## 5 Steps to a 5: AP Calculus AB 2017 (2016)

### STEP __4__

### Review the Knowledge You Need to Score High

### CHAPTER 10

### Big Idea 2: Derivatives

### 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 AB exam. 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 apply derivatives to solve rectilinear motion problems.

**Key Ideas**

Tangent and Normal Lines

Linear Approximations

Motion Along a Line

**10.1 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 10.1-1__ .)

**Figure 10.1-1**

Vertical Tangents: ( does not exist but = 0). (See __Figure 10.1-2__ .)

**Figure 10.1-2**

Parallel Tangents: . (See __Figure 10.1-3__ .)

**Figure 10.1-3**

**Example 1**

Write an equation of the line tangent to the graph of *y* = –3 sin 2*x* at . (See __Figure 10.1-4__ .)

**Figure 10.1-4**

*y* = –3 sin 2*x* ;

Slope of tangent .

Point of tangency:

Therefore, is the point of tangency.

Equation of tangent: *y* – 0 = 6(*x* – *π* /2) or *y* = 6*x* – 3*π* .

**Example 2**

If the line *y* = 6*x* + *a* is tangent to the graph of *y* = 2*x* ^{3} , find the value(s) of *a* .

Solution:

(See __Figure 10.1-5__ .)

**Figure 10.1-5**

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

Since *y* = 6*x* + *a* is tangent to the graph of *y* = 2*x* ^{3} , thus for some values of *x* .

Set 6*x* ^{2} = 6 ⇒ *x* ^{2} = 1 or *x* = ±1.

At *x* = –1, *y* = 2*x* ^{3} = 2(–1)^{3} = –2;(–1, –2) is a tangent point. Thus, *y* = 6*x* + *a* ⇒ –2 = 6(–1) + *a* or *a* = 4.

At *x* = 1, *y* = 2*x* ^{3} = 2(1)^{3} = 2; (1, 2) is a tangent point.

Thus, *y* = 6*x* + *a* ⇒ 2 = 6(1) + *a* or *a* = –4.

Therefore, *a* = ±4.

**Example 3**

Find the coordinates of each point on the graph of *y* ^{2} – *x* ^{2} – 6*x* + 7 = 0 at which the tangent line is vertical. Write an equation of each vertical tangent. (See __Figure 10.1-6__ .)

**Figure 10.1-6**

Step 1: Find .

Step 2: Find .

Step 3: Find points of tangency.

At *y* = 0, *y* ^{2} – *x* ^{2} – 6*x* + 7 = 0 becomes –*x* ^{2} – 6*x* + 7 = 0 ⇒ *x* ^{2} + 6*x* – 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 equations 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*= 0 ⇒

^{x}*e*(1 +

^{x}*x*) = 0 ⇒

*x*= –1 but

*x*≥ 0, therefore, no solution.

If *x <* 0, set –*e ^{x} *–

*xe*= 0 ⇒ –

^{x}*e*(1 +

^{x}*x*) = 0 ⇒

*x*= –1.

Step 3: Find points of tangency.

At

Thus at the point (–1, 1*/e* ), the graph has a horizontal tangent. (See __Figure 10.1-7__ .)

**Figure 10.1-7**

**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 (*x* ^{2} + 3) is equal to . (See __Figures 10.1-8__ and __10.1-9__ .)

**Figure 10.1-8**

**Figure 10.1-9**

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

Step 2: Enter *y* 2 = *d* (*y* _{1} (*x* ), *x* ) and enter

Step 3: Using the [*Intersection* ] function of the calculator for *y* _{2} and *y* _{3} , 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* = 2*x* ^{2} and *y* = *e ^{x} *have parallel tangents.

Step 1: Find for both *y* = 2*x* ^{2} and *y* = *e ^{x} *.

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

Set 4*x* = *e ^{x} *⇒ 4

*x*–

*e*= 0.

^{x}Using the [*Solve* ] function of the calculator, enter [*Solve* ] (4*x* – *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 (*x* _{1} , *y* _{1} ) is the line perpendicular to the tangent line at (*x* _{1} , *y* _{1} ). (See __Figure 10.1-10__ .)

