Question

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

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Solution

## Step 1. Given data:Area $A=5{t}^{2}+4t+8$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=34{m}^{2}{s}^{-1}$Therefore, the rate at which area increases at $T=3s$ is $34{m}^{2}{s}^{-1}$.

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