PRACTICE EXAM 2 SOLUTIONS 1. Use the Limit Definition of the Derivative to Find F/(X) for the Following Functions F(X) (Sec

PRACTICE EXAM 2 SOLUTIONS

1. Use the limit deﬁnition of the derivative to ﬁnd f 0(x) for the following functions f(x) (Sec. 2.2): f(x) = 3x2 + 2x + 7

Solution.

3(x + h)2 + 2(x + h) + 7 − (3x2 + 2x + 7) f 0(x) = lim h→0 h

3h2 + 6xh + 2h f 0(x) = lim = 6x + 2 h→0 h

f(x) = x3

Solution.

(x + h)3 − x3 f 0(x) = lim h→0 h

3x2h + 3xh2 + h3 f 0(x) = lim = 3x2 h→0 h

1 f(x) = x+3

Solution.

1/(x + h + 3) − 1/(x + 3) f 0(x) = lim h→0 h

−h/(x + 3)(x + h + 3) −1 f 0(x) = lim = h→0 h (x + 3)2

1 d 2. Find dx for each of the following functions. Show all of your work. (Sec. 2.3-2.5) f(x) = x2 + 1/x Solution.f 0(x) = 2x − 1/x2

g(x) = (3x2 − 2x)12 Solution.g0(x) = 12(3x2 − 2x)11(6x − 2)

6(x−3) h(x) = x+3 0 6(x+3)−6(x−3) 36 Solution.h (x) = (x+3)2 = (x+3)2

m(x) = x cos(x2 + 3) Solution.m0(x) = cos(x2 + 3) − x sin(x2 + 3)(2x)

sin(2πx) r(x) = x 0 2πx cos(2πx)−sin(2πx) Solution.r (x) = x2

2 x v(x) = 3 sin ( cos x ) 0 x x cos x+x sin x Solution.v (x) = (3)2 sin( cos x ) cos( cos x )[ (cos x)2 ]

w(x) = 9(x2 + 3x + 2)(tan x + 1) Solution.w0(x) = 9[(x2 + 3x + 2)(sec2 x) + (2x + 3)(tan x + 1)]

√ z(x) = 3 x2 + 3 Solution.z0(x) = 1/3(x2 + 3)−2/3(2x)

θ(x) = √x−1 x2−1 √ 0 x2−1−(x−1)(x)(x2−1)−1/2 Solution.θ (x) = x2−1 0 x−1 θ (x) = (x2−1)3/2

q(x) = 1/3x2/3 + (1 − x)1/3 Solution.q0(x) = 2/9x−1/3 + (−1)(1/3)(1 − x)−2/3

2 3. Find the indicated derivative for each function. Show your work. (Sec 2.6)

Find y000, y = 4x4 − x2 + 7. Solution. y0 = 16x3 − 2x y00 = 48x2 − 2 y000 = 96x

d2f Find dx2 , where f(x) = x sin x. Solution. f 0(x) = x cos x + sin x f 00(x) = cos x − x sin x + cos x = 2 cos x − x sin x

Find r(4), where r(x) = 1/x. Solution. r0(x) = −1/x2 r00(x) = 2/x3 r000(x) = −6/x4 r(4)(x) = 24/x5

4. Find a tangent line through the curve 2x2 + 3y2 = 14 at the point (1, 2). (Sec. 2.7)

0 −2x Solution.y = 3y y0(1, 2) = −1/3 Tangent Line y − 2 = −1/3(x − 1)

dy 5. Find dx for the following curves (Sec. 2.7).

x3 + y3 = x3y3

3y2−3y2x3 Solution.dy/dx = 3x2y3−3x2

y2 = 4x

2 Solution.dy/dx = y

3 6. (Sec 2.8 and 2.9) A rectangular box which is initially a cube of side length 2 inches is growing. The length is increasing by 1 in/sec, the width is increasing by 2 in/sec, and the height is increasing by 3 in/sec.

a) How fast is the volume increasing after 3 seconds? Solution. V = lwh. Also given in the problem are dl/dt = 1, dw/dt = 2, dh/dt = 3. d dl dh dw dl dV/dt = l dt (wh) + dt (wh) = l(w dt + h dt ) + dt (wh) dh dw dl dV/dt = lw dt + lh dt + wh dt

At 3 seconds, l = 5, w = 8, h = 11. dV/dt(3) = 40(3) + 55(2) + 88(1) = 318 in3/sec. Remark. The same solution could have also been found by setting l = 2 + t, w = 2 + 2t, h = 2 + 3t and substituting these immediately before diﬀerentiating. This method will be used in part (d).

b) How fast is the surface area increasing after 3 seconds? Solution. SA = 2lw + 2lh + 2wh dw dl dh dl dh dw d(SA)/dt = 2l dt + 2w dt + 2l dt + 2h dt + 2w dt + 2h dt d(SA)/dt(3) = 2(5)(2)+2(8)(1)+2(5)(3)+2(11)(1)+2(8)(3)+2(11)(2) = 180 in2/sec.

c) Using diﬀerentials, estimate the volume at 3.1 seconds using your answer from part (a). Solution. At t = 3,V = (5)(8)(11) = 440. dV/dt(3) = 318 dV = 318dt dV = 318(0.1) = 31.8 Estimated Volume V (3.1) ≈ 471.8

d) Do parts (a) and (c) again, but assuming that the length of the box instead follows the equation l = 5 + 3 sin(t), where t is in seconds. Set l = 5 + 3 sin(t), w = 2 + 2t, h = 2 + 3t Then V = lwh = (5 + 3 sin(t))(2 + 2t)(2 + 3t) = (5 + 3 sin(t))(4 + 10t + 6t2) dV/dT = (5 + 3 sin(t))(10 + 12t) + (4 + 10t + 6t2)(3 cos t)

4 dV/dT (3) = 46(5 + 3 sin 3) + 88(3 cos 3) = −11.88 V (3) = 477.25 dV = −11.88dt = −11.88(0.1) = −1.188 Estimate for V (3.1) ≈ 476.06

7. (Sec 2.8 and 2.9) A lighthouse is 1 mile from shore and has a light that revolves at 2 revolutions per second.

a) How fast is the beam moving along the shoreline when it passes the point on the shore closest to the lighthouse? Solution. If θ is the angle the light makes with the line going directly to shore, then the following are true: θ = 0 is when the light passes the closest point on the shore. If x is the distance from the closest point on the shore to the light’s current position on the shore, then by setting up a triangle, tan(θ) = x/1 (opposite side over adjacent side). We know that the lighthouse revolves over an angle of 4π every second, that is, dθ/dt = 4π. We want to ﬁnd dx/dt. So diﬀerentiate the above equation with respect to t: sec2(θ)dθ/dt = dx/dt For θ = 0, dx/dt = sec2(0)4π = 4π(1) = 4π mi/sec.

b) How fast is the beam moving along the shoreline when it passes the point on the shore 1/2 mile from the closest point on the shoreline? p 2 √Solution. Same equation but when x = 1/2, sec(θ) = hypotenuse/adjacent = 1 + 1/2 /1 = 5/2. So dx/dt = sec2(θ)4π = 4π(5/4) = 5π mi/sec.

c) Use diﬀerentials to estimate where the beam of light will be shining 1/10 second after it passes the closest point on the shoreline, using your answer from part (a). Solution. When the beam of light is shining at the closest point, x = 0. dx = 4πdt = 4π(0.1) = 0.4π So we estimate the light is shining 0.4π or 1.26 miles away from the closest point to the shore 1/10 second later.