solutions arise all the time in physical and mathematical problems, so we need methods of finding roots even without a general formula.

3 Excruciating Practice Final 3 (i) Let p(x) = 7x 5 7x 2 +. Prove that there exists a root of p somewhere in the interval [, 2]. (ii) Call the root from (i) c. Figure out an estimation x n of c so that p(x n ) is within 2 decimal places of p(c) (you should use a calculator to help with the tedious calculations for this part). (c) You are tasked with the job of building a gated area for kids to play in, but you only have 00 ft. of fencing material available. The fence will be rectangular, and luckily (due to the limited amount of fencing material available) one of side of the play area will be against a building wall, i.e. one side of the rectangle will not need to be fenced off. You want to be able to maximize the amount of area that will be enclosed. (i) Draw a picture, labeling all relevant variables. Write down the equation for area, and any constraints the variables may be subject to. (ii) Re-write the area formula in terms of one variable. Figure out of the dimension of that variable corresponds to maximum area. (iii) Figure out the other dimension corresponding to maximum area, and finally, compute the maximum area. (d) You just bought a brand-new flatscreen TV that you have hung-up 5 ft. off the ground in your apartment. The TV screen is 3 ft. high. Being that you re broke now, you don t have a couch, so you re sitting on the floor. Being that you have purchased this nice new TV, you want to make sure you are getting a maximum viewing angle of the TV; in other words, you want to know exactly how far a distance x from the horizontal position of the TV you should sit down so that your viewing angle of the TV is as large as possible. In other words, maximize θ in the following picture, where x is the only other variable: 3 You α θ x 5 (e) Suppose that the volume of a sphere (V = 4 3 πr3 ) is changing at a rate of 2 ft.3 sec.. Calculate the rate-of-change of the diameter of the sphere when the surface area of the sphere (S = 4πr 2 ) is equal to 5 ft. 2.

4 Excruciating Practice Final 4 Solutions. Part One. (a) (x 7 + )(x 8 6x 2 + x) lim x 7x 2 + 9x 7 4x 5 x x 5 6x 9 + 2x 8 6x 2 + x 7x 2 + 9x 7 4x 5 x 5 x x x 6 x 7 x 3 x x 3 x x = 4. Note: you *could* do l Hôpital on this but it would take 8 derivatives to get a workable form, so best just to remember the original way to take limits. (b) (c) (d) lim x 0 cos 2 (x) ( = [ 0 sin(x) 0 ]) l Hôp x 0 2 cos(x) ( sin(x)) 0 cos(x) x 0 2 sin(x) = 0. x 2 8 (x + 9)(x 9) lim x 9 x 3 x 9 x 3 x 9 (x + 9)( x + 3)( x 3) x 3 x 9 (x + 9)( x + 3) = (9 + 9)( 9 + 3) = 08. lim x( x 2 + x) x ( x2 + x) x2 + + x x x x2 + + x x ((x2 + ) (x 2 )) x x2 + + x x x x2 + + x x x x + + x 2 = = 2.

5 Excruciating Practice Final 5 (e) Recall from my previous practice midterm that the derivative of x x is x x (ln(x)+). You should know how to obtain this derivative manually, via logarithmic differentiation, or the e ln -trick. (f) x x lim ( = [ 0 x 0 + x 0 ]) l Hôp x x (ln(x) + ) x x 2 ln(x) + x 0 + x x x 2 2ln(x) + x 0 + x x 2 I pulled this problem from a previous Berkeley final, and though it can be finished with one more l Hôpital, it is unnecessarily long and unintuitive, so let s just say I don t expect anything this annoying to be on the final. Kudos to those of you who feel like finishing this one. If you DO care, the answer ends up being 0 (Wolfram Mathemtica confirmed that for me). Part Two. (a) f(x) = ln(ln(sec x)): x lim x x f (x) = = = ln(sec x) ln(sec x) ln(sec x) x x x x x (4) (3 + x) x (x )( x) x x (x )( x) x (x ) (x )( x) x x = 4. ( ) ln(sec x) d dx sec(x) d dx sec(x) sec(x) tan(x) = tan(x) sec(x) ln(sec(x).