**Figure 10.1-10**

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 10.1-11__ .)

**Figure 10.1-11**

At these points, *m* _{tangent} = 0; but *m* _{normal} does not exist.

(See __Figure 10.1-12__ .)

**Figure 10.1-12**

At these points, *m* _{tangent} does not exist; however *m* _{normal} = 0.

**Example 1**

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

Step 1: Find *m* _{tangent} .

Step 2: Find *m* _{normal} .

⇒ *x* = cos^{–1} (–1) or *x* = *π* . (See __Figure10.1-13__ .)

**Figure 10.1-13**

Step 3: Write equation of normal line.

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

Since , equation of normal is:

or .

**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 the point on the 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 10.1-14__ .)

**Figure 10.1-14**

**Example 3**

Given the curve : (a) write an equation of the normal to the curve at the point (2, 1/2), 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.

*m* _{normal} = 4; (2, 1/2)

Equation of normal: , or .

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 and .

**10.2 Linear Approximations**

**Main Concepts:** Tangent Line Approximation, Estimating the *n* th 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 10.2-1__ .) 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 10.2-1**

**Example 1**

Write an equation of the tangent line to *f* (*x* ) = *x* ^{3} 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* ) = 3*x* ^{2} ; *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 + 12*x* – 24 = 12*x* – 16*f* (1.9) ≈ 12(1.9) – 16 = 6.8*f* (2.01) ≈ 12(2.01) – 16 = 8.12. (See __Figure 10.2-2__ .)

**Figure 10.2-2**

**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 ;

(c) the equation of the tangent is or ;

(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).

(a) Let *y* = *f* (*x* ), then

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

(b) *f* (3.1) ≈ –2(3.1) + 8 ≈ 1.8

**Estimating the n th Root of a Number**

Another way of expressing the tangent line approximation is: *f* (*a* + Δ*x* ) ≈ *f* (*a* ) + *f* ′(*a* )Δ*x* , where Δ*x* is a relatively small value.

**Example 1**

Find the approximate value of using linear approximation.

Using and Δ*x* = 1.

Thus,

**Example 2**

Find the approximate value of using linear approximation.

Let *f* (*x* ) = *x* ^{1/3} , *a* = 64, Δ*x* = – 2. Since and , you can use *f* (*a* + Δ*x* ) ≈ *f* (*a* ) + *f* ′(*a* )Δ*x* . Thus, *f* (62) =

• Use calculus notations and not calculator syntax, e.g., write and not .

**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 .

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

**10.3 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* ) = 2*t* ^{3} – 10*t* ^{2} + 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* ) = 6*t* ^{2} – 20*t* *v* (1) = 6(1)^{2} – 20(1) = –14

(c) *a* (*t* ) = *v* ′(*t* ) = 12*t* – 20*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 10.3-1__ .) The graph of *a* (*t* ) indicates that:

**Figure 10.3-1**

(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 10.3-2__ .

**Figure 10.3-2**

(a) When is the acceleration 0?

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

(c) When is the speed the greatest?

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.

(c) Speed = |*v* (*t* )| at *t* = 1, *v* (*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* ) = –16*t* ^{2} + 400, *t* ≥ 0 is in seconds. Find:

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

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

(c) when the ball will hit the ground.

Solution:

(a) *v* (*t* ) = *s* ′(*t* ) = –32*t* *v* (2) = 32(2) = –64 ft/sec

(b) Average velocity .

(c) When the ball hits the ground, *s* (*t* ) = 0.

Thus, set *s* (*t* ) = 0 ⇒ –16*t* ^{2} + 400 = 0; 16*t* ^{2} = 400; *t* = ±5.

Since *t* ≥ 0, *t* = 5. The ball hits the ground at *t* = 5 sec.

• Remember that the volume of a sphere is and the surface area is *s* = 4*πr* ^{2} . Note that *v* ′ = *s* .

**Horizontal Motion**

**Example**

The position function of a particle moving in a straight line is *s* (*t* ) = *t* ^{3} – 6*t* ^{2} + 9*t* – 1, *t* ≥ 0. Describe the motion of the particle.

