The area ' $A$ ' of a blot of ink is growing such that after t sec. its area is given by $A = 5 t^{2} + 9 {cm}^{2}$ . Calculate the rate of increase of area at $t = 5 \sec$ .

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$30 {cm}^{2} / \sec$

$134 {cm}^{2} / \sec$

$50 {cm}^{2} / \sec$

$40 {cm}^{2} / \sec$

answer is C.

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Detailed Solution

$\begin{array}{rcl} A & = & (5 t^{2} + 9) {cm}^{2} \\ \frac{dA}{dt} & = & 10 t \\ at t & = & 5 \sec \\ \frac{dA}{dt} & = & 10 \times 5 = 50 {cm}^{2} / \sec \end{array}$

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