More Applications of Derivatives - Review the Knowledge You Need to Score High

5 Steps to a 5 AP Calculus AB & BC, 2012-2013 Edition (2011)

STEP 4. Review the Knowledge You Need to Score High

Chapter 9. More Applications of Derivatives

IN THIS CHAPTER

Summary: Finding an equation of a tangent is one of the most common questions on the AP Calculus exams. In this chapter, you will learn how to use derivatives to find an equation of a tangent, and to use the tangent line to approximate the value of a function at a specific point. You will also learn to find derivatives of parametric, polar, and vector functions, and to apply derivatives to solve rectilinear motion problems.

Key Ideas

Tangent and Normal Lines

Linear Approximations

Motion Along a Line

Parametric, Polar, and Vector Derivatives

9.1 Tangent and Normal Lines

Main Concepts: Tangent Lines, Normal Lines

Tangent Lines

If the function y is differentiable at x = a, then the slope of the tangent line to the graph of y at x = a is given as

Types of Tangent Lines

Horizontal Tangents: . (See Figure 9.1-1.)

Figure 9.1-1

Vertical Tangents: . (See Figure 9.1-2.)

Figure 9.1-2

Parallel Tangents: . (See Figure 9.1-3.)

Figure 9.1-3

Example 1

Write an equation of the line tangent to the graph of y = −3 sin 2x at . (See Figure 9.1-4 on page 178.)

Figure 9.1-4

y = −3 sin 2x;

Slope of tangent : .

Point of tangency: At , y = −3 sin(2x)

= −3 sin[2(π/2)] = −3 sin (π) = 0.

Therefore, is the point of tangency.

Equation of Tangent: y − 0 = 6(x − π/2) or y = 6x − 3π.

Example 2

If the line y = 6x + a is tangent to the graph of y = 2x³, find the value(s) of a.

Solution:

. (See Figure 9.1-5.)

Figure 9.1-5

The slope of the line y = 6x + a is 6.

Since y = 6x + a is tangent to the graph of y = 2x³, thus for some values of x.

Set 6x² = 6 ⇒ x² = 1 or x = ± 1.

At x = −1, y = 2x³ = 2(−1)³ = −2; (−1, −2) is a tangent point. Thus, y = 6x + a ⇒ −2 = 6(−1) + a or a = 4.

At x = 1, y = 2x³ = 2(1)³ = 2; (1, 2) is a tangent point. Thus, y = 6x + a ⇒ 2 = 6(1) + a or a = −4.

Therefore, a = ± 4.

Example 3

Find the coordinates of each point on the graph of y² − x² − 6x + 7 = 0 at which the tangent line is vertical. Write an equation of each vertical tangent. (See Figure 9.1-6.)

Figure 9.1-6

Step 1: Find .

Step 2: Find .

Step 3: Find points of tangency.

At y = 0, y² − x² − 6x + 7 = 0 becomes − x² − 6x + 7 = 0 ⇒ x² + 6x − 7 = 0 ⇒ (x + 7)(x − 1) = 0 ⇒ x = −7 or x = 1.

Thus, the points of tangency are (− 7, 0) and (1, 0).

Step 4: Write equation for vertical tangents: x = −7 and x = 1.

Example 4

Find all points on the graph of y = |xe^x| at which the graph has a horizontal tangent.

Step 1: Find .

Step 2: Find the x-coordinate of points of tangency.

Horizontal Tangent .

If x ≥ 0, set e^x + xe^x = 0 ⇒ e^x(1 + x) = 0 ⇒ x = −1 but x ≥ 0, therefore, no solution.

If x < 0, set − e^x − xe^x = 0 ⇒ − e^x(1 + x) = 0 ⇒ x = −1.

Step 3: Find points of tangency.

At x = − 1, y = −xe^x = − (− 1)e⁻¹ = .

Thus at the point (− 1, 1/e), the graph has a horizontal tangent. (See Figure 9.1-7.)

Figure 9.1-7

Example 5

Using your calculator, find the value(s) of x to the nearest hundredth at which the slope of the line tangent to the graph of y = 2 ln(x² + 3) is equal to . (See Figures 9.1-8 and 9.1-9.)