Step 1: Find *v* (*t* ) and *a* (*t* ).*v* (*t* ) = 3*t* ^{2} – 12*t* + 9

*a* (*t* ) = 6*t* – 12

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

Step 3: Determine the directions of motion. (See __Figure 10.3-3__ .)

**Figure 10.3-3**

Step 4: Determine acceleration. (See __Figure 10.3-4__ .)

**Figure 10.3-4**

Step 5: Draw the motion of the particle. (See __Figure 10.3-5__ .) *s* (0) = –1, *s* (1) = 3, *s* (2) = 1 and *s* (3) = –1

**Figure 10.3-5**

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.)

**10.4 Rapid Review**

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

*x*= 0.

*Answer* :

At *x* = 0, *y* = *e* ^{0} = 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* = –*x* ^{2} + 3*x* 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* ) = *x* ^{3} 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* ) = 3*x* ^{2} ⇒ *f* ′(1) = 3.*y* – 1 = 3(*x* – 1) or *y* = 3*x* – 2.*f* (1.1) ≈ 3(1.1) – 2 ≈ 1.3

4. (See __Figure 10.4-1__ .)

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

**Figure 10.4-1**

*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* ) = *t* ^{2} + 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.

**10.5 Practice Problems**

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

__1__ . Find the linear approximation of *f* (*x* ) = (1 + *x* )^{1/4} at *x* = 0 and use the equation to approximate *f* (0.1).

__2__ . Find the approximate value of using linear approximation.

__3__ . Find the approximate value of cos 46° using linear approximation.

__4__ . Find the point on the graph of *y* = |*x* ^{3} | such that the tangent at the point is parallel to the line *y* – 12*x* = 3.

__5__ . Write an equation of the normal to the graph of *y* = *e ^{x} *at

*x*= ln 2.

__6__ . If the line *y* – 2*x* = *b* is tangent to the graph *y* = –*x* ^{2} + 4, find the value of *b* .

__7__ . If the position function of a particle , find the velocity and position of particle when its acceleration is 0.

__8__ . The graph in __Figure 10.5-1__ represents the distance in feet covered by a moving particle in *t* seconds. Draw a sketch of the corresponding velocity function.

**Figure 10.5-1**

__9__ . The position function of a moving particle is shown in __Figure 10.5-2__ .

**Figure 10.5-2**

For which value(s) of *t* (*t* _{1} , *t* _{2} , *t* _{3} ) is:

(a) the particle moving to the left?

(b) the acceleration negative?

(c) the particle moving to the right and slowing down?

__10__ . The velocity function of a particle is shown in __Figure 10.5-3__ .

**Figure 10.5-3**

(a) When does the particle reverse direction?

(b) When is the acceleration 0?

(c) When is the speed the greatest?

__11__ . A ball is dropped from the top of a 640-foot building. The position function of the ball is *s* (*t* ) = –16*t* ^{2} + 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.

(c) The average velocity for the first 4 seconds.

(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 10.5-4__ .

**Figure 10.5-4**

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

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

(c) When is the particle standing still?

(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* ) = *t* ^{3} – 3*t* ^{2} + 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 *y* ^{2} = 4 – 4*x* ^{2} 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* = *x* ^{2} + 3 have parallel tangents.

__18__ . The position functions of two moving particles are *s* _{1} (*t* ) = ln *t* and *s* _{2} (*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* ≤ 2*π* . 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* ) = –16*t* ^{2} – *v* _{0} *t* + *s* _{0} , where *s* is measured in feet, *t* in seconds, and *v* _{0} is the initial velocity and *s* _{0} is the initial position. Find the approximate height of the building to the nearest foot.

**10.6 Cumulative Review Problems**

**(Calculator) indicates that calculators are permitted.**

__21__ . Find if *y* = *x* sin^{–1} (2*x* ).

__22__ . Given *f* (*x* ) = *x* ^{3} – 3*x* ^{2} + 3*x* – 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).

__23__ . Evaluate .

__24__ . A function *f* is continuous on the interval (–1, 8) with *f* (0) = 0, *f* (2) = 3, and *f* (8) = 1/2 and has the following properties:

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

(b) Find where *f* has its absolute extrema.

