Question

A metallic disc is being heated. Its area A (in $m^2$) at any time T (in second) is given by $A=5t^2+4t+8$. Calculate the rate of increase in area at $T=3s$

Answer 1

Step 1. Given data:

Area $A=5t^2+4t+8$

Time =$3s$

Step 2. Calculating rate of increase in area:

The rate of increase in area = $\frac{dA}{dt}=10t+4$

At time $T=3s$

$\frac{dA}{dt}=10\left(3\right)+4=34m^2{s}^{-1}$

Therefore, the rate at which area increases at $T=3s$ is $34m^2{s}^{-1}$.