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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