Question

A Metal ring is heated such that its area in m2 at any time t is given by A=3t2+pie
The rate of increase area at t=5 sec is

Answer 1

Question 32302

Area in $m^2$ as a function of time is $A(t) = 3t^2 + \pi$ . The rate of increase of area is by definition the derivative of $A(t)$ with respect to $t$ . Hence, $rate = \frac{dA}{dt} = 6t$ .