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.
 Equation of tangent line:
 f (a + Δx) ≈ f(a) + f '(a)Δx
 f(a + Δx) ≈ f(a) + f'(a)Δx Convert to radians:
 Step 1: Find m_{tangent.}
 Step 1: Find m_{tangent.}
 Step 1: Find m_{tangent.}
 (x – ln 2) or y = –
 v(t) = s'(t) = t^{2} – 6t;
 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 9.82.

 At t = t_{2, the slope of the tangent is negative. Thus, the particle is moving to the left.}
 _{At t = t1, and at t = t2, the curve is concave downward = acceleration is negative.}
 _{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.}

 _{At t = 2, v(t) changes from positive to negative, and thus the particle reverses its direction.}
 _{At t = 1, and at t =3, the slope of the tangent to the curve is 0. Thus the acceleration is 0.}
 _{At t = 3, speed is equal to  &38211; 5 =5 and 5 is the greatest speed.}

 _{s (4)= – 16(4)2 + 640 = 384 ft}
 _{v(t) = s'(t) = – 32t}
 _{}
 _{}
 _{}
_{v(4) = – 32(4) ft/s = – 128 ft/s}

 _{At t = 5, s(t)=1.}
 _{For 3 < t < 4, s(t) decreases. Thus, the particle moves to the left when 3 < t < 4.}
 _{When 4 < t < 6, the particle stays at 1.}
 _{When 6 < t < 7, speed=2 ft/s, the greatest speed, which occurs where s has the greatest slope.}
 _{Step 1: v(t) = 3t2 – 6t}
 _{Linear approximation:}
 _{y = f (a) + f'(a)(x – a)}
 _{}
 _{Step 1: Find for y = ln x and y = x2 + 3.}
 _{y = ln x; }
 _{y = x2 +3; = 2x}
 _{s1(t) = ln t and s'1(t) = ;1 ≤ t ≤ 8}
 _{Step 1: s(t) = sin t}
 _{v(t) = cos t}
 _{a(t) = – sin t}
 _{s(t) = – 16t2 + v0t + s0 s0 = height of building and v0 =0.}
 _{}
 _{}
 _{}
 _{}
 _{}
Step 2: Set m_{tangent.} =slope of line y – 12x = 3.
Since y – 12x = 3 y = 12x +3, then m =12. Set 3x^{2} = 12 x = ± 2 since x ≥ 0, x = 2. Set –3x^{2} = 12 x^{2} = – 4.
Step 3: Find the point on the curve. (See Figure 9.81.)
 At x = 2, y = x^{3} = 2^{3} =8. Thus the point is (2, 8).
Step 2: Find m_{normal.}
Step 3: Write equation of normal
 At x = ln 2, y =e^{x} = e^{ln2} = 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.
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 m_{tangent.} =slope of line
 y – 2x = b –2x = 2 x = – 1.
 At x = – 1, y = – x^{2} + 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.
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.
_{Part B Calculators are allowed.}
_{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 9.83.}
_{}
_{Step 4: Determine acceleration. See Figure 9.84.}
_{}
_{Step 5: Draw the motion of the particle. See Figure 9.85. 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.}
_{y = f (a) + f'(a)(x – a) 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.}
_{}
_{f (x) = ln (1+x) and f(2)= ln (1+2)= ln 3}
_{}
_{}
_{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}
_{Step 2: Find the xcoordinate of point(s) of tangency.}
 _{}
_{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.}
_{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 9.86.}
_{}
_{Step 4: Determine acceleration. See Figure 9.87.}
_{}
_{Step 5: Draw the motion of the particle. See Figure 9.88.}
_{}
_{}
_{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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