Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : 2014

If $y=0 x=-7$ or $x=-4$ the loop

Let s be the total length of loop

$S=2\int\limits_{-7}^{-4}\sqrt{1+(\dfrac{dy}{dx})^2}.$dx

Now

$18y\dfrac{dy}{dx}=(x+7)2(x+4)+(x+4)^2$

$\dfrac{dy}{dx}=\dfrac{(x+4)(2x+14+x+4)}{18y}$

$\dfrac{dy}{dx}=\dfrac{(x+4)(x+6)}{6y}$

$1+\Bigg(\dfrac{dy}{dx}\Bigg)^2=1+\dfrac{(x+4)^2(x+6)^2}{36y^2}$

$1+\Bigg(\dfrac{dy}{dx}\Bigg)^2=1+\dfrac{(x+4)^2(x+6)^2}{4(x+7)(x+4)^2}$

$(ay^2=(x+7)(x+4)^2)$

$\therefore 1+\Bigg(\dfrac{dy}{dx}\Bigg)^2=1+\dfrac{x^2+12x+36}{4x+28}$

$\therefore 1+\Bigg(\dfrac{dy}{dx}\Bigg)^2=1+\dfrac{4x+28+x^2+12x+36}{4(x+7)}$

$\therefore 1+\Bigg(\dfrac{dy}{dx}\Bigg)^2=1+\dfrac{(x+8)^2}{4(x+7)}$

$S=2\int\limits_{-7}^{-4}\dfrac{x+8}{2\sqrt{x+7}}$dx

put $x+7=t^2$ when $x=-4 \space \space t=\sqrt3$

$dx=2t\space dt\hspace {2cm } X=-7\space \space t=0$

$S=2\int\limits_0^{\sqrt3}\dfrac{t^2+1}{2t}.2t$dt

$S=2\int\limits_0^{\sqrt3}(t^2+1)$dt

$=2\Bigg[\dfrac{t^3}3 + t\Bigg]_0^{\sqrt3}$

$=2\Bigg[\dfrac{3\sqrt3}3+\sqrt3\Bigg]$

$=4\sqrt3$