(c) Find where *f* has the points of inflection.

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

(e) Sketch a possible graph of *f* .

__25__ . The graph of the velocity function of a moving particle for 0 ≤ *t* ≤ 8 is shown in __Figure 10.6-1__ . Using the graph:

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

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

**Figure 10.6-1**

**10.7 Solutions to Practice Problems**

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

__1__ . Equation of tangent line:

__2__ . *f* (*a* + Δ*x* ) ≈ *f* (*a* ) + *f* ′(*a* )Δ*x*

Let and *f* (28) = *f* (27+1).

__3__ . *f* (*a* + Δ*x* ) ≈ *f* (*a* ) + *f* ′(*a* ) Δ*x* Convert to radians:

Let *f* (*x* ) = cos *x* and *f* (45°) =

Then *f* ′(*x* ) = – sin *x* and

__4__ . Step 1: Find *m* _{tangent} .

Step 2: Set *m* _{tangent} = slope of line *y* – 12*x* = 3.

Since *y* – 12*x* = 3 ⇒ *y* = 12*x* + 3, then *m* = 12.

Set 3*x* ^{2} = 12 ⇒ *x* = ±2

since *x* ≥ 0, *x* = 2.

Set –3*x* ^{2} = 12 ⇒ *x* ^{2} = –4. Thus ∅.

Step 3: Find the point on the curve. (See __Figure 10.7-1__ .)

**Figure 10.7-1**

At *x* = 2, *y* = *x* ^{3} = 2^{3} = 8.

Thus, the point is (2, 8).

__5__ . Step 1: Find *m* _{tangent} .

Step 2: Find *m* _{normal} .

At *x* = ln 2, *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 2: Find the slope of line *y* – 2*x* = *b* *y* – 2*x* = *b* ⇒*y* = 2*x* + *b* or *m* = 2.

Step 3: Find point of tangency.

Set *m* _{tangent} = slope of line *y* – 2*x* = *b* ⇒ – 2*x* = 2 ⇒ *x* = –1.

At *x* = –1, *y* = –*x* ^{2} + 4 = –(–1)^{2} + 4 = 3; (–1, 3).

Step 4: Find *b* .

Since the line *y* – 2*x* = *b* passes through the point (–1, 3), thus 3 – 2(–1) = *b* or *b* = 5.

__7__ . *v* (*t* ) = *s* ′(*t* ) = *t* ^{2} – 6*t* ;*a* (*t* ) = *v* ′(*t* ) = *s* ″(*t* ) = 2*t* – 6

Set *a* (*t* ) = 0 ⇒ 2*t* – 6 = 0 or *t* = 3. *v* (3) = (3)^{2} – 6(3) = – 9;

.

__8__ . On the interval (0, 1), the slope of the line segment is 2. Thus the velocity *v* (*t* ) = 2 ft/sec. On (1, 3), *v* (*t* ) = 0 and on (3, 5), *v* (*t* ) = –1. (See __Figure 10.7-2__ .)

**Figure 10.7-2**

__9__ . (a) At *t* = *t* _{2} , the slope of the tangent is negative. Thus, the particle is moving to the left.

(b) At *t* = *t* _{1} , and at *t* = *t* _{2} , the curve is concave downward acceleration is negative.

(c) At *t* = *t* _{1} , 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.

(c) At *t* = 3, speed is equal to |– 5| = 5 and 5 is the greatest speed.

__11__ . (a) *s* (4) = –16(4)^{2} + 640 = 384 ft

(b) *v* (*t* ) = *s* ′(*t* ) = –32*t* *v* (4) =–32(4) ft/s = –128 ft/sec

(c) Average velocity =

(d) Set *s* (*t* ) = 0 ⇒ –16*t* ^{2} + 640 = 0 ⇒ 16*t* ^{2} = 640 or *t* = ± .

Since *t* ≥ 0, or *t* = ± or *t* ≈ 6.32 sec.

(e) |*v* ( )| = |–32( | = | – 64 | ft/s or ≈ 202.39 ft/sec

__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.

(c) When 4 < *t* < 6, the particle stays at 1.

