Question

The radius of the base of a cylinder is increasing at a rate of \[7\] millimeters per hour. the height of the cylinder is fixed at \[1.5\] millimeters. at a certain instant, the radius is \[12\] millimeters. what is the rate of change of the volume of the cylinder at that instant (in cubic millimeters per hour)?

Answer 1

The rate of change of the volume of the cylinder at the instant when the radius is 12 mm and the rate of radius increase is 7 mm/hr, with a fixed height of 1.5 mm, is 1,257.36 mm³/hr.

Step-by-step explanation:

The rate of change of the volume of a cylinder can be determined by differentiating the volume formula with respect to time. The volume V of a cylinder with radius r and height h is given by the formula V = πr²h. Given that the radius is increasing at a rate of 7 millimeters per hour (dr/dt = 7 mm/hr) and the height is constant at 1.5 millimeters, we can find the rate of change of volume dV/dt at the instant when the radius is 12 millimeters.

To calculate this, we use the chain rule: dV/dt = π * 2r * (dr/dt) * h. Substituting the given values, we get dV/dt = π * 2 * 12 mm * 7 mm/hr * 1.5 mm = 3.142 * 2 * 12 * 7 * 1.5 mm³/hr, which simplifies to dV/dt = 1,257.36 mm³/hr. Therefore, the rate of change of the volume of the cylinder at that instant is 1,257.36 cubic millimeters per hour.