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A metallic disc is being heated. Its are...

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`.

Text Solution

Verified by Experts

The correct Answer is:
`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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