FIGURE N4–24
BC ONLY
SOLUTIONS:
(a) Use r = 2(1 + cos θ), x = r cos θ, y = r sin θ, and r ′ = −2 sin θ; thenAt
(b) Since the cardioid is symmetric to θ = 0 we need consider only the upper half of the curve
for part (b). The tangent is horizontal where (provided ). Since factors into 2(2
cos θ − 1)(cos θ + 1), which equals 0 for cos or −1, or π. From part (a),
does equal 0 at π. Therefore, the tangent is horizontal only at (and, by
symmetry, at ).
It is obvious from Figure N4–24 that r ′(θ) does not give the slope of the cardioid. As θ varies
from 0 to the slope varies from −∞ to 0 to +∞ (with the tangent rotating counterclockwise),
taking on every real value. However, r ′(θ) equals −2 sin θ, which takes on values only between
−2 and 2!Chapter Summary
In this chapter we reviewed many applications of derivatives. We’ve seen how to find slopes of
curves and used that skill to write equations of lines tangent to a curve. Those lines often provide
very good approximations for values of functions. We have looked at ways derivatives can help us
understand the behavior of a function. The first derivative can tell us whether a function is increasing
or decreasing and locate maximum and minimum points. The second derivative can tell us whether the
graph of the function is concave upward or concave downward and locate points of inflection. We’ve
reviewed how to use derivatives to determine the velocity and acceleration of an object in motion
along a line and to describe relationships among rates of change.