Figure 9.1-8

Figure 9.1-9

Step 1: Enter y 1 = 2 * ln (x^∧2 + 3).

Step 2: Enter y 2 = d(y₁(x), x) and enter .

Step 3: Using the [Intersection] function of the calculator for y₂ and y₃, you obtain x = −7.61 or x = − 0.39.

Example 6

Using your calculator, find the value(s) of x at which the graphs of y = 2x² and y = e^x have parallel tangents.

Step 1: Find for both y = 2x² and y = e^x.

Step 2: Find the x-coordinate of the points of tangency. Parallel tangents ⇒ slopes are equal.

Set 4x = e^x ⇒ 4x − e^x = 0.

Using the [Solve] function of the calculator, enter [Solve] (4x − e^∧(x) = 0, x) and obtain x = 2.15 and x = 0.36.

• Watch out for different units of measure, e.g., the radius, r, is 2 feet, find in inches per second.

Normal Lines

The normal line to the graph of f at the point (x₁, y₁) is the line perpendicular to the tangent line at (x₁, y₁). (See Figure 9.1-10.)

Figure 9.1-10

Note that the slope of the normal line and the slope of the tangent line at any point on the curve are negative reciprocals, provided that both slopes exist.

(m_{normal line})(m_{tangent line}) = − 1.

Special Cases:

(See Figure 9.1-11.)

At these points, m_tangent = 0; but m_normal does not exist.

Figure 9.1-11

(See Figure 9.1-12.)

At these points, m_tangent does not exist; however m_normal = 0.

Figure 9.1-12

Example 1

Write an equation for each normal to the graph of y = 2 sin x for 0 ≤ x ≤ 2π that has a slope of .

Step 1: Find m_tangent.

Step 2: Find m_normal.

Figure 9.1-13

Step 3: Write equation of normal line.

At x = π, y = 2 sin x = 2(0) = 0; (π, 0).

Since m = , equation of normal is: =

Example 2

Find the point on the graph of y = ln x such that the normal line at this point is parallel to the line y = −ex − 1.

Step 1: Find m_tangent.

Step 2: Find m_normal.

Slope of y = −ex − 1 is − e.

Since normal is parallel to the line y = −ex − 1, set m_normal = −e ⇒ − x = −e or x = e.

Step 3: Find point on graph. At x = e, y = ln x = ln e = l. Thus the point of the graph of y = ln x at which the normal is parallel to y = −ex − 1 is (e, 1). (See Figure 9.1-14.)

Figure 9.1-14

Example 3

Given the curve : (a) write an equation of the normal to the curve at the point (2, ½), and (b) does this normal intersect the curve at any other point? If yes, find the point.

Step 1: Find m_tangent.

Step 2: Find m_normal.

Step 3: Write equation of normal.

Step 4: Find other points of intersection.

Using the [Intersection] function of your calculator, enter and and obtain x = −0.125 and y = −8. Thus, the normal line intersects the graph of at the point (−0.125, − 8) as well.

• Remember that ∫1dx = x + C and .

9.2 Linear Approximations

Main Concepts:

Tangent Line Approximation, Estimating the nth Root of a Number, Estimating the Value of a Trigonometric Function of an Angle

Tangent Line Approximation (or Linear Approximation)

An equation of the tangent line to a curve at the point (a, f(a)) is: y = f(a) + f′ (a)(x − a), providing that f is differentiable at a. (See Figure 9.2-1.) Since the curve of f(x) and the tangent line are close to each other for points near x = a, f (x) ≈ f(a) + f′ (a)(x − a).

Figure 9.2-1

Example 1

Write an equation of the tangent line to f (x) = x³ at (2, 8). Use the tangent line to find the approximate values of f (1.9) and f (2.01).

Differentiate f(x): f′(x) = 3x²; f′ (2) = 3(2)² = 12. Since f is differentiable at x = 2, an equation of the tangent at x = 2 is:

y = f(2) + f′(2)(x − 2)

y = (2)³ + 12(x − 2) = 8 + 12x − 24 = 12x − 16

f (1.9) ≈ 12(1.9) − 16 = 6.8

f (2.01) ≈ 12(2.01) − 16 = 8.12. (See Figure 9.2-2.)

