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(x1, y1) is
y − y1 = f ′(x1)(x − x1).
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.
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).
Find the equations of the tangent and normal to the curve of f (x) = x3 − 3x2 at the point (1, −2).
SOLUTION: Since f ′(x) = 3x2 − 6x and f ′(1) = −3, the equation of the tangent is
y + 2 = −3(x − 1) or y + 3x = 1,
and the equation of the normal is
or 3y − x = −7.
Find the equation of the tangent to x2 y − x = y3 − 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 12y + x = 24.
Find the coordinates of any point on the curve of y2 − 4xy = x2 + 5 for which the tangent is horizontal.
SOLUTION: Since and the tangent is horizontal when then x = −2y. 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).
Find the x-coordinate of any point on the curve of y = sin2 (x + 1) for which the tangent is parallel to the line 3x − 3y − 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
Find the equation of the tangent to F(t) = (cos t, 2 sin2 t) at the point where
SOLUTION: Since we see that