Solutions to Math 41 Second Exam November 4, 2010

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1 Solutions to Math 41 Secon Exam November 4, (13 points) Differentiate, using the metho of your choice. (a) p(t) = ln(sec t + tan t) + log 2 (2 + t) (4 points) Using the rule for the erivative of a sum, p (t) = t (ln(sec t + tan t) + log 2(2 + t)) = t ln(sec t + tan t) + t log 2(2 + t) Now, we o each separate erivative using the Chain Rule: an t ln(sec t + tan t) = 1 sec t + tan t = sec t tan t + sec2 t sec t + tan t (sec t + tan t) t = sec t t log 2(2 + t) = ln(2 + t) = 1 ln(2 + t) t ln 2 ln 2 t 1 = (ln 2)(2 + t) t (2 + t) = 1 (ln 2)(2 + t) Combining, we get that p (t) = sec t + 1 (ln 2)(2 + t) (b) f(x) = 10 sin(πx) + x + x (4 points) Again, we ecompose using the rule for a sum: f (x) = 10sin(πx) + x + x For the first term, using the chain rule an the fact that (ax ) = ln a a x, we get 10sin(πx) = ln sin(πx) sin(πx) = ln 10 π 10 sin(πx) cos(πx) Furthermore, using the chain rule for the secon term, x + x = (x + x 1/2) 1/2 = 1 ( x + x 1/2) 1/2 (x + x 1/2) 2 = 1 (x + x 1/2) ( 1/ ) 2 2 x 1/2 = 1 2( x + x) ( ) 2 x Combining the above, we get that f (x) = ln 10 π 10 sin(πx) 1 cos(πx) + 2( x + x) ( ) 2 x

2 Math 41, Autumn 2010 Solutions to Secon Exam November 4, 2010 Page 2 of 12 (c) g(x) = arctan(x cos x ) ( points) Again using the chain rule, arctan(xcos x ) = = (x cos x ) 2 (xcos x ) x 2 cos x (xcos x ) Now we nee to calculate (xcos x ). Note that we can t use the rule that xn = nx n 1, which only works when n is constant; an we can t use the rule that ax = ln a a x, which only works when a is constant. Thus, we o the following: (xcos x ) = ( ) e ln(xcos x ) = cos x ln (e x) cos x ln x = e (cos x ln x) ( = e cos x ln x cos x ) x sin x ln x = x cos x ( cos x x sin x ln x ) (We coul have also set y = x cos x, calculate that ln y = cos x ln x an use implicit ifferentiation.) Combining, we fin g (x) = xcos x 1 + x 2 cos x ( cos x x sin x ln x )

3 Math 41, Autumn 2010 Solutions to Secon Exam November 4, 2010 Page 3 of (12 points) Consier the curve with equation x 2 y 2 + xy = 2. (a) Fin an expression for y. (Your answer can be in terms of both x an y.) (4 points) We ifferentiate both sies with respect to x: So solving for y, we get ( x 2 y 2 + xy ) = (2) = 2xy 2 + 2x 2 y y + y + x y = 0 = x(2xy + 1) y = y(2xy + 1) y = y x (b) There is a unique point on the curve with x = 1 an positive y; fin this point an write the equation of the tangent line to the curve at this point. (4 points) Substituting x = 1 into x 2 y 2 + xy = 2, we get y 2 y = 2 = (y 2)(y + 1) = 0 So there are two points with x-coorinate equal to 1 on the curve. The only one with positive y-coorinate is (x, y) = ( 1, 2). The slope of the curve at this point is given by So the equation of the tangent line is m = 2 1 = 2 y 2 = 2(x + 1)

4 Math 41, Autumn 2010 Solutions to Secon Exam November 4, 2010 Page 4 of 12 For easy reference, the curve has equation x 2 y 2 + xy = 2. (c) Fin all points on the curve where the slope of the tangent line is 1. (4 points) The slope conition implies y x = 1; hence the points must satisfy y = x. Substituting back to the original equation, one gets x 4 + x 2 = 2 = (x 2 1)(x 2 + 2) = 0 So the only real solutions are x = 1 an x = 1. Now using the conition y = x again, we fin that (x, y) = (1, 1) or ( 1, 1) are the only points satisfying the requirement of (c).