Figure 9.2-2

Example 2

If f is a differentiable function and f(2) = 6 and , find the approximate value of f (2.1).

Using tangent line approximation, you have

(a) f (2) = 6 ⇒ the point of tangency is (2, 6);

(b) the slope of the tangent at x = 2 is ;

(d) thus, .

Example 3

The slope of a function at any point (x, y) is . The point (3, 2) is on the graph of f. (a) Write an equation of the line tangent to the graph of f at x = 3. (b) Use the tangent line in part (a) to approximate f (3.1).

Equation of tangent: y − 2 = −2(x − 3) or y = −2x + 8.

(b) f (3.1) ≈ − 2(3.1) + 8 ≈ 1.8

Estimating the nth Root of a Number

Another way of expressing the tangent line approximation is: f(a + Δx) ≈ f(a) + f′(a)Δx; where Δx is a relatively small value.

Example 1

Find the approximation value of using linear approximation.

Using f(a + Δx) ≈ f(a) + f′(a)Δx, let ; a = 49 and Δx = 1.

Thus, .

Example 2

Find the approximate value of using linear approximation.

Let f (x) = x^⅓, a = 64, Δx = −2. and , you can use f(a + Δx) ≈ f(a) + f′(a)Δx. Thus, .

• Use calculus notations and not calculator syntax, e.g., write ∫ x²dx and not ∫ (x^∧2, x).

Estimating the Value of a Trigonometric Function of an Angle Example

Approximate the value of sin 31°.

Note: You must express the angle measurement in radians before applying linear approximations. radians and radians.

Let f(x) = sin x, and Δx = .

Since f′(x) = , you can use linear approximations:

9.3 Motion Along a Line

Main Concepts:

Instantaneous Velocity and Acceleration, Vertical Motion, Horizontal Motion

Instantaneous Velocity and Acceleration

Example 1

The position function of a particle moving on a straight line is s(t) = 2t³ − 10t² + 5. Find (a) the position, (b) instantaneous velocity, (c) acceleration, and (d) speed of the particle at t = 1.

Solution:

(a) s (1) = 2(1)³ − 10(1)² + 5 = −3

(b) v(t) = s′(t) = 6t² − 20t

v(1) = 6(1)² − 20(1) = −14

a(1) = 12(1) − 20 = −8

(d) Speed = |v(t)| = |v(1)| = 14.

Example 2

The velocity function of a moving particle is for 0 ≤ t ≤ 7.

What is the minimum and maximum acceleration of the particle on 0 ≤ t ≤ 7?

(See Figure 9.3-1.) The graph of a(t) indicates that:

Figure 9.3-1

(1) The minimum acceleration occurs at t = 4 and a(4) = 0.

(2) The maximum acceleration occurs at t = 0 and a(0) = 16.

Example 3

The graph of the velocity function is shown in Figure 9.3-2.

Figure 9.3-2

(a) When is the acceleration 0?

(b) When is the particle moving to the right?

Solution:

(a) a(t) = v′(t) and v′(t) is the slope of tangent to the graph of v. At t = 1 and t = 3, the slope of the tangent is 0.

(b) For 2 < t < 4, v(t) > 0. Thus the particle is moving to the right during 2 < t < 4.

Thus, speed at t = 1 is |−4| = 4 which is the greatest speed for 0 ≤ t ≤ 4.

• Use only the four specified capabilities of your calculator to get your answer: plotting graphs, finding zeros, calculating numerical derivatives, and evaluating definite integrals. All other built-in capabilities can only be used to check your solution.

Vertical Motion

Example

From a 400-foot tower, a bowling ball is dropped. The position function of the bowling ball s(t) =- 16t² + 400, t ≥ 0 is in seconds. Find:

(a) the instantaneous velocity of the ball at t = 2s.

(b) the average velocity for the first 3 s.

Solution:

(a) v(t) = s′ (t) = −32t

v(2) = 32(2) = −64 ft/s

Thus, set s(t) = 0 ⇒ − 16t² + 400 = 0; 16t² = 400; t = ± 5. Since t ≥ 0, t = 5. The ball hits the ground at t = 5s.

• Remember that the volume of a sphere is v = and the surface area is s = 4πr². Note that v′ = s.

