Question

The volume of a cone is increasing at the rate of 28pi cubic units per second. At the instant when the radius of the cone is 3 units, its volume is 12pi cubic units and the radius is increasing at 1/2 unit per second. a. At the instant when the radius of the cone is 3 units, what is the rate of change of the area of its base? b. At the instant when the radius of the cone is 3 units, what is the rate of change of height? c. At the instant when the radius of the cone is 3 units, what is the instantaneous rate of change of the area of its base with respect to its height?

Answer 1

Answer:

(a) 3π

(b) 28/3

(c) 9π/28

Step-by-step explanation:

The volume of a cone is given by

$V = \frac{1}{3} \pi r^2 h$

r is the radius, h is the height.

From the question,

$\dfrac{dV}{dt} = 28\pi$

When r = 3, V = 12π and

$\dfrac{dr}{dt} = 0.5$

(a)

The area of the base (a circle) = $\pi r^2$

$\dfrac{dA}{dr} = 2\pi r$

$\dfrac{dA}{dt} = \dfrac{dA}{dr} \cdot\dfrac{dr}{dt}$

$\dfrac{dA}{dt} = 2\pi \times 3 \times0.5= 3\pi$

(b)

$\dfrac{dV}{dh} = \frac{1}{3} \pi r^2$

$\dfrac{dV}{dt} = \dfrac{dV}{dh} \cdot\dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{dV}{dt}\div \dfrac{dV}{dh} = \dfrac{28\pi}{\frac{1}{3}\pi\times3^2 } = \dfrac{28}{3}$

(c)

$\dfrac{dA}{dh} = \dfrac{dA}{dt}\div\dfrac{dh}{dt} = \dfrac{3\pi}{28/3} = \dfrac{9\pi}{28}$