HOMEWORK HELP 2 FOR MATH 151

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1 HOMEWORK HELP 2 FOR MATH 151 Here we go; te second round of omework elp. If tere are oters you would like to see, let me know! 2.4, 43 and 44 At wat points are te functions f(x) and g(x) = xf(x)continuous, were 1, x Q f(x) = 0, x / Q? f(x) is not continuous anywere; for ɛ = 1/2, it is clear tat no matter wat you coose as a limit (L) for f at any point x tat for all δ tere are y witin δ of x suc tat f(y) is not witin 1/2 of L, since tere are bot rational and irrational numbers in any open interval. On te oter and, g is continuous precisely at x = 0. It is not continuous anywere else by using te same argument as above, using ɛ = x/2 at te point x. But at x = 0, if we coose δ = ɛ, ten if we place y witin δ of 0, certainly g(y) will be witin ɛ of 0 as well (since it will be eiter zero already or smaller tan ɛ at any rate). 2.6, 2 Use te intermediate value teorem to sow tat tere is a solution of x 4 x 1 = 0 in te interval [ 1, 1]. Consider tat, if f(x) = x 4 x 1, ten f( 1) = 1 and f(1) = 1. So by te IVT, since f is continuous (a polynomial) we know tat if K is between 1 and 1, tere is some c ( 1, 1) suc tat f(c) = K. Coose K = 0. Ten c for tis is precisely te desired root. 2.6, 32 Let n be a positive integer. Prove tat if 0 a < b, ten a n < b n. Furter sow tat every nonnegative real number x as a unique nonnegative nt root x 1/n. We assume tat 0 a < b. We wis to sow a n < b n for all positive n. For n = 1 tis is given. Furter, if we assume it is true for n = k, ten a k < b k, so as a < b we ave a k+1 < b k+1 (tis does not ave to be proved, it would only ave to be in 160s). So by induction, since we ave sown tat te statement for n = k + 1 follows from n = k, we are done. Now consider te function f(t) = t n x. Ten f(0) 0 and f(x + 1) > 0 (tis is true by te fact tat 1 < 1 + x and te ting sown in te first part). Tus, as f(0) 0 < f(x + 1) and f is continuous (a polynomial), by te intermediate value teorem tere is some t 0 [0, x + 1] suc tat f(t 0 ) = 0. Tus t n 0 = x, so at least t 0 is a positive nt root. But suppose tere were anoter nt root not equal to t 0 ; call it s. Ten eiter s < t or s > t. But ten eiter t n 0 > s n or s n > t n 0 ; eiter of tese cases implies tat in fact s is not an nt root, toug. So t 0 is unique. 2.6, 35 Prove tat te cubic equation x 3 + ax 2 + bx + c = 0 as at least one real root. Te key is to see tat tere is a negative and a positive value of tis equation f(x) = 0; ten as it is a polynomial, ence continuous, by te intermediate value 1

2 2 HOMEWORK HELP 2 FOR MATH 151 teorem it as a zero. We rewrite f as x 3 (1 + a x + b x + c 2 x ). Ten, by te teorem 3 on products of limits, we see tat te limit of te cubic as x approaces ± is in fact te same as te limit of x 3 (te ting in parenteses certainly goes to 1). Since tese limits we know are in fact ±, we know tat somewere tere ten as to be x, y suc tat f(x) > 0 and f(y) < 0; oterwise te limits couldn t tend to infinity. But tat is wat was desired to use te IVT. 3.1, 11 Differentiate f(x) = 1/x 2. f(x+) f(x) We form lim 0. Te fraction is = 2x 2 x 2 (x+). 2 x 2 (x 2 +2x+ 2 ) x 2 (x+) 2 1 (x+) 2 1 x 2, wic simplifies to Taking te limit and canceling we get f (x) = 2x x = 2 4 x , 14 Find f (2) using te definition of derivative, for f(x) = 7x x 2. f(x+) f(x) 7(2+) (2+) We form lim 0. Tis is, for x = 2, lim 2 7(2) wic simplifies to lim = = , 49 Sow tat f(x) = x 2, x 1 2x, x > 1 is not differentiable at x = 1. Note tat f(x) = 1, but lim x 1 + = 2 by continuity of 2x. Tus f is not continuous, so by Teorem it certainly is not differentiable. 