Horizontal Motion

Example

The position function of a particle moving in a straight line is s (t) = t³ − 6t² + 9t − 1, t ≥ 0. Describe the motion of the particle.

Step 1: Find v(t) and a(t).

v(t) = 3t² − 12t + 9

a(t) = 6t − 12

Step 2: Set v(t) and a(t) = 0.

Set v(t) = 0 ⇒ 3t² − 12t + 9 = 0 ⇒ 3(t² − 4t + 3) = 0

⇒ 3(t − 1)(t − 3) = 0 or t = 1 or t = 3.

Set a(t) = 0 ⇒ 6t − 12 = 0 ⇒ 6(t − 2) = 0 or t = 2.

Step 3: Determine the directions of motion. (See Figure 9.3-3.)

Figure 9.3-3

Step 4: Determine acceleration. (See Figure 9.3-4.)

Figure 9.3-4

Step 5: Draw the motion of the particle. (See Figure 9.3-5.) s (0) = −1, s (1) = 3, s (2) = 1 and s (3) = −1

Figure 9.3-5

At t = 0, the particle is at − 1 and moving to the right. It slows down and stops at t = 1 and at t = 3. It reverses direction (moving to the left) and speeds up until it reaches 1 at t = 2. It continues moving left but slows down and stops at − 1 at t = 3. Then it reverses direction (moving to the right) again and speeds up indefinitely. (Note: “Speeding up” is defined as when |v(t)| increases and “slowing down” is defined as when |v(t)| decreases.)

9.4 Parametric, Polar, and Vector Derivatives

Main Concepts:

Derivatives of Parametric Equations; Position, Speed, and Acceleration; Derivatives of Polar Equations; Velocity and Acceleration of Vector Functions

Derivatives of Parametric Equations

If a function is defined parametrically, you can differentiate both x(t) and y (t) with respect to t, and then .

Example 1

A curve is defined by x(t) = t² − 3t and y (t) = 5 cos t. Find .

Step 1: Differentiate x (t) and y (t) with respect to t. and .

Step 2:

Example 2

A function is defined by x (t) = 5t − 2 and y (t) = 9 − t² when − 5 ≤ t ≤ 5. Find the equation of any horizontal tangent lines to the curve.

Step 1: Differentiate x (t) and y (t) with respect to t. and .

Step 2:

Step 3: In order for the tangent line to be horizontal, must be equal to zero, therefore t = 0, x = −2 and y = 9.

Step 4: The equation of the horizontal tangent line at (−2, 9) is y = 9.

Example 3

A curve is defined by x(t) = t² − 5t + 2 and for 0 ≤ t ≤ 3. Find the equation of the tangent line to the curve when t = 1.

Step 1: and .

Step 2:

Step 3: At t = 1, , and .

Step 4: The equation of the tangent line is .

Position, Speed, and Acceleration

When the motion of a particle is defined parametrically, its position is given by (x (t), y(t)). The speed of the particle is and its acceleration is given by the vector .

Example 1

Find the speed and acceleration of a particle whose motion is defined by x = 3t and y = 9t − 3t² when t = 2.

Step 1: Differentiate and . When t = 2, and .

Step 2: Calculate the speed.

Step 3: Determine second derivatives. and . The acceleration vector is ⟨0, − 6⟩.

Example 2

A particle moves along the curve ln x so that x = and t > 0. Find the speed of the particle when t = 1.

Step 1: Substitute x = in ln x to find

Step 2: and . Evaluated at t = 1, and .

Step 3: The speed of the particle is .

Derivatives of Polar Equations

For polar representations, remember that r = f(θ), so x = r cos θ = f(θ) cos θ and y = r sinθ = f (θ) sinθ. Differentiating with respect to θ requires the product rule. and . Dividing by gives .

Example

Find the equation of the tangent line to the curve r = 2 + 2 sinθ when .

Step 1:

Step 2: By the Pythagorean identity, 2(cos² θ − sin² θ − sinθ) = 2(1 − sin² θ − sin² θ − sinθ) = 2(1 − sinθ − 2 sin² θ) = 2(1 − 2 sinθ)(1 + sinθ). Also, .

