Question

the volume of a cone of radius r and height h is given by v=1/3pir^2h. If the radius and height is both increasing at a constant rate of 1/2 centimeter per second, at what rate, in cubic centimeters per second, is the volume increasing when the height is 9 centimeters and the radius is 6 centimeters

Answer 1

Answer:

The rate of change of the volume $\frac{dV}{dt}$ when the height is 9 centimeters and the radius is 6 centimeters is $24\pi \:\frac{cm^3}{s}$

Step-by-step explanation:

This is a related rate problem because you know a rate and want to find another rate that is related to it. If 2 variables both vary with respect to time and have a relation between them, we can express the rate of change of one in terms of the other.

From the information given we know:

• $\frac{dr}{dt}=\frac{1}{2}\:\frac{cm}{s}$
• $\frac{dh}{dt}=\frac{1}{2}\:\frac{cm}{s}$
• The volume of a cone of radius r and height h is given by $V=\frac{1}{3} \pi r^2 h$

We want to find the rate of change of the volume $\frac{dV}{dt}$ when the height is 9 centimeters and the radius is 6 centimeters.

Applying implicit differentiation to the formula of the volume of a cone we get

$\frac{dV}{dt}=\frac{1}{3}\pi [r^2\frac{dh}{dt}+2rh\frac{dr}{dt} ]$

Substituting the values we know into the above formula:

$\frac{dV}{dt}=\frac{1}{3}\pi [(6)^2\frac{1}{2}+2(6)(9)\frac{1}{2} ]\\\\\frac{dV}{dt}=\frac{1}{3}\pi[18+54]\\\\\frac{dV}{dt}=\frac{72\pi}{3}=24\pi \:\frac{cm^3}{s}$

Answer 2

The volume of the cone is increasing at the rate of $24 \pi cm^3/s$

Given the volume of a cone expressed as;

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

If the radius and height are both increasing at a constant rate of 1/2 centimeter per second, then:

$\frac{dr}{dt} = 0.5cm/s\\\frac{dh}{dt} =0.5cm/s$

Get the rate of change in volume by differentiating the formula implicitly

$\frac{dV}{dt} = \frac{1}{3}\pi (r^2\frac{dh}{dt} + 2rh\frac{dr}{dt} )$

Given the following:

radius r = 6cm

height h = 9cm

Substitute the given parameters into the formula above;

$\frac{dV}{dt} = \frac{1}{3}\pi ((6)^2(0.5)+ 2(6)(9)(0.50 ))\\\frac{dV}{dt} =\frac{1}{3}\pi [36 \times 0.5 + 108 \times 0.5]\\\frac{dV}{dt} =\frac{1}{3}\pi(18+54)\\\frac{dV}{dt} =\frac{1}{3}\pi(72)\\\frac{dV}{dt} = 24\pi cm^2/s$

Hence the volume of the cone is increasing at the rate of $24 \pi cm^3/s$

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