The volume of a cone of radius r and height h is given by V=1/3pir^2h. If the radius and the height both increase 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 centimeter and the radius is 6 centimeters?
explain please
annnnd....
a)1/2pi
b) 10pi
c) 24pi
d) 54pi
e) 108pi
To find the rate at which the volume of the cone is increasing, we need to take the derivative of the volume equation with respect to time. Let's break it down step by step:
Given:
- Volume of the cone, V = (1/3)πr^2h
- Rate of increase of radius = dr/dt = 1/2 cm/s
- Rate of increase of height = dh/dt = 1/2 cm/s
- Height, h = 9 cm
- Radius, r = 6 cm
We want to find:
- The rate at which the volume is increasing, dV/dt (in cubic centimeters per second) when h = 9 cm and r = 6 cm.
Step 1: Find the expressions for dr/dt and dh/dt in terms of r and h.
Since both the radius and height increase at a constant rate of 1/2 cm/s, we know:
dr/dt = 1/2
dh/dt = 1/2
Step 2: Use the volume equation, V = (1/3)πr^2h, to express V in terms of r and h.
V = (1/3)πr^2h
Step 3: Differentiate V with respect to time (t) using the product rule.
dV/dt = (1/3)π * (2rh * dr/dt + r^2 * dh/dt)
Step 4: Substitute the given values into the expression for dV/dt.
dV/dt = (1/3)π * (2(6)(9) * (1/2) + (6^2) * (1/2))
= (1/3)π * (108 + 18)
= (1/3)π * 126
= 42π
Therefore, when the height is 9 centimeters and the radius is 6 centimeters, the volume of the cone is increasing at a rate of 42π cubic centimeters per second.