The volume of a cone of a radius r and height h is given by V=(3.14/3)(r^2)(h). If the radius and the height both increase at a constant rate of .5 cm per second, at what rate, in cubic cm per second, is the volume increasing when the height is 9cm and the radius is 6cm.
dV/dt= PI/3(2rh)dr/dt+PI/3(r^2)dh/dt
you are given, r, h,dr/dt, dh/dt
To find the rate at which the volume is increasing, we need to use the concept of related rates.
Let's start by taking the derivative of the volume equation with respect to time (t):
dV/dt = (3.14/3) * 2r * dr/dt * h + (3.14/3) * r^2 * dh/dt
Here, dV/dt represents the rate of change of volume, dr/dt represents the rate of change of radius, and dh/dt represents the rate of change of height.
Given that both the radius and height are increasing at a constant rate of 0.5 cm/s, we can substitute these values into the equation:
dV/dt = (3.14/3) * 2 * 6 * 0.5 * 9 + (3.14/3) * 6^2 * 0.5
Simplifying this equation will give us the rate at which the volume is increasing:
dV/dt = 56.52 cm^3/s
Therefore, when the height is 9 cm and the radius is 6 cm, the volume is increasing at a rate of 56.52 cubic cm per second.