FIGURE N4–23
SOLUTION: Liquid flowing in at a constant rate means the change in volume is constant per
unit of time. Obviously, the depth of the liquid increases as t does, so h ′(t) is positive
throughout. To maintain the constant increase in volume per unit of time, when the radius grows,
h ′(t) must decrease. Thus, the rate of increase of h decreases as h increases from 0 to a (where
the cross-sectional area of the vessel is largest). Therefore, since h ′(t) decreases, h ′′(t) < 0
from 0 to a and the curve is concave down.
As h increases from a to b, the radius of the vessel (here cylindrical) remains constant, as do the
cross-sectional areas. Therefore h ′(t) is also constant, implying that h(t) is linear from a to b.
Note that the inflection point at depth a does not exist, since h ′′(t) < 0 for all values less than a
but is equal to 0 for all depths greater than or equal to a.BC ONLYN. SLOPE OF A POLAR CURVE
We know that, if a smooth curve is given by the parametric equations
x = f (t) and y = g(t),
then
provided that f ′(t) ≠ 0.
To find the slope of a polar curve r = f (θ), we must first express the curve in parametric form.
Since
x = r cos θ and y = r sin θ,
therefore,
x = f (θ) cos θ and y = f (θ) sin θ.
If f (θ) is differentiable, so are x and y; then
Also, if then
In doing an exercise, it is often easier simply to express the polar equation parametrically, then
find dy/dx, rather than to memorize the formula.
EXAMPLE 39
(a) Find the slope of the cardioid r = 2(1 + cos θ) at See Figure N4–24.
(b) Where is the tangent to the curve horizontal?