## Calculus AB and Calculus BC

## CHAPTER 4 Applications of Differential Calculus

**Concepts and Skills**

In this chapter, we review how to use derivatives to

• find slopes of curves and equations of tangent lines;

• find a function’s maxima, minima, and points of inflections;

• describe where the graph of a function is increasing, decreasing, concave upward, and concave downward;

• analyze motion along a line;

• create local linear approximations;

• and work with related rates.

For BC Calculus students, we also review how to

• find the slope of parametric and polar curves

• and use vectors to analyze motion along parametrically defined curves.

### A. SLOPE; CRITICAL POINTS

**Slope of a curve**

If the derivative of *y* = *f* (*x*) exists at *P*(*x*_{1}, *y*_{1}), then the *slope* of the curve at *P* (which is defined to be the slope of the tangent to the curve at *P*) is *f* *′*(*x*_{1}), the derivative of *f* (*x*) at *x* = *x*_{1}.

Any *c* in the domain of *f* such that either *f* *′*(*c*) = 0 or *f* *′*(*c*) is undefined is called a *critical point* or *critical value* of *f*. If *f* has a derivative everywhere, we find the critical points by solving the equation *f* *′*(*x*) = 0.

**Critical point**

**EXAMPLE 1**

For *f* (*x*) = 4*x*^{3} − 6*x*^{2} − 8, what are the critical points?

**SOLUTION:** *f* *′*(*x*) = 12*x*^{2} − 12*x* = 12*x*(*x* − 1),

which equals zero if *x* is 0 or 1. Thus, 0 and 1 are critical points.

**EXAMPLE 2**

Find any critical points of *f* (*x*) = 3*x*^{3} + 2*x*.

**SOLUTION:** *f* *′*(*x*) = 9*x*^{2} + 2.

Since *f* *′*(*x*) never equals zero (indeed, it is always positive), *f* has no critical values.

**EXAMPLE 3**

Find any critical points of *f* (*x*) = (*x* − 1)^{1/3}.

**SOLUTION:**

Although *f* *′* is never zero, *x* = 1 is a critical value of *f* because *f* *′* does not exist at *x* = 1.

AVERAGE AND INSTANTANEOUS RATES OF CHANGE.

Both average and instantaneous rates of change were defined in Chapter 3. If as *x* varies from *a* to *a* + *h*, the function *f* varies from *f* (*a*) to *f* (*a* + *h*), then we know that the difference quotient

is the average rate of change of *f* over the interval from *a* to *a* + *h.*

Thus, the *average velocity* of a moving object over some time interval is the change in distance divided by the change in time, the average rate of growth of a colony of fruit flies over some interval of time is the change in size of the colony divided by the time elapsed, the average rate of change in the profit of a company on some gadget with respect to production is the change in profit divided by the change in the number of gadgets produced.

The (instantaneous) rate of change of *f* at *a*, or the derivative of *f* at *a*, is the limit of the average rate of change as *h* → 0:

On the graph of *y* = *f* (*x*), the rate at which the *y*-coordinate changes with respect to the *x*-coordinate is *f* *′*(*x*), the slope of the curve. The rate at which *s*(*t*), the distance traveled by a particle in *t* seconds, changes with respect to time is *s* *′*(*t*), the velocity of the particle; the rate at which a manufacturer’s profit *P*(*x*) changes relative to the production level *x* is *P* *′*(*x*).

**EXAMPLE 4**

Let *G* = 400(15 − *t*)^{2} be the number of gallons of water in a cistern *t* minutes after an outlet pipe is opened. Find the average rate of drainage during the first 5 minutes and the rate at which the water is running out at the end of 5 minutes.

**SOLUTION:** The average rate of change during the first 5 min equals

The average rate of drainage during the first 5 min is 10,000 gal/min.

The instantaneous rate of change at *t* = 5 is *G* *′*(5). Since

*G* *′*(*t*) = −800(15 − *t*),

*G* *′*(5) = −800(10) = −8000 gal/min. Thus the rate of drainage at the end of 5 min is 8000 gal/min.