Suppose that the radius r of a right circular cone of fixed height h = 20 cm is increasing at a rate 4 cm/s. How fast is the volume increasing when r = 10? Give your answer correct to two decimal places.
V = (pi/3) h r^2
dV/dt = 2*(pi/3)*h*r*(dr/dt)
You have all the terms you need to calculate that.
To find how fast the volume is increasing, we need to differentiate the volume formula with respect to time. The volume (V) of a right circular cone can be calculated using the formula:
V = (1/3)πr²h
Since the height (h) is fixed at 20 cm, we can substitute this value into the equation:
V = (1/3)πr²(20)
Now, we will differentiate both sides of the equation with respect to time (t) using the chain rule. Let's denote the rate of change of radius (dr/dt) as 4 cm/s:
dV/dt = (1/3)π(2r)(20)(dr/dt)
We can simplify this to:
dV/dt = (2/3)π(20r)(dr/dt)
Substituting the given values, r = 10 and (dr/dt) = 4:
dV/dt = (2/3)π(20 * 10)(4)
dV/dt = (2/3)π(200)(4)
dV/dt = (8/3)π(200)
Finally, calculating the value:
dV/dt = (1600/3)π ≈ 1677.11 cm³/s
Therefore, the volume is increasing at a rate of approximately 1677.11 cm³/s when the radius is 10 cm.
To find the rate at which the volume of the cone is changing with respect to time (dv/dt), we can use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
Given that the height of the cone, h, is fixed at 20 cm, we can express V solely in terms of the radius, r:
V = (1/3) * π * r^2 * 20
Now, let's differentiate both sides of the equation with respect to time, t:
dv/dt = (1/3) * π * 2r * dr/dt * 20
We are given that dr/dt = 4 cm/s and we need to find dv/dt when r = 10 cm. Let's substitute these values into the equation:
dv/dt = (1/3) * π * 2(10) * 4 * 20
Simplifying the expression:
dv/dt = (1/3) * π * 2 * 10 * 4 * 20
dv/dt = (8/3) * π * 20 * 10
dv/dt = (1600/3) * π
Now, we can calculate this value using a calculator:
dv/dt ≈ 1675.52 cm^3/s
Therefore, when the radius is 10 cm, the volume is increasing at a rate of approximately 1675.52 cm^3/s.