# A Metallic Disc Is Being Heated. Its Area A (In M2) At Any Time T (In Second) Is Given By A = 5t2 + 4t + 8. Calculate The Rate Of Increase In Area At T = 3s.

Given:

A =$$5t^{2} + 4t + 8$$

t = 3 seconds

The rate of increase in area = $$\frac{dA}{dt} = 10t + 4$$ $$\Rightarrow \frac{dA}{dt} = 10 * 3 + 4 = 34 ms^{-2}$$

Therefore, the rate of increase in area at t =$$3s is 34 ms^{-2}.$$

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