5 Math 41, Autumn 2010 Solutions to Secon Exam November 4, 2010 Page of (9 points) (a) Use linear approximation to estimate e , showing your reasoning. Express your answer as a simplifie rational multiple of e; that is, give your answer in the form a e, where a an b are whole b numbers. (6 points) We use linear approximation to estimate the value e Inee, efining the function then the value we re approximating is f(x) = ex x, f(1.1) = f( ) = f(a + x), where a = 1 an x = 0.1. We nee to compute f (1). Differentiating, we fin f (x) = ex x e x /(2 x) ( x) 2 = e (x x 1/2 12 ) x 3/2. Substituting in x = 1, we obtain f (1) = e1 1 e 1 /(2 1) ( 1) 2 = e 2. We can now procee by either taking the linearization of f or estimating using ifferentials, observing that f(a) = f(1) = e. Using the first metho, we obtain the linearization L(x) of f; that is, the line of slope f (a) = e 2 passing through the point (a, f(a)) = (1, e): L(x) = f (a)(x a) + f(a) = e (x 1) + e. 2 Hence, we get L(1.1) = e 2 (1.1 1) + e = e e = ( ) 21 e. 20 Alternatively, we can use ifferentials: f(1.1) = f(a + x) = f(a) + y f(a) + y x = e + e = ( ) 21 e. 20 (b) Is your estimate in part (a) larger or smaller than the actual value? Justify your answer. (3 points) We calculate the secon erivative of f: f (x) = f (x) = (x ex 1/2 12 ) x 3/2 = e (x x 1/2 12 )+e ( x 3/2 x 12 x 3/ ) x /2. Hence, f (1) = 3e 4 > 0, so f opens concave-upwars; thus the tangent line y = L(x) will lie below the curve y = f(x), an so the above estimate is an uner-approximation.

6 Math 41, Autumn 2010 Solutions to Secon Exam November 4, 2010 Page 6 of (10 points) A caffeinate beverage is brewe by placing water in an inverte cone-shape filter of raius 3 cm an height cm, an allowing it to leak through the cone s vertex into a cylinrical cup of the same raius an height. (You may assume the cone is suspene well above the top of the cup.) The leaking is such that the water level in the cone filter ecreases at a rate of 0. cm/sec. How fast is the water level in the cup rising when the water level in the cone filter is 2 cm above the vertex? Formulas for reference: the volumes of a circular cone an a circular cyliner, each of height h an base raius r, are given, respectively, by V cone = 1 3 πr2 h, V cyl = πr 2 h. Let V 1 be the volume of water in the cone an V 2 the volume of water in the cyliner. Let h be the water level in the cone an r the raius of the surface of water in the cone. Finally, let y be the water level in the cyliner. We re given that h y t = 0. cm/sec, an we nee to fin t when h = 2 cm. Let us fin the volume of water insie the cone as a function of h. Using similar triangles, we see that r = 3h, an plugging this back in the formula for the volume of a cone we get V 1 = π ( ) 3h 2 h = π 9h h = 3π 2 h3. The raius of the surface of the water in the cyliner is constant an equal to 3 cm, so V 2 = 9πy. Taking the erivatives of these volumes with respect to t, we fin V 1 t = 9π h h2 2 t an V 2 t = 9π y t Since the water leaking from the cone is falling into the cyliner, we have V 1 + V 2 = constant V 2 t = V 1 t Now, plugging h = 2 an h t = 0., we get: ( ) V 2 t = V 1 9π t = 2 4 ( 0.) = 18π 2 cm3 /sec 18π 2 Therefore, the water level in the cup is rising at a rate of 2 2 cm/sec. = 9π y t y t = 2 2

7 Math 41, Autumn 2010 Solutions to Secon Exam November 4, 2010 Page 7 of 12. (13 points) Fin each of the following its, with justification. If there is an infinite it, then explain whether it is or. f(x) 1 (a), where f is such that the tangent line to f at x = 4 has equation 2x 4y = 4. x 4 x 4 (You can assume that f an all of its erivatives are continuous near x = 4.) ( points) The tangent line to the graph of y = f(x) at x = 4 passes through the point (4, f(4)). Therefore we can fin f(4) from the equation of the tangent line: f(4) = 4 f(4) = 1 Then the given it turns out to be ineterminate of the form 0 0, an since both functions are ifferentiable, we can use L Hospital s rule: f(x) 1 f (x) = = f (x) = f (4) x 4 x 4 x 4 1 x 4 where the last equality follows because we ll assume f (x) is continuous. So we only nee to fin f (4). Remember that f (4) is the slope of the tangent line at x = 4, an this is 2x 4y = 4, which can be rewritten as y = 2x 4 4 = 1 2 x 1. Therefore, f (4) = 1 2. Hence: f(x) 1 = 1 x 4 x 4 2 Alternate solution: Define g(x) = f(x) 1 an h(x) = x 4, so that the it we seek is f(x) 1 g(x) L = = x 4 x 4 x 4 h(x). Observe that the tangent line to g(x) at x = 4 has equation 2x 4y = 8; to see this, notice that f(x) passes through the point (0, 1), whereas g(x) passes through the point (0, 2). Hence, It is easily verifie that g(x) = g(4) = 0 an x 4 g (4) = 1/2. h(x) = h(4) = 0 an x 4 h (4) = 1. Hence, by l Hôpital s rule applie to the ineterminate form 0/0, L = g (4) h (4) = 1/2 1 = 1 2.

