## Calculus AB and Calculus BC

## CHAPTER 4 Applications of Differential Calculus

### L. TANGENT-LINE APPROXIMATIONS

**Local linear approximation**

If *f* *′*(*a*) exists, then the *local linear approximation* of *f(x)* at *a* is

*f* (*a*) + *f* *′*(*a*)(*x* − *a*).

Since the equation of the tangent line to *y* = *f* (*x*) at *x* = *a* is

*y* − *f* (*a*) = *f* *′*(*a*)(*x* − *a*),

we see that the *y* value on the tangent line is an approximation for the actual or true value of *f*. Local linear approximation is therefore also called *tangent-line approximation*.^{†} For values of *x* close to *a*, we have

**FIGURE N4–20**

where *f* (*a*) + *f* *′*(*a*)(*x* − *a*) is the linear or tangent-line approximation for *f* (*x*), and *f* *′*(*a*)(*x* − *a*) is the approximate change in *f* as we move along the curve from *a* to *x*. See Figure N4–20.

In general, the closer *x* is to *a*, the better the approximation is to *f* (*x*).

**EXAMPLE 30**

Find tangent-line approximations for each of the following functions at the values indicated:

(a) sin *x* at *a* = 0 (b) cos *x* at *a* =

(c) 2*x*^{3} − 3*x* at *a* = 1 (d) at *a* = 8

**SOLUTIONS:**

**(a)** At *a* = 0, sin *x* sin (0) + cos (0)(*x* − 0) 0 + 1 · *x* *x*

**(b)**

**(c)** At *a* = 1, 2*x*^{3} − 3*x* − 1 + 3(*x* − 1) 3*x* − 4

**(d)**

^{†} Local linear approximation is also referred to as “local linearization” or even “best linear approximation” (the latter because it is better than any other linear approximation).

**EXAMPLE 31**

Using the tangent lines obtained in Example 30 and a calculator, we evaluate each function, then its linear approximation, at the indicated *x*-values:

Example 31 shows how small the errors can be when tangent lines are used for approximations and *x* is near *a*.

**EXAMPLE 32**

A very useful and important local linearization enables us to approximate (1 + *x*) * ^{k}* by 1 +

*kx*for

*k*any real number and for

*x*near 0. Equation (1) yields

Then, near 0, for example,

**EXAMPLE 33**

Estimate the value of at *x* = 0.05.

**SOLUTION:** Use the line tangent to at *x* = 0; *f* (0) = 3.

so *f* *′*(0) = 6 ; hence, the line is *y* = 6*x* + 3.

Our tangent-line approximation, then, is

At *x* = 0.05, we have *f* (0.05) ≈ 6(0.05) + 3 = 3.3.

The true value, to three decimal places, of when *x* = 0.05 is 3.324; the tangent-line approximation yields 3.3. This tells us that the curve is concave up, lying above the tangent line to the curve near *x* = 0. Graph the curve and the tangent line on [−1, 1] × [−1, 6] to verify these statements.

**Approximating the Change in a Function.**

Equation (1) above for a local linear approximation also tells us by about how much *f changes* when we move along the curve from *a* to *x*: it is the quantity *f* *′*(*a*)(*x* − *a*). (See Figure N4–20.)

**EXAMPLE 34**

By approximately how much does the area of a circle change when the radius increases from 3 to 3.01 inches?

**SOLUTION:** We use the formula *A* = π*r*^{2}. Then Equation (1) tells us that the local linear approximation for *A*(*r*), when *A* is near 3, is

*A*(3) + *A* *′*(3)(*r* − 3).

Here we want only the change in area; that is,

*A* *′*(3)(*r* − 3) when *r* = 3.01.

Since *A* *′*(*r*) = 2π*r*, therefore *A* *′*(3) = 6π; also, (*r* − 3) = 0.01, so the approximate change is (6π)(0.01) 0.1885 in.^{2} The true increase in area, to four decimal places, is 0.1888 in.^{2}

**EXAMPLE 35**

Suppose the diameter of a cylinder is 8 centimeters. If its circumference is increased by 2 centimeters, how much larger, approximately, are

(a) the diameter, and

(b) the area of a cross section?

**SOLUTIONS:**

**(a)** Let *D* and *C* be respectively the diameter and circumference of the cylinder. Here, *D* plays the role of *f*, and *C* that of *x*, in the linear approximation equation (1) a previous page. The approximate increase in diameter, when *C* = 8π, is therefore equal to *D* *′*(*C*) times (the change in *C*). Since *C* = π*D*, and (which is constant for *all C*). The change in *C* is given as 2 cm; so the increase in diameter is equal approximately to 2. 0.6366 cm.

**(b)** The approximate increase in the area of a (circular) cross section is equal to

*A* *′*(*C*) · (change in *C*),

where the area Therefore,

Since the change in *C* is 2 cm, the area of a cross section increases by approximately 4 · 2 = 8 cm^{2}.