Step 3:

Step 4: When ,

Evaluating, .

Step 5: When , , so and .

Step 6: The equation of the tangent line is or .

Velocity and Acceleration of Vector Functions

A vector-valued function assigns a vector to each element in a domain of real numbers. If r = x, y is a vector-valued function, exists only if and y (t) exist. . A vector-valued function is continuous at c if its component functions are continuous at c. The derivative of a vector-valued function is .

If r = x, y is a vector-valued function that represents the path of an object in the plane, and x and y are both functions of a variable t, x = f (t) and y = g (t), then the velocity of the object is . Speed is the magnitude of velocity, so . The direction of v is along the tangent to the path. The acceleration vector is and the magnitude of acceleration is . The vector T tangent to the path at t is and the normal vector at t is .

Example 1

The position function describes the path of an object moving in the plane. Find the velocity and acceleration of the object at the point (8, 4).

Step 1: The velocity . At the point (8, 4), t = 2. Evaluated at t = 2, the velocity v = ⟨12, 4⟩. The speed .

Step 2: The acceleration vector . Evaluated at t = 2, the acceleration is ⟨12, 2⟩. The magnitude of the acceleration is .

Example 2

The left field fence in Boston’s Fenway Park, nicknamed the Green Monster, is 37 feet high and 310 feet from home plate. If a ball is hit 3 feet above the ground and leaves the bat at an angle of , write a vector-valued function for the path of the ball and use the function to determine the minimum speed at which the ball must leave the bat to be a home run. At that speed, what is the maximum height the ball attains?

Step 1: The horizontal component of the ball’s motion, the motion in the “x” direction, is . The vertical component follows the parabolic motion model , where g is the acceleration due to gravity. The path of the ball can be represented by the vector-valued function .

Step 2: In order for the ball to clear the fence, its height must be greater than 37 feet when its distance from the plate is 310 feet. , solved for t, gives t = seconds. At this time, , and this value must exceed 37 feet. Setting and solving gives s ≈ 105.556. The ball must leave the bat at 105.556 feet per second in order to clear the wall.

Step 3: Since r = , the derivative is r′ = , and the ball will attain its maximum height when the vertical component is equal to zero. Since s ≈ 105.556, produces t ≈ 2.462 seconds. For that value of t, . The ball will reach a maximum height of 89.779 feet, when it is 183.762 feet from home plate.

Example 3

Find the velocity, acceleration, tangent and normal vectors for an object on a path defined by the vector-valued function r(t) = ⟨e^t cos t, e^t sin t⟩ when .

Step 1: v(t) = r′(t) = ⟨e^t(cos t − sin t), e^t(sin t + cos t)⟩. When evaluated at , . The velocity vector is ⟨−4.810, 4.810.⟩.

Step 2: a(t) = ⟨−2e^t sin t, 2e^t cos t⟩. Evaluated at , this is ≈ ⟨−9.621, 0.⟩.

Step 3: The tangent vector is given by . Since , the tangent vector becomes , which simplifies to . When t = , the tangent vector is .

Step 4: The normal vector At . Check to be certain the tangent and normal vectors are orthogonal.

9.5 Rapid Review

1. Write an equation of the normal line to the graph y = e^x at x = 0.

Answer: At x = 0, y = e⁰ = 1 ⇒ you have the point (0, 1). Equation of normal: y − 1 = −1(x − 0) or y = −x + 1.

2. Using your calculator, find the values of x at which the function y = − x² + 3x and y = ln x have parallel tangents.

Answer:

Set . Using the [Solve] function on your calculator, enter [Solve] and obtain x = 1 or .

3. Find the linear approximation of f(x) = x³ at x = 1 and use the equation to find f (1. 1).

Answer: f (1) = 1 ⇒ (1, 1) is on the tangent line and f′ (x) = 3x² ⇒ f′(1) = 3. y − 1 = 3(x − 1) or y = 3x − 2. f (1.1) ≈ 3(1.1) − 2 ≈ 1.3

4. (See Figure 9.5-1.)

(a) When is the acceleration zero? (b) Is the particle moving to the right or left?

Figure 9.5-1

Answer: (a) a(t) = v′(t) and v(t) is the slope of the tangent. Thus, a(t) = 0 at t = 2.

