The diagonal of a square is changing at the rate of 0.5 cm / . Then the rate of change of area, when the area is 400 cm^ 2 , is equal to

Correct answer: B. 10 2 cm^ 2 / Diagonal D= 2 a Differentiating w.r.t. t: dD dt = 2 da at or da dt = 1 2 da dt = 1 2 × 0.5 cm / s Let Area is denoted by A d A dt =2 a da dt when area A is 400 cm^ 2 then a=20 d A dt =2 × 20 × 0.5 2 =10 2 cm^ 2 /

BITSAT 2012

Exam: BITSAT
Year: 2012
Subject: Mathematics
Chapter: Differentiation

BITSAT 2012 · Mathematics · Differentiation

The diagonal of a square is changing at the rate of 0.5 cm / . Then...

The diagonal of a square is changing at the rate of

0.5 \mathrm{ cm} / \mathrm\sec

. Then the rate of change of area, when the area is

400 \mathrm{ cm}^{2}

, is equal to

Options

A $20 \sqrt{2} \mathrm{ cm}^{2} / \mathrm\sec$
B $10 \sqrt{2} \mathrm{ cm}^{2} / \mathrm\sec$
C $\frac{1}{10 \sqrt{2}} \mathrm{ cm}^{2} / \mathrm\sec$
D $\frac{10}{\sqrt{2}} \mathrm{ cm}^{2} / \mathrm\sec$

Correct answer

B. $10 \sqrt{2} \mathrm{ cm}^{2} / \mathrm\sec$

Step-by-step solution

Diagonal

\mathrm{D}=\sqrt{2}

a Differentiating w.r.t. t:

\frac{\mathrm{dD}}{\mathrm{dt}}=\sqrt{2} \frac{\mathrm{da}}{\text{ at }}

\frac{\mathrm{da}}{\mathrm{dt}}=\frac{1}{\sqrt{2}} \frac{\mathrm{da}}{\mathrm{dt}}=\frac{1}{\sqrt{2}} \times 0.5 \mathrm{ cm} / \mathrm{s}

Let Area is denoted by

\underline{A}\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=2 \mathrm{a} \frac{\mathrm{da}}{\mathrm{dt}}

when area

A

400 \mathrm{ cm}^{2}

then

\mathrm{a}=20\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=2 \times 20 \times \frac{0.5}{\sqrt{2}}=10 \sqrt{2} \mathrm{ cm}^{2} / \mathrm\sec

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