6 Excruciating Practice Final 6 (b) f(x) = x5 +sin( x ) e x2 = x5 +sin(x ) f (x) = e x2 : ) (5x 4 + cos(x ) ( )x )(e 2 x2 (e x2 ) 2 ( )( ) x 5 + sin(x ) e x2 2x (this answer is fine; for derivatives, we prefer an answer that is not reduced). ( ) /3 csc(2x+) (c) f(x) = 3 7 = (csc(2x+)) /2 = (csc(2x+))/6. I m going to use logarithmic (x 7 5x) 2 (x 7 5x) 2/7 (x 7 5x) 4/7 differentiation, so I m going to write y = (csc(2x+))/6 (x 7 5x) 4/7 : ( ) (csc(2x + )) /6 ln(y) = ln = (x 7 5x) 4/7 6 ln(csc(2x + )) 4 7 ln(x7 5x) dy y dx = 6 csc(2x + ) ( csc(2x + ) cot(2x + )) (2) 4 7 x 7 5x (7x6 5) cot(2x + ) = 4 ( ) 7x x 7 5x dy ( dx = cot(2x + ) 4 ( )) 7x 6 5 y 3 7 x 7 5x ( cot(2x + ) = 4 ( )) 7x 6 5 (csc(2x + ))/ x 7 5x (x 7 5x) 4/7 (d) xy = x y. I don t like the exponent on the right, so I m going to take the natural log of both sides to remove it. The equation becomes ln(xy) = ln(x y ), which is equivalently ln(x) + ln(y) = y ln(x): ln(x) + ln(y) = y ln(x) x + dy y dx = (dy dx ) ln(x) + y( x ) = ln(x)dy dx + y x ( y ln(x))dy dx = y x x = y x dy ( )/( ) y dx = x y ln(x) y 2 y = x( y ln(x)).

7 Excruciating Practice Final 7 (e) sin(x + 3y) = cos(3x y): cos(x + 3y) ( + 3 dy dy ) = sin(3x y) (3 dx dx ) cos(x + 3y) + 3 cos(x + 3y) dy = sin(3x y)dy 3 sin(3x y) dx ( dx ) dy cos(x + 3y) + 3 sin(3x y) = sin(3x y) 3 cos(x + 3y) dx cos(x + 3y) + 3 sin(3x y) sin(3x y) 3 cos(x + 3y) = dy dx. (x (f) f(x) = 7 5x) 4 e 2x+7 (x+x 2 +x 3 ) 4. There s no way I (or you) should try quotient rule on this. x 5 x 3 Instead, use logarithmic differentiation: (x 7 5x) 4 e 2x+7 y = (x + x 2 + x 3 ) 4 x 5 x 3 ( ) (x 7 5x) 4 e 2x+7 ln(y) = ln (x + x 2 + x 3 ) 4 x 5 x 3 = 4 ln(x 7 5x) + (2x + 7) 4 ln(x + x 2 + x 3 ) 2 ln(x5 x 3 ) dy y dx = 4 7x6 5 x 7 5x + 2 4( + 2x + 3x2 ) 5x4 3x 2 x + x 2 + x 3 2(x 5 x 3 ) ( dy dx = (x 7 5x) 4 e 2x+7 (x + x 2 + x 3 ) 4 x 5 x 3 4 7x6 5 x 7 5x + 2 4( + 2x + ) 3x2 ) 5x4 3x 2 x + x 2 + x 3 2(x 5 x 3 ). Part Three. (a) or equivalently, For f(x) = e x, we see: f f(a + h) f(a) (a), h 0 h f f(x) f(a) (a). x a x a f (a) h 0 e a+h e a h h 0 e a eh h l Hôp = e a lim h 0 e h = e a = e a. (= [ 0 0 ])

8 Excruciating Practice Final 8 (b) The IVT: Let f(x) be continuous on the interval [a, b]. The for any value y between f(a) and f(b), there exists (at least one) point c in (a, b), such that f(c) = y. The problem: we want to know wen p(x) has a root, so in the definition above we set y = 0 (since we wish to know when p(x) = 0). Then we want to show that either p(a) < 0 < p(b) or p(a) > 0 > p(b) for some a and b in [0, ]. Just take a = 0 and b =, since p(a) = = < 0 p(b) = 5 + = + > 0 so (this is the most crucial part of the answer) by the IVT, there exists some point c in (0, ) such that p(c) = 0, so we have proven that a root c exists. (c) The MVT: Let f(x) be continuous and differentiable on the interval [a, b], and let m be the slope of the line connecting the points (a, f(a)) and (b, f(b)). Then there is some point c in (a, b) such that f (c) = m = f(b) f(a) b a. The Problem: We have a = 6, f(a) = 2, b = 5, and f (x) 0 for all x in [6, 5]. Just playing around with what we know: for some c in (a, b), f (c) = f(5) f(6) = f(5) f(5) = 9 f (c) 2 and since we know that f (x) 0 for all x in (6, 5), in particular, f (c) 0, so f(5) = 9 f (c) 2 9 (0) 2 = 88 so that f(5) can never be any larger than 88. (d) Definition: We say that lim x x0 f(x) = L if for all ɛ > 0 there is some δ > 0 such that 0 < x x 0 < δ implies that f(x) L < ɛ. The Problem: We have f(x) = ax + b, x 0 =, and L = a + b. As my students know, I like to take the stupid-proof approach, doing all my δ stuff first, all by ɛ stuff on the other, then combining in the end: x x 0 {}}{ < δ δ < x < δ f(x) δ < x < + δ {}}{{}}{ ( ax + b) ( a + b) < ɛ L ax a < ɛ ɛ < ax a < ɛ ɛ a < x < + ɛ a.

