The volume of an ice cube as it melts is given by the equation: V = 2.1x10^-5 t^3 - 6.5x10^-4 t^2 - 0.346 t + 21.31

Determine the rate at which the sides of the ice cube melt at both 10 and 75 minutes after the beginning of the melting process.

You may assume the the ice cube keeps its cubic shape throughout the melting process.

You have to define if the variable t is in minutes, hours, etc

find the derivative dV/dt
sub in the values of 10 if in minutes, 1.6 if in hours etc

To determine the rate at which the sides of the ice cube melt, we need to calculate the derivative of the volume equation with respect to time. The derivative will give us the rate of change of volume with respect to time, which is equivalent to the rate at which the sides of the ice cube melt.

First, let's start by finding the derivative of the volume equation:
V = 2.1x10^-5 t^3 - 6.5x10^-4 t^2 - 0.346 t + 21.31

To find the derivative, we differentiate each term separately while keeping constants. The derivative of t^n (where n is any real number) is n*t^(n-1).

Differentiating each term:
dV/dt = 2.1x10^-5 * 3t^(3-1) - 6.5x10^-4 * 2t^(2-1) - 0.346 * 1 * t^(1-1)
dV/dt = 6.3x10^-5 t^2 - 1.3x10^-3 t - 0.346

Now we have the derivative of the volume equation, which represents the rate at which the volume (sides) of the ice cube is changing with respect to time.

To find the rate at which the sides of the ice cube melt at 10 and 75 minutes, we substitute the values of t into the derivative equation.

For t = 10 minutes:
dV/dt = 6.3x10^-5 * (10^2) - 1.3x10^-3 * 10 - 0.346

For t = 75 minutes:
dV/dt = 6.3x10^-5 * (75^2) - 1.3x10^-3 * 75 - 0.346

By evaluating these equations, we can determine the rates at which the sides of the ice cube melt at 10 and 75 minutes after the beginning of the melting process.