A square with side x cm, is changing with time where x(t) = 3t + 1. What is the rate of change of the area of the square when t = 2 seconds?
A = x^2 = (3t+1)^2 = 9 t^2 + 6 t + 1
dA/dt = 18 t + 6
when t = 2
dA/dt = 36 + 6 = 42
To find the rate of change of the area of the square, we need to take the derivative of the area function with respect to time.
First, let's find the area function of the square. The area of a square is calculated by squaring the length of one side, so if the side length is x cm, the area A(x) is given by A(x) = x^2.
Now, we have x(t) = 3t + 1, representing the side length of the square changing with time. We can substitute this into our area function to express the area A as a function of time t.
A(t) = (3t + 1)^2.
To find the rate of change of A with respect to t (dA/dt), we need to take the derivative of A(t) with respect to t.
Using the chain rule, dA/dt = 2(3t + 1) * (3).
Simplifying this expression, we have dA/dt = 6(3t + 1).
Now, to find the rate of change of the area when t = 2 seconds, substitute t = 2 into the derivative expression:
dA/dt = 6(3(2) + 1) = 6(6 + 1) = 6(7) = 42.
Therefore, the rate of change of the area of the square when t = 2 seconds is 42 square cm per second.