(b) Since v(t) ≥ 0, the particle is moving to the right.

5. Find the maximum acceleration of the particle whose velocity function is v(t) = t² + 3 on the interval 0 ≤ t ≤ 4.

Answer: a(t) = v′(t) = 2(t) on the interval 0 ≤ t ≤ 4, a(t) has its maximum value at t = 4. Thus a(t) = 8. The maximum acceleration is 8.

6. Find the slope of the tangent to the curve defined by x = 3t − 5, y = t² − 9 when t = 3.

Answer: and , so .

7. Find the slope of the tangent line to the graph of r = −3 cos θ.

Answer:

8. Find for the vector function r(t) = 3ti − 2tj = ⟨3t, − 2t⟩.

Answer: and , so .

9.6 Practice Problems

Part A—The use of a calculator is not allowed.

1. Find the linear approximation of f(x) = (1 + x)^¼ at x = 0 and use the equation to approximate f(0.1).

2. Find the approximate value of 3 using linear approximation.

3. Find the approximate value of cos 46^° using linear approximation.

4. Find the point on the graph of y = |x³| such that the tangent at the point is parallel to the line y − 12x = 3.

5. Write an equation of the normal to the graph of y = e^x at x = ln 2.

6. If the line y − 2x = b is tangent to the graph y = −x² + 4, find the value of b.

7. If the position function of a particle is = , find the velocity and position of particle when its acceleration is 0.

Figure 9.6-1

9. The position function of a moving particle is shown in Figure 9.6-2.

Figure 9.6-2

8. The graph in Figure 9.6-1 represents the distance in feet covered by a moving particle in t seconds. Draw a sketch of the corresponding velocity function.

For which value(s) of t(t₁, t₂, t₃) is:

(a) the particle moving to the left?

(b) the acceleration negative?

10. The velocity function of a particle is shown in Figure 9.6-3.

Figure 9.6-3

(a) When does the particle reverse direction?

(b) When is the acceleration 0?

11. A ball is dropped from the top of a 640-foot building. The position function of the ball is s(t) = −16t² + 640, where t is measured in seconds and s (t) is in feet. Find:

(a) The position of the ball after 4 seconds.

(b) The instantaneous velocity of the ball at t = 4.

(d) When the ball will hit the ground.

(e) The speed of the ball when it hits the ground.

12. The graph of the position function of a moving particle is shown in Figure 9.6-4.

Figure 9.6-4

(a) What is the particle’s position at t = 5?

(b) When is the particle moving to the left?

(d) When does the particle have the greatest speed?

Part B—Calculators are allowed.

13. The position function of a particle moving on a line is s(t) = t³ − 3t² + 1, t ≥ 0 where t is measured in seconds and s in meters. Describe the motion of the particle.

14. Find the linear approximation of f(x) = sin x at x = π. Use the equation to find the approximate value of .

15. Find the linear approximation of f(x) = ln (1 + x) at x = 2.

16. Find the coordinates of each point on the graph of y² = 4 − 4x² at which the tangent line is vertical. Write an equation of each vertical tangent.

17. Find the value(s) of x at which the graphs of y = ln x and y = x² + 3 have parallel tangents.

18. The position functions of two moving particles are s₁(t) = ln t and s₂(t) = sin t and the domain of both functions is 1 ≤ t ≤ 8. Find the values of t such that the velocities of the two particles are the same.

19. The position function of a moving particle on a line is s(t) = sin(t) for 0 ≤ t ≤ 2π. Describe the motion of the particle.

20. A coin is dropped from the top of a tower and hits the ground 10.2 seconds later. The position function is given as s(t) = −16t² − v₀t + s₀, where s is measured in feet, t in seconds and v₀ is the initial velocity and s₀ is the initial position. Find the approximate height of the building to the nearest foot.

21. Find the equation of the tangent line to the curve defined by x = cos t - 1, y = sin t + t at the point where .

22. An object moves on a path defined by x = e²t + t and y = 1 + e^t. Find the speed of the object and its acceleration vector with t = 2.

23. Find the slope of the tangent line to the curve r = 3 sin 4θ at .

24. The position of an object is given by . Find the velocity and acceleration vectors, and determine when the magnitude of the acceleration is equal to 2.

