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MATH 1914 Homework 9 Solutions Section 3.5 Problems 6. Use the guidelines of this section to sketch the curve. y = 2x − tan x, −π/2 < x < π/2 y 0 = 2 − sec2 x y 00 = −2 sec x(sec x tan x) = −2 sec2 x tan x Critical points are the values of x so that y 0 = 0 Or 2 − sec2 x = 0 sec2 x = 2 1 =2 cos2 x 1 cos2 x = 2 1 cos x = √ 2 π π x=− , 4 4 Where the solutions are over the interval −π/2 < x < π/2. By either plugging test values or considering the fact that cos x increases on the interval (− π2 , 0) and decreases on the interval (0, π2 ), the signs for y 0 is as follows: x y0 (− π2 , − π4 ) (− π4 , π4 ) ( π4 , π2 ) − + − The function achieves local minimum at x = − π4 , y = − π2 + 1, and local maximum at x = π4 , y = π2 − 1 For the concavity, set y 00 = 0. This implies either sec x = 0 or tan x = 0. Since sec x = cos1 x can never be zero, the only solution comes from tan x = 0, or x = 0. The table for concavity 1 is as follows: x y 00 (− π2 , 0) (0, π2 ) + − Before drawing the graph, the function y has the following vertical assymptotes: when x → − π2 + , tan x → −∞ and when x → π2 − , tan x → ∞. Also, notice when replacing x with −x, the y values also gets negated. This implies that the function is symmetric about the origin. The graph is as follows. 8 Use the guidelines of this section to sketch the curve. In guideline D find an equation of the slant asymptote. 8. 2x2 y= x−1 First perform long division to simplify the calculation: y = 2x + 2 + 2 x−1 2 (x − 1)2 4 y 00 = (x − 1)3 y0 = 2 − 2 (*) The critical points are where 2− 2 =0 (x − 1)2 (x − 1)2 = 1 x − 1 = ±1 x = 0, 2 The derivative is also undefined at x = 1. The table for the first derivative is as follows: x y0 (−∞, 0) (0, 1) (1, 2) (2, ∞) + − − + with local minimum at (0, 0) and local maximum at (2, 8). Similarly, inflection points are the x-values such that y 00 = 0, which is never satisfied for all values of x. However, y 00 is undefined at x = 1. The table for concavity is as follows: x y 00 (−∞, 1) (1, ∞) − + In addition, the funciton also has a vertical asymptote at x = 1. y → −∞ as x → 1− and y → +∞ as x → 1+ . For the slant assymptote, notice in the (*) expression of y, the 2 term x−1 has limit 0 as x → ∞. Therefore the slant assymptote is y = 2x + 2. The graph is as follows. 3 Section 3.7 Problems 11. Find two positive numbers whose product is 200 and whose sum is a minimum. Let the two numbers be x and y. Since xy = 200, y = function of x, the sum can be written as s(x) = x + 200 . x To minimize the sum as a 200 x Its extreme is achieved at the critical point, when s0 (x) = 1 − 200 = 0 or equivalently x2 √ x = 200. Using√either the first or second derivative test one can conclude this is a local minimum. x = y 200. 16. Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r. Let θ be the angle OAB as drawn in the graph. Let a(θ) be the area of the isoscele triangle ACD as a function of the angle θ. Since the radius of the circle is r, OA = OD = OC = r AB = OA · cos θ = r cos θ OB = OA · sin θ = r sin θ Therefore the area can be written as 1 a(θ) = AD · CB 2 1 = · 2AB · (CO + OB) 2 = r cos θ(r + r sin θ) = r2 cos θ(1 + sin θ) where r is a constant. To maximize the area is to find the critical point of this function, or a0 (θ) = r2 ((− sin θ)(1 + sin θ) + cos2 θ) = r2 (− sin θ − sin2 θ + cos2 θ) 4 To solve the equation a0 (θ) = 0, rewrite the second factor as − sin θ − sin2 θ + (1 − sin2 θ) = 0 −2 sin2 θ − sin θ + 1 = 0 This is equivalent to solving the equation −2u2 −u+1 = 0 when u = sin θ. The quadratic formula yields u = 21 or u = −1 (discard). Note this is when θ = π6 = 30◦ . To intepret the result, notice the angle AOB equals 90◦ − 30◦ = 60◦ . Consequently, the angle DOB equals 60◦ and the total angle AOD equals 120◦ . The surrounding two angles DOC and AOC are equal based the symmetry, and each should be 120◦ . This implies that ACD is in fact an equilateral triangle. 18. Two vertical poles at P and Q are secured by a rope PRQ going from the top of the first pole to a point R on the ground between the poles and then to the top of the second pole as in the figure. Show that the shortest length of such a rope occurs when θ1 = θ2 . Let x be the distance between M and R as shown in the graph. Let H be the fixed height of the corridor and L be the distance between the two poles P and Q. The length of the rope s(x) can be written as a function in x: s(x) = P R + QR p √ = H 2 + x2 + (L − x)2 + H 2 Its minimum is achieved at is critical point where s0 (x) = 0: 2x −2(L − x) s0 (x) = √ +p =0 2 H 2 + x2 (L − x)2 + H 2 5 This is equivalent to 2(L − x) 2x √ =p 2 2 2 H +x (L − x)2 + H 2 (L − x) x √ =p 2 H 2 + x2 (L − x)2 + H 2 Without solving for actual values of x, this equation is the exact statement for RN MR = PR RQ cos θ1 = cos θ2 Since the cosine function is one-to-one on the interval (0, π2 ), the minimum is achieved when θ1 = θ2 . 6