9 Excruciating Practice Final 9 Match the left and right sides of both sets of inequalities δ < x < + δ and ɛ a < x < + ɛ a, so δ = ɛ a δ = ɛ a or or +δ = + ɛ a δ = ɛ a so we take the smaller of the two options of δ (lucky us, they are the same), so we take δ = ɛ a. Part Four. (a) We want to know when S = x. By squaring both sides, this is equivalent to solving when S = x 2, or equivalently, when x 2 S = 0. So if we write f(x) = x 2 S, we can see that we are trying to solve for a root (AKA a zero) of f(x), so Newton s method is immediately applicable. Let y = x in the Newton s method algorithm. Then y = x 0 f(x 0) f (x 0 ) = x 0 x2 0 S 2x 0 = 2x2 0 (x 2 0 S) 2x 0 = ( ) x S 2 x 0 = ) (x 0 + Sx0. 2 (b.i) This is just using the intermediate value theorem, exactly like we did in 3(b). We see that p() = 9 < 0 and p(2) = 57 > 0, so by the intermediate value theorem (since p(x) is a polynomial and all polynomials are continuous), there exists some c in (, 2) such that p(c) = 0. (b.ii) This is a perfect time to apply Newton s method, which gives you a really quick way of estimating zeros of ANY function (not just polynomials) if you have a decent starting point. I will pick my starting point as x 0 =.5 just because I feel like it; you could pick x 0 = or x 0 = 2 or some other point in-between if you like. x n+ = x n 7x5 n 7x 2 n + 35x 4 n 34x n x = (.5) 7(.5)5 7(.5) (.5) 4 34(.5) p(x) x p(x 2 ) x p(x 3 )

10 Excruciating Practice Final 0 so my approximation is x 3 =.329, since p(x 3 ) is accurate up to the two decimals of p(c) = 0 that I wanted. (c.i) Given the picture below, we will have the constraint 00 = x + 2y, where we don t see 2x appear in the constraint because we don t need to worry about fencing off the wall. Of course, we necessarily have that A = xy. y Wall x y (c.ii) In the constraint, we can easily solve x = 00 2y, so the area equation A = xy can be re-written as A = (00 2y)y = 00y 2y 2. Since we are optimizing, we set the derivative equal to zero and solve for y: 0 = da = 00 4y dy y = 25. Of course, this doesn t prove that y = 0 is a maximum, only that it is a critical point, so to show that it is a maximum, I m going to use the second derivative test. I compute d 2 A = 4, so when y = 25, A (y) = 4 < 0, so the area function is concave down at dy 2 y = 25. This looks like a frowny face (concave up = smiley, concave down = frowney), so that y corresponds to a maximum indeed. This gives us y = 25 as out first dimension corresponding to maximum area. (c.iii) Go back to the constraint 00 = x + 2y. Plugging=in y = 25, we solve for x to get x = 50, our other dimension corresponding to maximum area. Therefore, the maximum area is A = xy = (50)(25) =, 250 ft. 2. (d) You ll notice that I drew in a little angle α in the picture as well. There doesn t seem like a lot of information to go off here, since these triangles aren t similar, and the Pythagorean (if you play around with it a while) will get you nowhere. However, notice that tan(α) = 5 x tan(θ + α) = 8 x so that α = tan ( 5 x ) θ + α = tan ( 8 x )

11 Excruciating Practice Final so that θ = tan ( 8 x ) α = tan ( 8 x ) tan ( 5 x ) which gives us our all-important relationship between θ and x. We are trying to maximize θ with respect to x, so we will compute dθ and set it equal to 0 to optimize: dx 0 = dθ dx = + ( 8 x )2 ( 8 x 2 ) + ( 5 x )2 ( 5 x 2 ) = 8 x 2 ( + 64 ) x 2 5 x 2 ( + 25 x 2 ) = 8 x x I m going to clear the denominators by multiplying both sides by (x )(x ) (luckily, 0 times anything is 0 still), so this tells me 5(x ) 8(x ) = 0 3x = 0 x 2 = 40 x = ±2 0. Of course, we don t want negative values for x since we are talking about a physical distance, so the answer is x = ft.. Of course, we should check that this IS a maximum, but I ll leave that to you. To find θ, we of course plug x into: θ = tan 8 ( 2 0 ) 5 tan ( 2 0 ) (e) We are given 5 = dv and we can calculate that dv = 4 dr dr π 3r2 = 4πr2. Notice that dt dt 3 dt dt the formula for surface area is sitting in the equation (S = 4πr 2 ), so dv = S dr = 5 dr, dt dt dt so 5 = 5 dr dt dr dt =. But I asked you for the rate-of-change of the diameter, so use the equation D = 2r to get dd = 2 dr ft. = 2() = 2, so the diameter is changing at a rate of 2. dt dt sec.