Question: Show that the length of the curve $9ay^2 = x(x-3a)^2$ is $4\sqrt3a$
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Subject : Applied Mathematics 2

Topic : Differentiation under Integral sign, Numerical Integration and Rectification

Difficulty: High

mumbai university • 22 views
 modified 3 days ago  • written 4 days ago by Ankit Pandey ♦♦ 10
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Solution:

Curve: $9 \mathrm{ay}^{2}=x(x-3 a)^{2}$-----------------(1)

The given curve is strophoid.

\begin{aligned} \text { Differentiate eqn }(1) \text { w.r.t } x \\ 18 a y \frac{d y}{d x}=2 x(x-3 a)+(x-3 a)^{2} \end{aligned}

$\therefore 18 a y \frac{d y}{d x}=3(x-3 a)(x-a)$

$\therefore \frac{d y}{d x}=\frac{(x-3 a)(x-a)}{6 a y}$

\begin{aligned} \text { Squaring both the sides, } \\ \left(\frac{d y}{d x}\right)^{2}=\frac{(x-3 a)^{2}(x-a)^{2}}{36 a^{2} y^{2}} \\ \therefore\left(\frac{d y}{d x}\right)^{2} =\frac{(x-3 a)^{2}(x-a)^{2}}{4 a x(x-3 a)^{2}} \end{aligned}

The perimeter of given curve is

\begin{aligned} \mathrm{S} &=\int_{0}^{3 a} \sqrt{1+\left(\frac{d y}{d x}\right)^{2} d x}=\int_{0}^{3 a} \sqrt{1+\frac{(x-a)^{2}}{4 a x}} d x=\int_{0}^{3 a} \sqrt{\frac{(x+a)^{2}}{4 a x}} d x \\ \therefore \mathrm{S} &=\int_{0}^{3 a} \frac{x+a}{2 \sqrt{x} \sqrt{a}} d x \\ \therefore \mathrm{S} &=\frac{1}{2 \sqrt{a}} \int_{0}^{3 a}\frac{ x+a}{\sqrt{x}} d x \end{aligned}

$=\frac{1}{2 \sqrt{a}}\left[\frac{2 x \sqrt{x}}{3}+2 \sqrt{x}\right]_{0}^{3 a}$

$=\frac{1}{2 \sqrt{a}}\left(\frac{2 a \sqrt{3 a}}{1}+2 \sqrt{3 a}\right)$

$\therefore S=2 \sqrt{3}$---------------(Half Curve length)

$\therefore$ The totallength of given curve $=2 \mathrm{S}=4 \sqrt{3}$ units.