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

**Part B—Calculators are allowed.**

__13__ . Step 1: *v* (*t* ) = 3*t* ^{2} – 6*t* *a* (*t* ) = 6*t* – 6

Step 2: Set *v* (*t* ) = 0 ⇒ 3*t* ^{2} – 6*t* = 0 ⇒ 3*t* (*t* – 2) = 0, or *t* = 0 or *t* = 2 Set *a* (*t* ) = 0 ⇒ 6*t* – 6 = 0 or *t* = 1.

Step 3: Determine the directions of motion. (See __Figure 10.7-3__ .)

**Figure 10.7-3**

Step 4: Determine acceleration. (See __Figure 10.7-4__ .)

**Figure 10.7-4**

Step 5: Draw the motion of the particle. (See __Figure 10.7-5__ .) *s* (0) = 1, *s* (1) = –1, and *s* (2) = –3.

**Figure 10.7-5**

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:

or ≈ –0.0175.

__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: Find .*y* ^{2} = 4 – 4*x* ^{2}

Step 2: Find .

Set or *y* = 0.

Step 3: Find points of tangency.

At *y* = 0, *y* ^{2} = 4 – 4*x* ^{2} becomes 0 = 4 – 4*x* ^{2}

⇒ *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: Find for *y* = ln *x* and *y* = x^{2} + 3.

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

Parallel tangents ⇒ slopes are equal. Set .

Using the [*Solve* ] function of your calculator, enter [*Solve* ] and obtain or . Since for *y* = ln *x* , .

__18__ . *s* _{1} (*t* ) = ln *t* and .

*s* _{2} (*t* ) = sin(*t* ) and*s* ′_{2} (*t* ) = cos(*t* ); 1 ≤ *t* ≤ 8.

Enter and *y* ^{2} = 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: Set *v* (*t* ) = 0 ⇒ cos *t* = 0;

and .

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

Step 3: Determine the directions of motion. (See __Figure 10.7-6__ .)

**Figure 10.7-6**

Step 4: Determine acceleration. (See __Figure 10.7-7__ .)

**Figure 10.7-7**

Step 5: Draw the motion of the particle. (See __Figure 10.7-8__ .)

**Figure 10.7-8**

The particle is initially at 0, *s* (0) = 0. It moves to the right but slows down to a stop at 1 when = 1. 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 . It then turns around again, moving to the right, speeding up to 0 when *t* = 2π, *s* (2π) = 0.

__20__ . *s* (*t* ) = – 16*t* ^{2} + *v* _{0} *t* + *s* _{0} *s* _{0} = height of building and *v* _{0} = 0.

Thus, *s* (*t* ) =– 16*t* ^{2} + *s* _{0} .

When the coin hits the ground, *s* (*t* ) = 0, *t* = 10.2. Thus, set *s* (*t* ) = 0 ⇒ – 16*t* ^{2} + *s* _{0} = 0 ⇒ – 16(10.2)^{2} + *s* _{0} = 0 *s* _{0} = 1664.64 ft. The building is approximately 1665 ft tall.

**10.8 Solutions to Cumulative Review Problems**

__21__ . Using product rule, let *u* = *x* ; *v* = sin^{–1} (2*x* ).

__22__ . Let *y* = *f* (*x* ) ⇒ *y* = *x* ^{3} – 3*x* ^{2} + 3*x* – 1. To find *f* ^{–1} (*x* ), switch *x* and*y* : *x* = *y* ^{3} – 3*y* ^{2} + 3*y* – 1.

__23__ . Substituting *x* = 100 into the expression would lead to . Apply *L’Hôpital’s* Rule, and you have or . Another approach to solve the problem is as follows. Multiply both numerator and denominator by the conjugate of the denominator ( + 10):

An alternative solution is to factor the numerator:

__24__ . (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) = 1/2, *f* has an absolute minimum at *x* = –1.

(c) At *x* = 5, *f* has a change of concavity and *f* ′ exists at *x* = 5.

(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 10.8-1__ .

**Figure 10.8-1**

__25__ . (a) *v* (*t* ) = 3 ft/sec 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/sec^{2} .

(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.