18.01SC Exam 2 Solutions

1. (a) Give a general expression for the quadratic approximation to a twice differentiable function f (x) at x = a. f (x) = f (a) + f 0 (a)(x − a) + f 00 (a) (x − a)2 2 (b) Use your answer from part (a) to give an approximate value for ln(1.2), where ln(x) is the natural log function. (ln x)0 = ln(1.2) ≈ = = ln(1.2) ≈ 1 , x (ln x)00 = − 1 x2 ln00 (1) ln 1 + (ln (1))(1.2 − 1) + (1.2 − 1)2 2 1 0 + 0.2 − (0.2)2 2 0.04 0.2 − 2 0.18 0 2. Salt is poured from a conveyer belt at a rate of 30 ft3 /min, forming a conical pile with a circular base whose height and diameter of base are always equal. How fast is the height of the pile increasing when the pile is 10 ft. high? h = 2r V 1 2 πr h 3  2 1 h = π h 3 2 1 3 = πh 12 = dV 1 2 dh = πh dt 4  dt  dh dV 4 ⇒ = dt dt πh2   4 = (30) 102 π 120 100π 12 ft = 10π min = 3. Draw a careful picture of the graph of the function f (x) = x − 3x1/3 . Be sure to indicate the coordinates of any local maxima and minima, the intervals on which the function is increasing and decreasing, and asymptotes (if any of these features occur). Computing inflection points may help you draw an accurate picture, but is not necessary. 1/3 When f (x) = 0, x − 3x√ = 0,√so x1/3 (x2/3 − 3) = 0. f (x) = 0 when x = 0, 3 3, −3 3. f 0 (x) = 1 − 3 · 31 x−2/3 = 1 − x−2/3 : critical points are at x = ±1. Critical points are (1, −2) (minimum) and (1, 2) (maximum). The tangent line to the graph is vertical at x = 0. f (x) is increasing on (−∞, −1) and (1, ∞) and decreasing on (−1, 1). y 4. A metal storage tank is to be made in the shape of a cylinder with a circular base (−1,and 2) a hemispherical top. Find the dimensions of the tank which require the least amount of metal used to hold a fixed constant volume V . √ −3 3 Start by drawing a diagram.
Constant V = 12 43 πr3 +√ (πr2 h) x= 23 πr3 + πr2 h. 3 3 (1, −2) (Assuming) the amount of metal used is proportional to the surface area, we want to minimize surface area SA. 1 (4πr2 ) + (2πr)h + πr2 2 = 2πr2 + 2πrh + πr2 SA = 3πr2 + 2πrh SA = V − 23 πr3 h = πr2 V − 23 πr3 S A = 3πr + 2πr πr2 2V 4 = 3πr2 + − πr2 r 3 5 2 2V = πr + . 3 r 2 d SA dr = 10 πr 3   + (−1) 2V . Set this equal to 0, assuming r 6= 0. r2 10 2V πr + (−1) 2 = 0 3 r 5 V πr = 2 3 r 5 3 V = πr 3 This tells us that r = q 3 3V 5π and: h= 5 πr3 3 − 23 π r3 πr3 = = r. πr2 πr2 q is a critical point. Show that it is indeed a minimum by checking h = r = 3 3V 5π the boundaries r = 0 and h = 0. When r = 0, SA = 0 + ∞. This is not a minimum. 2 2 πr3 . 3 When h = 0, SA = 3πr and V = Therefore r = when h = 0. q
3V 2/3 , SA = 5π . When h = r = 3 3V 5π 5π q 3 3V 2π and SA = 3π
3V 2/3 2π Comparing the surface areas when h = r and when h = 0 by cubing both sides, we see that: 125 · 9/25 = 45 < 27 · 9/4 = 60.75. We conclude that the critical point represents the minimum surface area, and so r 3 3V the answer to the question is h = r = 5π 5. Explain why Newton’s method eventually fails when finding zeroes of f (x) = x3 − 3x + 7 with a starting value x1 = 2. Newton’s method says that xn = xn−1 − For x1 = 2, x2 = 2 − x3 = 1 − f (1) f 0 (1) =1− f (2) f 0 (2) 5 0 =2− 9 9 f (xn−1 ) . f 0 (xn−1 ) Here f 0 (x) = 3x2 − 3. = 1. is undefined. In particular, f 0 (x2 ) = 0. The problem is that the tangent line to the graph of f at x = x2 = 1 is horizontal. Thus, it does not intersect the x-axis and we cannot continue Newton’s method. 6. Prove that √ Let g(x) = 1 + 12 − √ 1 1 + x < 1 + x, 2 if x > 0. 1 + x. We need to show that g(x) > 0 for x > 0. g(0) = 1 + 0 − 1 = 0, so by the mean value theorem, it’s sufficient to show that g 0 (x) > 0 for all x > 0. g 0 (x) = 0 + 1 1 1 1 1 = − √ . − √ 2 2 1+x 2 2 1+x Because x > 0, √ √ 1+x > 1 = 1, so 1 1 √ < = 1, and 1 1+x 1 1 1 − √ > − ·1=− . 2 2 2 1+x Therefore g 0 (x) = 1 1 1 1 − √ > − = 0. 2 2 1+x 2 2 We have√shown g(0) = 0 and g 0 (x) > 0 for x > 0, so g(x) > 0 for x > 0, which implies 1 + x < 1 + 12 x if x > 0.