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Practice Problems for Exam #2 Solutions 1. a) Evaluate lim x→0 tan x 4x Use your limit laws: tan x 1 tan x = lim x→0 4x 4 x sin x 1 = lim cos x 4 x→0 x 1 limx→0 sinx x = 4 limx→0 cos x 1 11 = . = 41 4 Here we also use the fact that the cosine function is continuous and cos 0 = 1, so limx→0 cos x = lim x→0 cos 0 = 1. b) Find dy/dx and d2 y/dx2 if y = x tan x − sin2 x. Nothing tricky here; just use our derivative rules and formulas. On the exam BE SURE TO SHOW ALL OF THE STEPS!! dy dx tan x d sin2 x = − dx dx dx dx d tan x du2 d sin x = tan x +x − dx dx du dx (u = sin x) = tan x + x sec2 x − 2u cos x = tan x + x sec2 x − 2 sin x cos x. d2 y d tan x dx sec2 x d2 sin x cos x = + − dx2 dx dx dx 2 dx d sec x d sin x cos x = sec2 x + sec2 x +x −2 dx dx dx 2 d sin x d cos x du d sec x 2 2 = sec x + sec x + x − 2 cos x + sin x du dx dx dx (u = sec x) = 2 sec2 x + x(2u) sec x tan x − 2(cos2 x − sin2 x) = 2 sec2 x + x2 sec2 x tan x − 2 cos2 x + 2 sin2 x. 2. A 13 foot ladder leaning up against a wall is sliding down the wall at 1 foot per second. The ladder makes an angle θ with the ground. How is this angle θ changing when the top of the 1 ladder is five feet from the ground? Draw an appropriate picture, which in this case will be a right triangle. The base x is the distance from the bottom of the ladder to the wall, the opposite side y is the wall (just from the ground up to the top of the ladder), and the hypotenuse 13 is the ladder. Then sin θ = y/13. Differentiating with respect to time t gives dy/13 d sin θ = dt dt d sin θ dθ 1 dy = dθ dt 13 dt dθ −1 cos θ = , dt 13 since we’re told that dy/dt is −1 foot per second (negative since y is getting smaller with time). Now when y = 5, then by the Pythagorean Theorem x = 12 (using x2 + y 2 = 132 ). Since cos θ = x/13, we get cos θ = 12/13 when y = 5. Plugging that in above, we see that dθ/dt = −1/12 when y = 5. This is in radians per second, and is negative since this angle is getting smaller with time. 3. Find an equation for the tangent line to the curve y 2 = 2x2 − xy at the point (1, 1) As always, the derivative gives the slope of the tangent line. To find dy/dx we use implicit differentiation: dy 2 d2x2 − xy = dx dx 2 dy dy d2x2 dxy = − dy dx dx dx 2 dx dx dy 0 2yy = 2 − y +x dx dx dx 2yy 0 = 4x − (y + xy 0 ). At the point x = 1, y = 1 we have 2y 0 = 4 − (1 + y 0 ), so y 0 = 1. That’s the slope of the tangent line, so in point-slope form an equation for the tangent line is (y − 1) = 1(x − 1). 4. Use the linearization to approximate sin(π/2 − .01). 2 Here we are approximating the graph of y = sin x when x is near π/2. So first we find the linearization of sin x at x = π/2. The derivative is d sin x/dx = cos x, which is cos(π/2) = 0 at x = π/2. The function value is sin(π/2) = 1 at x = π/2. Thus, an equation for the tangent line at x = π/2 is (y − 1) = 0(x − π/2), so the linearization (just solve for y) is L(x) = 1. This means that sin(π/2 − .01) ≈ L(π/2 − .01) = 1. 5. Locate the absolute extrema for f (x) = x3 + 3x2 − 24x + 4 on [−5, 5]. We first find any critical points within the interval. For this function, f 0 (x) = 3x2 + 6x − 24 = 3(x2 + 2x − 8) = 3(x − 2)(x + 4). There are two critical points; one at x = 2 and the other at x = −4. The absolute extrema thus occur at x = ±5, 2 or −4. Plugging these values into our function, we have f (−5) = −125 + 75 + 120 + 4 = 74, f (2) = 8 + 12 − 48 + 4 = −24, f (5) = 125 + 75 − 120 + 4 = 84, f (−4) = −64 + 48 + 96 + 4 = 84. The absolute minimum is −24 and occurs at x = 2. The absolute maximum is 84 and occurs at x = −4 and at x = 5. 3