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Explain solution RD Sharma class 12 chapter Derivative as a Rate Measure exercise 12.1 question 4 maths

Answers (1)

Answer: 3cm


We have to differentiate area of circular disc (A) with respect to its circumference (L)

Area of circular disc, A=\pi r^{2}

Circumference of circular disc, L=2\pi r


Radius of a circular disc, r=3 cm


Here we have,

Radius of disc, r=3 cm

Circumference of disc, L=2\pi r

Let’s differentiate with the circumference with respect to radius r


\therefore \frac{dr}{dL}=\frac{1}{2\pi }\: \; \; \; \; \; \; \; \; .....(i)            \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]

Area of circular disc, A=\pi r^{2}.

Let’s differentiate with the area with respect to L

\therefore \frac{d A}{d L}=2 \pi r\frac{dr}{d L}

\therefore \frac{d A}{d L}=\frac{2 \pi r}{2 \pi }                                    {From eq.(i)}

\therefore \frac{d A}{d L}=r

Putting value of r=3cmr = 3cm


Here we cannot write directly,

A=\frac{L}{2}r            It's wrong

Posted by

Gurleen Kaur

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