Winter 2016 - Q.3 - Grad Plus

# Winter 2016 – Q.3

3. a) Trace the curve $latex a^2x^2=y^3\left(2a-y\right)$ and show that its area is equal to $latex \pi a^2$. [6M]

The given equation of the curve is

$latex a^2x^2=y^3\left(2a-y\right)—–\left(i\right)$

I) Symmetry:- The equation (i) has even power of x Hence the curve is symmetry about y-axis

II) Origin:- The curve passes through the Origin tangent at origin is given by $latex x^2=0$ i.e. x = 0, 0

∴ origin is a cusp.

III) Intercepts:- put x = 0 in equation (i) we get,

y = 0, 2a

∴ (0, 0), (0, 2a) are the y intercepts.

IV) Region:- When y > 2a, x is imaginary and the curve does not exist when y > 2a.

When y > 0, x is imaginary and the curve does not exist when y < 0

V) The Table of values of x and y is given below

 y 0 $latex \frac a2$ a $latex \frac{3a}2$ 2a $latex x=\frac{\pm\left[y^3\left(2a-y\right)\right]^{\displaystyle\frac12}}a$ 0 $latex \pm0.44a$ $latex \pm a$ $latex \pm1.3a$ 0

The curve is given below

$latex \therefore Required\;area=2\int_0^{2a}x\;dy$

$latex \begin{array}{l}=2\int_0^{2a}\frac{\left[y^3\left(2a-y\right)\right]^{\displaystyle\frac12}}a\;dy\\\\=2\int_0^{2a}\frac{y^{\displaystyle\frac32}\left(2a-y\right)^{\displaystyle\frac12}}a\;dy\\\\=\frac2a\int_0^{2a}y^\frac32\left(2a\right)^\frac12\cdot\left(1-\frac y{2a}\right)^\frac12\;dy\end{array}$

put

$latex \begin{array}{l}\frac y{2a}=t\\\\\therefore y=2at\end{array}$

Differentiate w r to t

$latex \begin{array}{l}\therefore\frac{dy}{dt}=2a\\\\\therefore dy=2adt\end{array}$

when y→0, t→0 and y→2a, t→1

$latex =\frac2a\int_0^1\left(2at\right)^\frac32\left(2a\right)^\frac12\left(1-t\right)^\frac122a\;dt$

$latex \begin{array}{l}=\frac2a\int_0^1\left(2at\right)^\frac32\left(2a\right)^\frac12\left(1-t\right)^\frac122a\;dt\\\\=\frac2a\int_0^1\left(2a\right)^\frac32\left(2a\right)^\frac12\left(t\right)^\frac32\left(1-t\right)^\frac122a\;dt\\\\=\frac2a\int_0^1\left(2a\right)^2\;t^\frac32\left(1-t\right)^\frac122a\;dt\\\\=\frac2a\int_0^1\left(2a\right)^3\;t^\frac32\left(1-t\right)^\frac12\;dt\\\\=\frac{16a^3}a\int_0^1t^\frac32\left(1-t\right)^\frac122a\;dt\\\\\left(\because By\;defferentiate\;of\;Beta\;function\right)\\\\=16a^2\beta\left(\frac52,\;\frac32\right)\\\\=16a^2\frac{\left|\overline{\displaystyle\frac52}\;\left|\overline{\displaystyle\frac32}\right.\right.}{\left|\overline4\right.}\\\\=\frac{16a^2{\displaystyle\frac32}\;{\displaystyle\frac12}\;\left|\overline{\displaystyle\frac12}\;{\displaystyle\frac12}\;\left|\overline{\displaystyle\frac12}\right.\right.}{1\times2\times3}\\\\\left(\because\left|\overline{\frac12}=\sqrt\pi\right.\right)\\\\=a^2\pi\end{array}$

$latex \therefore area=a^2\pi\;sq.unit$

b) Find the perimeter of the asteroid $latex x^\frac23+y^\frac23=a^\frac23.$. [6M]

Given that,

$latex x^\frac23+y^\frac23=a^\frac23$

The curve is symmetrical about the axies it meets the x-axis in the first quadrant at x = a

Differentiate equation (i) w r to x we get

$latex \frac23x^\frac{-1}3+\frac{\displaystyle2}{\displaystyle3}y^\frac{\displaystyle-1}{\displaystyle3}\frac{dy}{dx}=0$

$latex \therefore\frac{dy}{dx}=-\left(\frac yx\right)^\frac13$

∴ perimeter of line astroid $latex =4\int_0^a\sqrt{1+\left(\frac{dy}{dx}\right)^2dx}$

$latex \begin{array}{l}=4\int_0^a\sqrt{1+\left(\frac yx\right)^\frac23}\;dx\\\\=4\int_0^a\sqrt{\frac{x^{\displaystyle\frac23}+y^{\displaystyle\frac23}}{x^{\displaystyle\frac23}}}\;dx\\\\\because from\;equation\;\left(i\right)\\\\=4\int_0^a\sqrt{\left(\frac{a^{\displaystyle\frac23}}{x^{\displaystyle\frac23}}\right)}\;dx\\\\=4\int_0^a\sqrt{\left(\frac ax\right)^\frac23}\;dx\\\\=4\int_0^aa^\frac13x^\frac{-1}3\;dx\\\\=4\;a^\frac13\frac32\left[x^\frac23\right]_0^a\\\\=6a^\frac13a^\frac23\\\\=6a\end{array}$

∴ perimeter of astroid = 6a units

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