25. Find the tangent vector to the path defined by r = ⟨In t, In (t + 4)⟩ at the point where t = 4.

9.7 Cumulative Review Problems

(Calculator) indicates that calculators are permitted.

26. Find if y = x sin⁻¹(2x).

27. Given f (x) = x³ − 3x² + 3x − 1 and the point (1, 2) is on the graph of f⁻¹(x). Find the slope of the tangent line to the graph of f⁻¹(x) at (1, 2).

28. Evaluate .

29. A function f is continuous on the interval (− 1, 8) with f (0) = 0, f (2) = 3, and f (8) = ½ and has the following properties:

(a) Find the intervals on which f is increasing or decreasing.

(b) Find where f has its absolute extrema.

(d) Find the intervals on which f is concave upward or downward.

(e) Sketch a possible graph of f.

30. The graph of the velocity function of a moving particle for 0 ≤ t ≤ 8 is shown in Figure 9.7-1. Using the graph:

(a) Estimate the acceleration when v(t) = 3 ft/s.

(b) Find the time when the acceleration is a minimum.

Figure 9.7-1

31. Find the Cartesian equation for the curve defined by r = 4 cos θ.

32. The motion of an object is modeled by x = 5 sin t, y = 1 − cos t. Find the y-coordinate of the object at the moment when its x-coordinate is 5.

33. Calculate 4u − 3v if u = 6, − 1 and v = −4, 3.

34. Determine the symmetry, if any, of the graph of r = 2 sin (4θ).

35. Find the magnitude of the vector 3i + 4 j.

9.8 Solutions to Practice Problems

Part A—The use of a calculator is not allowed.

1. Equation of tangent line:

4. Step 1: Find m_tangent.

Step 2: Set m_tangent = 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. Thus Ø.

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

Figure 9.8-1

At x = 2, y = x³ = 2³ = 8.

Thus, the point is (2, 8).

5. Step 1: Find m_tangent.

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:

6. Step 1: Find m_tangent.

Step 3: Find point of tangency. Set m_tangent = 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.

8. 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.)

Figure 9.8-2

9. (a) At t = t₂, the slope of the tangent is negative. Thus, the particle is moving to the left.

(b) At t = t₁, and at t = t₂, the curve is concave downward = acceleration is negative.

(c) 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.

10. (a) At t = 2, v(t) changes from positive to negative, and thus the particle reverses its direction.

(b) At t = 1, and at t = 3, the slope of the tangent to the curve is 0. Thus, the acceleration is 0.

11. (a) s (4) = −16(4)² + 640 = 384 ft

(b) v(t) = s′(t) = − 32t

v(4) = − 32(4) ft/s = − 128 ft/s

(c)

(d)

(e)

12. (a) At t = 5, s (t) = 1.

(b) For 3 < t < 4, s (t) decreases. Thus, the particle moves to the left when 3 < t < 4.

(d) When 6 < t < 7, speed = 2 ft/s, the greatest speed, which occurs where s has the greatest slope.

Part B—Calculators are allowed.

13. Step 1: v(t) = 3t² − 6t

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.)

Figure 9.8-3

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

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.

Figure 9.8-5

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.

14. Linear approximation: 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.

is approximately:

15. y = f(a) + f′(a)(x − a) f (x) = ln (1 + x) and f (2) = ln (1 + 2) = ln 3

and .

Thus, .

16. Step 1:

Step 2:

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.

17. Step 1:

Step 2: Find the x-coordinate of point(s) of tangency.

Parallel tangents ⇒ slopes are equal. Set = 2x.

Using the [Solve] function of your calculator, enter [Solve] and obtain Since for

18. s 1(t) = ln t and ; 1 ≤ t ≤ 8.

s₂(t) = sin(t) and

s₂′(t) = cos(t); 1 ≤ t ≤ 8.

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

19. Step 1: s(t) = sin t

v(t) = cos t

a(t) = − sin t

Step 2:

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

Figure 9.8-6

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

Figure 9.8-7

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

Figure 9.8-8

The particle is initially at 0, s(0) = 0. It moves to the right but slows down to a stop at 1 when , . It then turns and moves to the left speeding up until it reaches 0, when t = π, s (π) = 0 and continues to the left, but slowing down to a stop at − 1 when t = π, s(π) = 0 , . It then turns around again, moving to the right, speeding up to 0 when t = 2π, s (2π) = 0.

