## Calculus AB and Calculus BC

## CHAPTER 4 Applications of Differential Calculus

### B. TANGENTS AND NORMALS

**Tangent to a curve**

The *equation of the tangent* to the curve *y* = *f* (*x*) at point *P*(*x*_{1}, *y*_{1}) is

*y* − *y*_{1} = *f* *′*(*x*_{1})(*x* − *x*_{1}).

The line through *P* that is perpendicular to the tangent, called the *normal* to the curve at *P*, has slope Its equation is

If the tangent to a curve is horizontal at a point, then the derivative at the point is 0. If the tangent is vertical at a point, then the derivative does not exist at the point.

TANGENTS TO PARAMETRICALLY DEFINED CURVES.

**BC ONLY**

If the curve is defined parametrically, say in terms of *t* (as in Chapter 1), then we obtain the slope at any point from the parametric equations. We then evaluate the slope and the *x*- and *y*-coordinates by replacing *t* by the value specified in the question (see Example 9).

**EXAMPLE 5**

Find the equations of the tangent and normal to the curve of *f* (*x*) = *x*^{3} − 3*x*^{2} at the point (1, −2).

**SOLUTION:** Since *f* *′*(*x*) = 3*x*^{2} − 6*x* and *f* *′*(1) = −3, the equation of the tangent is

*y* + 2 = −3(*x* − 1) or *y* + 3*x* = 1,

and the equation of the normal is

or 3*y* − *x* = −7.

**EXAMPLE 6**

Find the equation of the tangent to *x*^{2} *y* − *x* = *y*^{3} − 8 at the point where *x* = 0.

**SOLUTION:** Here we differentiate implicitly to get

Since *y* = 2 when *x* = 0 and the slope at this point is the equation of the tangent is

or 12*y* + *x* = 24.

**EXAMPLE 7**

Find the coordinates of any point on the curve of *y*^{2} − 4*xy* = *x*^{2} + 5 for which the tangent is horizontal.

**SOLUTION:** Since and the tangent is horizontal when then *x* = −2*y*. If we substitute this in the equation of the curve, we get

Thus *y* = ±1 and *x* = ±2. The points, then, are (2, −1) and (−2, 1).

**EXAMPLE 8**

Find the *x*-coordinate of any point on the curve of *y* = sin^{2} (*x* + 1) for which the tangent is parallel to the line 3*x* − 3*y* − 5 = 0.

**SOLUTION:** Since = 2sin(*x* + 1) cos(*x* + 1) = sin2(*x* + 1) and since the given line has slope 1, we seek *x* such that sin 2(*x* + 1) = 1. Then

or

**BC ONLY**

**EXAMPLE 9**

Find the equation of the tangent to *F*(*t*) = (cos *t*, 2 sin^{2} *t*) at the point where

**SOLUTION:** Since we see that