Question

Find the length of the portion of the parabola x =3t^2,y=6t cut off by the line 3x+y-3=0

Answer 1

The length of the curve is calculated by the formula:

$$l=\int^{x_2}_{x_1}\sqrt{1+(f'(x))^2}dx$$

wrere $x_1$ , $x_2$ - curve borders.

Bring the parabola to the form $x=f(y)$:

$y=6t\rarr t=\frac{y}{6}$ ;

$x=3t^2=\frac{y^2}{12}$ .

Find the boundaries of the curve as the intersection of two functions.

$\begin{cases} 3x+y-3=0 \\ x=\frac{y^2}{12} \end{cases}\rarr \frac{y^2}{4}+y-3=0 \rarr y_1=-6, y_2=2$.

Calculate the derivative of the function $f(y)$ :

$f'(y)=(\frac{y^2}{12})'=\frac{y}{6}\rarr (f'(y))^2=\frac{y^2}{36}$ .

We calculate the resulting integral

$$\int^{2}_{-6}\sqrt{1+\frac{y^2}{36}}dy=\frac{1}{6}\int^{2}_{-6}\sqrt{36+y^2}dy$$

Use hyperbolic substitution

$y=6\sinh{u} , \sqrt{36+y^2}=6\cosh{u},$

$dy=6\cosh{u}\space du, u=arcsinh({\frac{y}{6}}),$

$u_1=arcsin({\frac{y_1}{6}})=-arcsinh(1),$$u_2=arcsin({\frac{y_2}{6}})=arcsinh(\frac{1}{3}).$

We have

$\int^{2}_{-6}\sqrt{1+\frac{y^2}{36}}dy=\frac{1}{6}\int_{-arcsin(1)}^{arcsin(\frac{1}{3})}36\cosh^2{u}\space du=6\int_{-arcsin(1)}^{arcsin(\frac{1}{3})}\frac{\cosh{2u}+1}{2}du=6(\frac{1}{4}\sinh{2u}+\frac{u}{2})|^{arcsin(\frac{1}{3})}_{-arcsin(1)}=(3\sinh{u}\space \cosh{u}+3u)|^{arcsin(\frac{1}{3})}_{-arcsin(1)}=\frac{\sqrt{10}}{3}+3\sqrt2+3arcsinh(\frac{1}{3})+3arcsinh(1)≈8.923$