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# Solutions to More Applications of Derivatives Practice Problems for AP Calculus

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By — McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these solutions can be found at: More Applications of Derivatives Practice Problems for AP Calculus

Part A The use of a calculator is not allowed.

1. Equation of tangent line:
2. f (a + Δx) ≈ f(a) + f '(a)Δx
3. f(a + Δx) ≈ f(a) + f'(a)Δx Convert to radians:
4. Step 1: Find mtangent.
5. Step 2: Set mtangent. =slope of line y – 12x = 3.

Since y – 12x = 3 y = 12x +3, then m =12. Set 3x2 = 12 x = ± 2 since x ≥ 0, x = 2. Set –3x2 = 12 x2 = – 4.

Step 3: Find the point on the curve. (See Figure 9.8-1.)

At x = 2, y = x3 = 23 =8. Thus the point is (2, 8).
6. Step 1: Find mtangent.
7. Step 2: Find mnormal.

Step 3: Write equation of normal

At x = ln 2, y =ex = eln2 = 2. Thus the point of tangency is (ln 2, 2).
The equation of normal:
y – 2 = – (x – ln 2) or y = – (x – ln 2) + 2.
8. Step 1: Find mtangent.
(x – ln 2) or y = –
9. Step 2: Find the slope of line y – 2x = b

y – 2x = b (x – ln 2) or y = –

Step 3: Find point of tangency.

Set mtangent. =slope of line
y – 2x = b –2x = 2 x = – 1.
At x = – 1, y = – x2 + 4 = – (–1)2 + 4 = 3; (–1, 3).

Step 4: Find b.

Since the line y – 2x = b passes through the point (–1, 3), thus 3 – 2(–1) = b or b = 5.
10. v(t) = s'(t) = t2 – 6t;
11. a(t) = v'(t) = s''(t) = 2t – 6

Set a(t) = 0 2t – 6 = 0 or t = 3.

v(3) = (3)2 – 6(3)= – 9;

s(3) = 3 – 3(3)2 + 4 = – 14.

12. On the interval (0, 1), the slope of the line segment is 2. Thus the velocity v(t)=2 ft/s. On (1, 3), v(t)=0 and on (3, 5), v(t) = – 1. See Figure 9.8-2.
13.
1. At t = t2, the slope of the tangent is negative. Thus, the particle is moving to the left.
2. At t = t1, and at t = t2, the curve is concave downward = acceleration is negative.
3. At t = t1, the slope > 0 and thus the particle is moving to the right. The curve is concave downward the particle is slowing down.
14.
1. At t = 2, v(t) changes from positive to negative, and thus the particle reverses its direction.
2. At t = 1, and at t =3, the slope of the tangent to the curve is 0. Thus the acceleration is 0.
3. At t = 3, speed is equal to | &38211; 5| =5 and 5 is the greatest speed.
15.
1. s (4)= – 16(4)2 + 640 = 384 ft
2. v(t) = s'(t) = – 32t
3. v(4) = – 32(4) ft/s = – 128 ft/s

16.
1. At t = 5, s(t)=1.
2. For 3 < t < 4, s(t) decreases. Thus, the particle moves to the left when 3 < t < 4.
3. When 4 < t < 6, the particle stays at 1.
4. When 6 < t < 7, speed=2 ft/s, the greatest speed, which occurs where s has the greatest slope.
17. Part B Calculators are allowed.

18. Step 1: v(t) = 3t2 – 6t
19. a(t) = 6t – 6

Step 2: Set v(t) = 0 3t2 – 6t = 0

3t(t – 2) = 0, or t = 0 or t = 2

Set a(t) = 0 6t – 6 = 0 or t = 1.

Step 3: Determine the directions of motion. See Figure 9.8-3.

Step 4: Determine acceleration. See Figure 9.8-4.

Step 5: Draw the motion of the particle. See Figure 9.8-5. s (0)=1, s (1)= – 1, and s (2) = – 3.

The particle is initially at 1 (t = 0). It moves to the left speeding up until t =1, when it reaches –1. Then it continues moving to the left, but slowing down until t = 2 at – 3. The particle reverses direction, moving to the right and speeding up indefinitely.

20. Linear approximation:
21. y = f (a) + f'(a)(xa) a = π

f (x ) = sin x and f'(π) = sin π = 0

f'(x) = cos x and f'(π)= cos π = – 1.

Thus, y = 0 + (–1)(x – π) or y = – x + π.

22. y = f (a) + f'(a)(xa)
23. f (x) = ln (1+x) and f(2)= ln (1+2)= ln 3

24. Step 3: Find points of tangency

At y = 0, y2 = 4 – 4x2 becomes 0 = 4 – 4x2 x = ± 1
Thus points of tangency are (1, 0) and (–1, 0).

Step 4: Write equations of vertical tangents x =1 and x = – 1

25. Step 1: Find for y = ln x and y = x2 + 3.
y = ln x;
y = x2 +3; = 2x
26. Step 2: Find the x-coordinate of point(s) of tangency.

27. s1(t) = ln t and s'1(t) = ;1 ≤ t ≤ 8
28. s2(t) = sin(t) and

s'2 (t) = cos(t); 1 ≤ t ≤ 8

Enter y 1 = and y2 = cos(x). Use the [Intersection] function of the calculator and obtain t = 4.917 and t =7.724.

29. Step 1: s(t) = sin t
v(t) = cos t
a(t) = – sin t
30. Step 2: Set v(t) = 0 cos t = 0; t = and .

Set a(t) = 0 – sin t =0; t = π and 2π.

Step 3: Determine the directions of motion. See Figure 9.8-6.

Step 4: Determine acceleration. See Figure 9.8-7.

Step 5: Draw the motion of the particle. See Figure 9.8-8.

31. s(t) = – 16t2 + v0t + s0 s0 = height of building and v0 =0.
32. Thus, s(t) = – 16t2 + s0. When the coin hits the ground, s(t)=0, t = 10.2. Thus, set s(t) = 0 –16t2 + s0 = 0 – 16(10.2)2 + s0 = 0

s0 = 1664.64 ft. The building is approximately 1665 ft tall.

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