3.1, 50 Find A and B given tat te function f(x) = x 3, x 1 Ax + B, x > 1 is differentiable at x = 1. We need to find out for wat values of A, B te function is continuous and as te same derivative from te rigt and te left (i.e. is differentiable) at x = 1. For continuity, clearly te condition (1) 3 = A + B suffices, since te component functions are continuous. For differentiability, te rigt and left derivatives at 1 would be A and 3(1) 2 respectively. So we need A = 3, wic means tat B = , 65 x sin(1/x), x 0 x 2 sin(1/x), x 0 Let f(x) = and g(x) =. Sow tat f 0, x = 0 0, x = 0 and g are bot continuous at 0, f is not differentiable at 0, and g is differentiable at 0 (find g (0)). For te first part, we consider tat x f(x) x and x 2 g(x) x 2 as sin is bounded by ±1; ence by te pincing teorem, since te limits of ±x and ±x 2 are zero, so are tose of f and g, wic proves continuity. To continue, we examine te defining limits for te two derivatives: f (x) = lim x 0 sin(1/) and g (x) = lim x 0 2 sin(1/)

3 HOMEWORK HELP 2 FOR MATH Te second one (after cancelling ) is simply te limit of f at 0, so it exists and in fact is zero. Te first one, owever, is purporting to be lim x 0 sin(1/x), wic we ave previously sown does not exist. So f is not continuous at zero. 3.2, We wis to prove tat if f i are differentiable at x for 1 i n, and α i R, ten α 1 f 1 + α n f n is differentiable at x and (α 1 f 1 + α n f n ) (x) = α 1 f 1(x) + α n f n(x). To do tis, we proceed by induction. For n = 1, tis is completely trivial as bot sides are te same already. Suppose we ave it for n = k. Ten consider some (α 1 f 1 + α k+1 f k+1 ) (x); if we construct its derivative to be wat we want, ten it is differentiable at x since we know its derivative. But let f(x) = (α 1 f 1 + α k f k )(x) and g(x) = α k+1 f k+1 (x). By Teorem 3.2.3, (f + g) (x) = f (x) + g (x). So ere (α 1 f 1 + α k+1 f k+1 ) (x) = (α 1 f 1 + α k f k ) (x) + α k+1 f k+1 (x). But using te induction ypotesis, we see tat in fact (α 1 f 1 + α k+1 f k+1 ) (x) = α 1 f 1(x) + α k+1 f k+1(x). So by induction, te statement is true for all n. 3.3, 56 Verify te identity f(x)g (x) f (x)g(x) = d [f(x)g (x) f (x)g(x)]. We simply differentiate on te rigt. Te rigt side is, by te addition and product rules, f (x)g (x) + f(x)g (x) f (x)g(x) f (x)g (x). But tis is te left side, after cancellation. Project 3.3 Sow tat if g(x) = [f(x)] 4, ten g (x) = 4f 3 (x)f (x). Sow in general tat if g(x) = [f(x)] n for any integer n, ten g (x) = kf k 1 (x)f (x). By writing f 4 (x) = f 2 (x)f 2 (x), we see tat by te product rule and information in te problem (f 4 (x)) = 2f(x)f (x)f 2 (x) + 2f(x)f (x)f 2 (x) wic is wat we wanted. For positive n, consider tat we already know tis for n = 1, 2, 3, 4 (in fact for n = 0, since ten it s just 0). Ten assume we ave it for n = k. Ten f k+1 (x) = f(x)f k (x), so by te product rule we ave (f k+1 (x)) = f (x)f k (x)+f(x)(f k (x)) = (1+k)f (x)f k (x) as desired. So by induction we ave it for positive n. For negative n, we use precisely te same induction argument, except instead of te product rule we use te quotient rule. So we ave it for n = 1, and ten assuming for n = k, we ave (f k 1 (x)) = ( f k (x) f(x) ) =, wic is ( k)f (x)f k 1 (x)f(x) f k (x)f (x) f 2 (x). We can simplify tis to ( k 1)f (x) f k (x) f 2 (x) = ( k 1)f (x)f k 2 (x) as desired. Tus by induction te negative integers also ave tis property, so tey all do.

