Answer to Question #346587 in Calculus for Likhu

Question

Find the perimeter of loop of the curve 3ay²=x²(a-x)

Answer 1

$3ay^2=x^2(a-x),$

$y^2=\frac{x^2(a-x)}{3a},$

$y=\pm \frac{x\sqrt{a-x}}{\sqrt{3a}}.$

x=0: y=0;

x=a: y=0.

The curve is symmetrical about the x-axis and meets it in points (0, 0) and (a, 0), so the loop lies between this points:

To find the perimeter of the loop, we can double its part above the x-axis.

$P=2 \int_0^a \sqrt{1+(\frac{dy}{dx})^2}dx.$

$\frac{dy}{dx}=\frac{d}{dx}(\frac{x\sqrt{a-x}}{\sqrt{3a}})=\frac{1}{\sqrt{3a}}(\sqrt{a-x}-\frac{x}{2\sqrt{a-x}})=$

$=\frac{1}{\sqrt{3a}}\frac{2(a-x)-x}{2\sqrt{a-x}}=\frac{2a-3x}{2\sqrt{3a}\sqrt{a-x}}.$

$1+(\frac{dy}{dx})^2=1+(\frac{2a-3x}{2\sqrt{3a}\sqrt{a-x}})^2=$

$=1+\frac{(2a-3x)^2}{12a(a-x)}=\frac{12a^2-12ax+4a^2-12a+9x^2}{12a(a-x)}=$

$=\frac{16a^2-24ax+9x^2}{12a(a-x)}=\frac{(4a-3x)^2}{12a(a-x)}.$

So we have

$P=2 \int_0^a \sqrt{\frac{(4a-3x)^2}{12a(a-x)}}dx=\frac{2}{2\sqrt{3a}} \int_0^a \frac{4a-3x}{\sqrt{a-x}}dx= \frac{1}{\sqrt{3a}} \int_0^a \frac{4a-3x}{\sqrt{a-x}}dx$

Substitution: $a-x=u$ , $du=-dx,$ $x=a-u.$

$x=0: u=a-x=a-0=a;$

$x=a: u=a-x=a-a=0.$

$P=\frac{1}{\sqrt{3a}} \int_0^a \frac{4a-3x}{\sqrt{a-x}}dx=-\frac{1}{\sqrt{3a}} \int_a^0\frac{4a-3(a-u)}{\sqrt{u}}du=$ s

$=-\frac{1}{\sqrt{3a}} \int_a^0\frac{a+3u}{\sqrt{u}}du=-\frac{1}{\sqrt{3a}} \int_a^0 (\frac{a}{\sqrt{u}}+3\sqrt{u})du=$

$=-\frac{1}{\sqrt{3a}} (2a\sqrt{u}+2u\sqrt{u})|_a^0=$

$-\frac{1}{\sqrt{3a}}[ (2a\sqrt{0}+2\cdot 0 \cdot \sqrt{0})- (2a\sqrt{a}+2a\sqrt{a})]=$

$=\frac{4a\sqrt{a}}{\sqrt{3a}}=\frac{4a}{\sqrt{3}}.$

Answer: $\frac{4a}{\sqrt{3}}.$