Barrons AP Calculus

(a)

(b)

BC  ONLY

SOLUTIONS:

Use r   =   2(1 +   cos θ), x   =   r   cos θ,  y   =   r   sin θ,  and r′  =   −2  sin θ;  then

.

At  θ   =    ,      =   −1.

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

θ   =       or  −1, θ   =       or  π.  From    part    (a),        at   ,  but     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!