4 4 HOMEWORK HELP 2 FOR MATH , 4 Find te rate of cange of y = 1/x wit respect to x at x = 1. We simply take te derivative! So y = 1/x 2, wic at te point indicated is , 8 Find te rate of cange of te surface area wit respect to te radius r. Wat is tis rate of cange wen r = r 0? How must r 0 be cosen so tat te rate of cange is 1? Te surface area is 4πr 2. So we take te derivative wit respect to r, wic is 8πr. So at r = r 0, we ave 8πr 0. For tis to be 1, we solve 8πr 0 = 1; tis yields r 0 = 1 8π. 3.4, 15 For wat value of x is te rate of cange of y = ax 2 + bx + c wit respect to x te same as te rate of cange of z = bx 2 + ax + c wit respect to x? Assume a b. We take te derivatives dy dz and. Tis is y = 2ax + b, and z = 2xb + a. For tese to be equal, 2ax + b = 2xb + a, wic implies x = a b 2b 2a = 1/2. Clearly, if a = b ten te derivatives are equal always. 3.4, 39 During wat time interval(s) is te particle in motion described by x(t) = 5t 4 t 5 speeding up? We ave seen in te capter tat tis is te cange of te speed wit respect to time, wic is te cange of te cange of position. So we take te second derivative. Tis is x (t) = 60t 2 20t 3. We are asked wen tis is positive. Since it is continuous and te zeros are t = 0 and t = 3, we use te intermediate value teorem to ceck te tree possible intervals. Since x ( 1) = 80, x (1) = 40, and x (10) = 14000, we see it is positive on (, 0) (0, 3), wic is te answer. 3.4, 54 To estimate te eigt of a bridge a man drops a stone into te water below. How ig is te bridge (a) if te stone its te water 3 seconds later? (b) if te man ears te splas 3 seconds later? We recall tat motion in te vertical direction is described by y(t) = 16t 2 +v 0 t+ y 0 (for feet and seconds). So, wit te eigt of te bridge being considered y 0 and te initial velocity clearly v 0 = 0, ten y(3) = 0 in part (a). Tus, 16(3) 2 +y 0 = 0, or y 0 = 144 feet. However, in part (b), te stone falls and ten te man ears te splas. Since we are told sound travels at te constant speed of 1080 feet per second, we know tat it traveled y 0 feet in a certain amount of time, say s seconds. Ten te stone actually fell in only 3 s seconds, and y(3 s) = 0. We wis to find y 0. Ten 0 = 16(3 s) 2 + y 0. But 1080s = y 0, so s = y Substituting we see tat y 0 = y y2 0 (1080). We solve for y 2 0 and, after using te quadratic formula, arrive at y 0 = 147/4 (it turns out te quadratic factors as a square). 3.4, 60 Find te marginal cost function at a production level of 100 units and compare wit te actual cost of te 101st unit for C(x) = x x x 3. We ave te marginal cost defined as C (100), wic is (100) (100) 2 = = 9. On te oter and, C(101) C(100) = (101)+0.02(101) (101) 3 (1000+2(100)+0.02(100) (100) 3 )