20. s(t) = −16t² + v₀t + s₀ s₀ = height of building and v₀ = 0. Thus, s (t) = −16t² + s₀. When the coin hits the ground, s(t) = 0, t = 10.2. Thus, set s (t) = 0 ⇒ − 16t² + s 0 = 0 ⇒ − 16(10.2)² + s₀ = 0 s₀ = 1664.64 ft. The building is approximately 1665 ft tall.

21. When , , and , and so . Find and , and divide to find . Evaluate at to find the slope . Therefore, the equation of the tangent line is , or simplifying, .

22. Differentiate to find and . The speed of the object is . When t = 2, . Find second derivatives and and evaluate at t = 2, to find the acceleration vector 4e²t, e^t ≈ 218.393, 7.389.

23. Since x = r cos θ and y = sinθ,

Find and substitute. . When so the functions of are equal to those . Evaluate at

24. If the position of the object is given by , then the velocity vector is , and the acceleration vector is . The magnitude of the acceleration is equal to When . Solve to find .

25. If r = ⟨ln t, ln(t + 4)⟩, then . Evaluate at t = 4 for , then find , which, at t = 4, is equal to . The tangent

9.9 Solutions to Cumulative Review Problems

26. Using product rule, let u = x;

27. Let y = f (x) ⇒ y = x³ − 3x² + 3x − 1. To find f⁻¹(x), switch x and y: x = y³ − 3y² + 3y − 1.

28. Substituting x = 100 into the expression would lead to . Multiply both numerator and denominator by the conjugate of the denominator :

An alternative solution is to factor the numerator:

29. (a) f′ > 0 on(− 1, 2), f is increasing on (−1, 2) f′ < 0 on (2, 8), f is decreasing on (2, 8).

(b) At x = 2, f′ = 0 and f″ < 0, thus at x = 2, f has a relative maximum. Since it is the only relative extremum on the interval, it is an absolute maximum. Since f is a continuous function on a closed interval and at its endpoints f (−1) < 0 and f (8) = ½, f has an absolute minimum at x = −1.

(d) f″ < 0 on(−1, 5), f is concave downward on (−1, 5). f′ > 0 on (5, 8), f is concave upward on (5, 8).

(e) A possible graph of f is given in Figure 9.9-1.

Figure 9.9-1

30. (a) v(t) = 3 ft/s at t = 6. The tangent line to the graph of v(t) at t = 6 has a slope of approximately m = 1. (The tangent passes through the points (8, 5) and (6, 3); thus m = 1.) Therefore the acceleration is 1 ft/s².

(b) The acceleration is a minimum at t = 0, since the slope of the tangent to the curve of v(t) is the smallest at t = 0.

31. To convert r = 4 cos θ to a Cartesian representation, recall that and tan . Then, . Since , the equation becomes . Multiply through by to produce x² + y² = 4x. Completing the square produces (x − 2)² + y² = 4.

32. When x = 5 sin t = 5, , so .

33. If u = ⟨6, − 1⟩ and v = ⟨− 4, 3⟩, 4 ⟨6, −1⟩ −3 ⟨−4, 3⟩ = ⟨24, −4⟩ + ⟨12, −9⟩ = ⟨36, − 13⟩.

34. Replace θ with − θ. 2 sin(−4θ) = − 2 sin(4θ) ≠ 2 sin(4θ), so the graph is not symmetric about the polar axis. Replace θ with π − θ. 2 sin(4(π − θ)) = 2 sin(4π − 4θ) = 2 [sin 4π cos 4θ − sin 4θ cos 4π] = − 2 sin 4θ = 2 sin 4θ, so the graph is not symmetric about the line . Replace θ with θ + π. 2 sin(4(θ + π)) = 2 sin(4θ + 4π) = 2 [sin 4θ cos 4π + cos 4θ sin 4π] = 2 sin 4θ, so the graph is symmetric about the pole.

35. The magnitude of the vector 3i + 4j is .