A metallic disc is being heated. Its area A `("in" m^2)` at any time t – InterviewSolution

1.	A metallic disc is being heated. Its area A `("in" m^2)` at any time t `("in" sec)` is given by `A =5 t^2 +4 t+8` Calculate the rate of increase of area at `t =3 s`.
Answer» Correct Answer - `34 m^2//s` `(dA)/(dt)=(d)/(dt)(5t^2 +4t +8) = 10t+4` when `t =3 s , (dA)/(dt) =10xx3+4 =34m^2//s`

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