8 Math 41, Autumn 2010 Solutions to Secon Exam November 4, 2010 Page 8 of 12 (b) x ln x (4 points) This is an ineterminate of the form ; if we try to factor this expression to write it as a prouct, we fin: ( (x ln x) = x 1 ln x ) x ln x an we observe that x is ineterminate of the form. Thus, by l Hôpital s Rule, ln x x = 1/x 1 = (x ln x) = x 1 x = 0, ( 1 ln x x so that ) = because the right factor in the last expression approaches 1 while the left factor approaches. Alternate solution: Using the fact that ln(e t ) = t, we can write [ (x ln x) = ln e (x ln x)]. By continuity of natural log, we can first analyze the behavior of the inner function u = e (x ln x) : ex ln x e x = x (an ineterminate, since each factor approaches ) = =. e x 1 (by l Hôpital s rule) Thus, Thus (x ln x) = ln e(x ln x) = ln u =. u (c) (cos x)21/ sin2 x x 0 + (4 points) Let exp(x) = e x. Using the fact that exp(ln x) = x, we obtain [ ] x 0 cos(x)1/ sin2 (x) = exp ln cos(x) 1/ sin2 (x) + x 0 + [ ] = exp ln x 0 cos(x)1/ sin2 (x) (by continuity of exp) [ + ] ln cos(x) = exp x 0 + sin 2 (a 0 0 ineterminate form) (x) [ ] tan x = exp (by l Hôpital s rule) x sin x cos x [ ] 1 = exp x cos 2 x [ = exp 1 ] = e 1/2. 2 (tan x = sin x cos x ) The rule of l Hôpital was applie to the ineterminate form 0 0, since ln cos x = ln 1 = 0 an x 0 + x 0 sin2 x = 0. +

9 Math 41, Autumn 2010 Solutions to Secon Exam November 4, 2010 Page 9 of (10 points) Let f(x) = 9x 2/3 x /3. (a) Fin the critical numbers of f, an characterize each as a point of local minimum, local maximum, or neither. Show all steps of your reasoning. (7 points) A value x = c in the omain of f is a critical number if an only if either f (c) = 0 or f is unefine at c; we first compute f (x) an simplify: f (x) = x 1/3 3 x2/3 = 6 x 3 1/3 x2/3 = 18 x 3x 1/3 Clearly f (x) = 0 if an only if 18 x = 0, an f is unefine if an only if x 1/3 = 0. Thus, there are two critical numbers, x = 18 an x = 0. To etermine the nature of each critical number, we next note that 18 x > 0 if an only if x < 18, an that x1/3 > 0 if an only if x > 0. This leas to the following chart: interval sign of 18 x sign of x 1/3 sign of f (x) x < < x < x > 18 + We ve foun that at x = 0, f is changing from negative to positive; an at x = 18, f is changing from positive to negative. Thus, by the First Derivative Test, f has a local minimum at x = 0, an a local maximum at x = 18. Note: We coul instea use the fact that the the secon erivative of f at x = 18 is negative to etermine the nature of this critical number, using the Secon Derivative Test. (Inee, f (x) = 2x 4/3 (10/9)x 1/3, which is negative for all positive x.) However, this test fails to give us any characterization of x = 0, since f is iscontinuous here (a fact we can anticipate, since f is unefine at 0); thus, we still have to use the First Derivative Test in this case. (b) Fin the absolute maximum an minimum values of f on the interval [1, 8]. Justify completely. (Hint: it might be useful to note that f(x) = x 2/3 (9 x).) (3 points) We can apply the Close Interval Metho. Since the critical number x = 0 is not in this interval, we nee only compute the values f(1), f(8), an f(18/); we fin f(1) = 8, f(8) = 8 2/3 (9 8) = 4, f ( ) 18 = ( ) 18 2/3 ( 9 18 ) = ( 18 ) 2/3 ( ) 27 Now although comparing f( 18 ) against the other two values initially seems aunting, recall that f is increasing for all x in [0, 18 ], an thus f(1) < f(18 ), which guarantees that f(18 ) is the largest of the three values above. The smallest is f(8) = 4, so we conclue that on [1, 8], the absolute maximum value of f is ( ) 18 2/3 ( 27 ), an the absolute minimum value of f is 4. Notes: One coul alternatively observe that c = 18 is the sole critical number in the interval [1, 8], an that we ve alreay etermine that f has a local maximum here. Thus, by the Local to Global principle, in fact f has its absolute maximum on [1, 8] here as well. Incientally, it s actually quite simple to irectly etermine that f( 18 ) is greater than f(1) = 8, without a calculator: since 18 > 3 an 27 >, we have f ( ) 18 = ( 18 ) 2/3 ( ) 27 > 3 2/3 = 9 1/3 > 8 1/3 = 2 = 10 > 8 = f(1)

