The amount of water,V cm3 , in a leaking tank at time t seconds is given by
V = (15-t)^3 for 0<=t<=15
Find the rate at which the water leaves the tank when t = 4.
To find the rate at which the water leaves the tank when t = 4, we need to find the derivative of the function V with respect to time t. The derivative represents the rate of change of the function.
Given the function V = (15 - t)^3, we can find the derivative dV/dt using the power rule for differentiation.
The power rule states that if we have a function f(x) = x^n, then the derivative of f(x) with respect to x is given by f'(x) = nx^(n-1).
Applying the power rule to the function V = (15 - t)^3, we get:
dV/dt = 3(15 - t)^(3-1) * (d/dt)(15 - t)
Simplifying this expression, we have:
dV/dt = 3(15 - t)^2 * (-1)
Evaluating this expression when t = 4, we substitute t = 4 into the derivative:
dV/dt = 3(15 - 4)^2 * (-1)
= 3(11)^2 * (-1)
= 3(121) * (-1)
= -363
Therefore, the rate at which the water leaves the tank when t = 4 is -363 cm^3/second.