BC ONLYIf 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 equation of the tangent to the curve of f (x) = x^3 − 3x^2 at the point (1,
−2).
SOLUTION: Since f ′(x) = 3x^2 − 6x and f ′(1) = −3, the equation of the tangent
is
y + 2 = −3(x − 1) or y + 3 x = 1.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
y − 2 = − x or 12 y + x = 24.Example 7 __
Find the coordinates of any point on the curve of y^2 − 4xy = x^2 + 5 for which the
tangent is horizontal.
SOLUTION: Since and the tangent is horizontal when = 0, then x
= − 2 y. If we substitute this in the equation of the curve, we get