10 Math 41, Autumn 2010 Solutions to Secon Exam November 4, 2010 Page 10 of (17 points) Consier the function g(x) = xe x. (a) Determine if g has any asymptotes (horizontal an vertical), with complete reasoning. If there is any vertical asymptote, compute both corresponing one-sie its. (4 points) Let us compute the horizontal an vertical asymptotes. Recall that a horizontal asymptote correspons to the its g(x) an g(x), if either exists, while the vertical x asymptotes correspon to values of a (if any) such that the it of g(x) as x approaches a from at least one sie is + or. Horizontal asymptotes: Let us first calculate the horizontal asymptotes. We have that x xe x = e x This is in the form, so we can use l Hôpital s rule. Thus, we get that xe x = x ex = 1 e x = 0 since ex =. Thus, we get the horizontal asymptote y = 0 as x. Let us now check for an asymptote at. First note that Thus, x e x = +. x xe x =, because the first factor approaches an the secon approaches +. Therefore, there s no horizontal asymptote at. Vertical Asymptotes: Since the function g(x) = xe x is efine an continuous for all values of x, it follows that there are no vertical asymptotes. Thus, the only asymptote is the horizontal asymptote y = 0 as x. (b) On what interval(s) is g increasing? ecreasing? Explain completely. (4 points) g is increasing whenever g (x) > 0, an is ecreasing whenever g (x) < 0. We have that g (x) = ( xe x ) = x ( ) e x + x e x = xe x + e x = e x (1 x) Thus, g (x) = e x (1 x). Since e x is always positive, we just nee to examine 1 x, which is clearly positive when x < 1 an negative when x > 1. Thus, we see that g(x) is increasing on (, 1) an ecreasing on (1, ).

11 Math 41, Autumn 2010 Solutions to Secon Exam November 4, 2010 Page 11 of 12 (c) On what interval(s) is g concave up? concave own? Explain completely. (4 points) g is concave up whenever g (x) > 0, an is concave own whenever g (x) < 0. Since g (x) = (1 x)e x, we get that g (x) = (1 x)e x = (1 x)( e x ) e x = (x 2)e x Similarly to above, we see that g (x) is positive whenever x 2 is positive, an is negative whenever x 2 is negative. Thus, we see that g(x) is concave own on (, 2) an concave up on (2, ). () Using the information you ve foun, sketch the graph y = g(x). Label an provie the (x, y) coorinates of any local extrema an inflection points. ( points)

12 Math 41, Autumn 2010 Solutions to Secon Exam November 4, 2010 Page 12 of (10 points) At any point near a fixe electric charge, the electrostatic potential ue to the charge is equal to the amount of charge (in coulombs) ivie by the istance to the charge (in centimeters). Suppose two charges, one measuring 80 coulombs an the other 20 coulombs, are place 10 cm apart. At what point on the line between the charges is the total electrostatic potential minimize? Justify completely. Let s call the charge with 80 coulombs A, an the other one B. Given a point on the line between A an B, let x be the istance from the point to A, an so the istance from the point to B is 10 x. Then the total electrostatic potential is given by C(x) = 80 x x The omain of C is (0, 10), because the point must lie on the line segment between A an B, an C(x) is unefine when x = 0 or x = 10. To fin the absolute minimum value of C, we first ientify all of its critical numbers: Setting C (x) = 0, we get C (x) = 80 x (10 x) 2 = 80(10 x)2 + 20x 2 x 2 (10 x) 2 80(10 x) x 2 = 0 = 3x 2 80x = 0 = (3x 20)(x 20) = 0 So x = 20/3 or 20, but only x = 20/3 is in the omain of C. Next, we look for points where C (x) is unefine. The only places where C is unefine is x = 0 or 10, neither of which is in the omain of C. So we have only one critical number, x = 20/3. Finally, we nee to verify x = 20/3 is the point of absolute minimum for C. First we use the Growth Criterion: C(x) = x 0 + C(x) = x 10 Hence by continuity, the function C must have an absolute minimum on its omain, which must occur at a critical number; this must be at x = 20/3, the sole critical number on the omain of C. Alternatively, we can use the First Derivative Test. When x < 20/3, (3x 20)(x 20) > 0, an when 20/3 < x < 10, (3x 20)(x 20) < 0. Since x 2 (10 x) 2 > 0 on the omain of C, we conclue C (x) < 0 when 0 < x < 20 3 C (x) > 0 when 20 3 < x < 10 As before, we see that C must achieve its absolute minimum at x = That is, the position on the line between A an B that is a istance 20 3 cm away from A has the smallest electrostatic potential.