5 HOMEWORK HELP 2 FOR MATH wic simplifies to (201) (30301) = So te accuracy of te estimate is to , wic is around a alf of a percent from te actual answer - not bad! 3.5, 12 Differentiate f(x) = (x + 1 x )3. By te cain rule, f (x) = 3(x + 1 x )(1 1 x 2 ). 3.5, 28 Find dy if y = 1 + u2, u = 1 7x 1+x 2, and x = 5t + 2. We use te cain rule; dy = dy du We ten write 3.6, 2 du. Tis is simply dy = 2u( (1+x2 )( 7) (1 7x)(2x)) (1+x 2 ) 2 )(5). 1 7(5t + 2) (1 + (5t + 2) 2 )( 7) (1 7(5t + 2))(2(5t + 2))) (5t + 2) 2 (1 + (5t + 2) 2 ) 2 Differentiate y = x 2 sec x. Using te quotient rule, since y = x2 cos x, we ave y (x) = 2x cos x+x2 sin x cos 2 x. 3.6, 58a Verify tat d (cot x) = csc2 x. We note tat cot x = cos x sin x. Ten by te quotient rule, d wic is just 1 sin 2 x = csc2 x. 3.7, 4 (cot x) = sin x sin x cos x cos x sin 2 x, Use implicit differentiation to obtain dy/ in terms of x and y from te equation x + y = 4. 1 Applying d/ to bot sides gives us 2 x dy y = 0. Ten solving for dy = x y. 3.8, 2 A particle is moving in te circular orbit x 2 + y 2 = 25. As it passes troug (3, 4), its y-coordinate is decreasing at te rate of 2 units per second. At wat rate is te x-coordinate canging? We are asked for /. By implicit differentiation, we ave 2x dy + 2y = 0; furter, we are given tat dy/ = 2 and (x, y) = (3, 4). Plugging it all in gives 16 = 0, or / = 8/ , 7 A eap of rubbis in te sape of a cube is being compacted. Given tat te volume decreases at a rate of 2 cubic meters per minute, find te rate of cange of an edge of te cube wen te volume is exactly 27 meters cubic meters. Wat is te rate of cange of te surface area of te cube at tat instant? We recall te relation between volume, surface area, and lengt of an edge: V = l 3 and S = 6l 2 if l is te lengt. So we are told dv/ = 2, and we want dl/ at V = 27. Clearly at tat point l = 3, since 3 3 = 27. Using te cain rule, we see tat dv dl = 3l2. Tat is, te edge of te cube is slowly getting smaller at tis rate. Since te surface area S = 6l 2, after anoter use of te cain rule (or implicit differentiation), ds dl = 8l = 8(3)( 2 27 ). Tus te surface area is canging at te rate of 16 9 square feet per minute.. Tus at tis time, dl = dv/ 3(3) 2, wic is 2 27

6 6 HOMEWORK HELP 2 FOR MATH , 17 A 13-foot ladder is leaning against a vertical wall. If te bottom of te ladder is being pulled away from te wall at te rate of 2 feet per second, ow fast is te area of te triangle formed by te wall, te ground and te ladder canging at te instant te bottom of te ladder is 12 feet from te wall. It ll be ard to do tis witout a diagram, but you all ave good imaginations, based on te crazy entrance essays for te U of C. We ave some information rigt off te bat. If we call te distance of te base from te wall x, ten we know / = 2. Very conveniently, wen x = 12, te eigt of te wall were te ladder touces it is precisely = 5 feet. Now te area A = 1 da 2x, were is te eigt. Tus we want = 1 2 ( + x d ). d In general, = 2x as = 169 x 169 x 2. At tis point, tat means tat 2 d = 24 da 5. So = 1 2 (5(2) + 12( 24 5 )) = Looks like te area is decreasing pretty rapidly. 3.8, 34 Te diameter and eigt of a rigt circular cylinder are found at a certain instant to be 10 centimeters and 20 centimeters, respectively. If te diameter is increasing at te rate of 1 centimeter per second, wat cange in eigt will keep te volume constant? Let s tabulate te information we ave tus far. Te volume of suc a cylinder is V = π(d/2) 2 ; we also know dd/ = 1. We want to keep V constant, so let s regard it as suc; ten = 4V d = 8(π(100)/4) 1000π D 2 π, so d = 8V D 3 π = 1 5. So a rate of cange of 1 5 will suffice.. At te point in question, ten,