Let C be the piece of the curve of intersection of the cylinder $y = x^2$
and the plane
$x + y + z = 0$ between the points $(0, 0, 0)$ and $(2, 4, −6)$.Parametrize C and compute $$\int _C\sqrt{2+4x+8y}\:ds$$
How would I do this?
Here's what I tried:
Let $x=t$
So then: $y=t^2$
Then I use a random vector function $r(t) = (t,t^2)$
So:
$r(0) = (0,0)$ and $r(2) = (2,4)$
And those equal my points. But how do I get the Z value? Do I just throw in another value like:
$r(t) = (t,t^2,-t^2-2)$
How does it work? I have no clue how to solve this. I'm guessing my bounds are $0->2$ when I differentiate in respect to dt.
So then:
$$ds = ||r'(t)|| = ||(1,2t,-2t)|| = \sqrt{1+8t^2}$$
Therefore my integral is:
$$\int _0^2\:\left(t^2\right)\left(\sqrt{1+8t^2}\right)dt$$
Is this right?
But like. Even if it IS right, how do I get the parametrization without guessing and checking? I feel like I just used random logic to get to my integral instead of actual steps. So essentially I'm wondering, how to parametrize the curve of intersection.
Like what if it wasn't just $x=t$? Do you just eyeball the parametrization?
Best Answer
Let us find a parametrisation $r(t)=(x(t),y(t),z(t))$ such that $r(0)=(0,0,0)$ and $r(1)=(2,4,-6)$. We also require that $$x(t)+y(t)+z(t)=0\quad\text{and}\quad y(t)=x(t)^2\quad\forall\ t\in[0,1].$$ If we set $x(t)=2t$, then the two equations above demand that $$y(t)=x(t)^2=4t^2\quad\text{and}\quad z(t)=-(x(t)+y(t))=-2t-4t^2.$$ So we get the parametrisation $$r(t)=(2t,4t^2,-2t-4t^2).$$ As for "how do I get the parmetrisation without guessing and checking" - well, there is a level of choice. I chose the limits of $t$, then I chose an $x(t)$ that works. From there $y(t)$ and $z(t)$ are determined by those choices.
I should add that you should check that $r'(t)\ne 0$ for all $t\in[0,1]$. Otherwise this would correspond to the curve having a 'cusp'.