analytic geometryarc lengthcurvesparametrization
If $A$ is the curve of intersection of the parabolic cylinder $x^2 = 2y$ and the
surface $3z = xy$, then what is the length of $A$ from the origin to the point (6, 18, 36)?
What I have done so far:
I know to parametrise the cylinder to get $x$ and $y$ in terms of $sin$ and $cos$ however I am unsure of how to do this.
Square root is not a good idea. You lose half curve
Solve $ \left\{ \begin{array}{l} z=4 x^2+y^2 \\ y=x^2 \\ \end{array} \right. $
$ \left\{ \begin{array}{l} x= t\\ y =t^2\\ z=4 t^2+t^4 \end{array} \right.$
I got this nice image,
From $8x^2 + 10z^2 = 1$,you get $z^2 = \frac{1}{10}.(1-8x^2)$. Substitute this in the other equation $ x^2+y^2+z^2 = \frac{1}{8}$ you get
$$x^2 + 5y^2 = \frac{1}{8}$$
This is the curve of intersection, now parameterize this ellipse with
$x = \frac{1}{\sqrt{8}} sint$
and
$ y = \frac{1}{\sqrt{40}} cost$
$ z = \frac{1}{\sqrt{10}} cost$
Now arc length $ L= \int_0^{2\pi} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 +(\frac{dz}{dt})^2} dt$
$$ = \int_0^{2\pi} \sqrt{\frac{1}{8} cos^2t + \frac{1}{40} sin^2t+\frac{1}{10} sin^2t}dt$$
$$ = \int_0^{2\pi} \sqrt{\frac{1}{40}}.\sqrt{ 5cos^2t + sin^2t +4sin^2t} dt$$
$$=\int_0^{2\pi} \sqrt{\frac{5}{40}} dt$$
$$L= \frac{1}{2\sqrt{2}}\int_0^{2\pi} dt\tag 1$$
$$ L = \frac{2\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}$$
Best Answer
$$x^2=2y\implies y=x^2/2$$ $$3z=xy=x^3/2$$ Let $x=t$. Then $A$ from the origin to $(6,18,36)$ is parametrised as $$(t,t^2/2,t^3/6)\qquad t\in[0,6]$$ The arc length is $$\int_0^6\sqrt{1^2+t^2+(t^2/2)^2}\,dt=\frac12\int_0^6\sqrt{t^4+4t^2+4}\,dt$$ $$=\frac12\int_0^6(t^2+2)\,dt=\frac12\left[\frac13t^3+2t\right]_0^6=42$$