The area,Acm^2, of a blot of ink is growing such that after t seconds,it is given by A=3t^2+1/5t. Find the rate at which the area is changing after 2 seconds.
dA/dt = 6t + 1/5
plug in t = 2 for dA/dt
To find the rate at which the area is changing after 2 seconds, we need to find the derivative of the area function with respect to time and evaluate it at t = 2.
Given that the area A (in cm^2) is given by A = 3t^2 + (1/5)t, we can find the derivative by differentiating the function:
dA/dt = d/dt (3t^2) + d/dt (1/5)t
To differentiate each term separately, we use the power rule of differentiation. The power rule states that for any term of the form x^n, where n is a constant, the derivative is given by nx^(n-1).
Differentiating the first term, we have:
d/dt (3t^2) = 2 * 3t^(2-1) = 6t
Differentiating the second term, we have:
d/dt (1/5)t = (1/5) * d/dt (t) = (1/5) * 1 = 1/5
Combining both derivatives, we get:
dA/dt = 6t + 1/5
Now we can evaluate the derivative at t = 2:
dA/dt(t=2) = 6(2) + 1/5 = 12 + 1/5 = 12.2
Therefore, the rate at which the area is changing after 2 seconds is given by 12.2 cm^2/s.