Given the planar curve, find the equation of the plane

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1. Oct 15, 2016

dlacombe13

1. The problem statement, all variables and given/known data
$r(t) = < 2e^t - 5 , e^t +3t^2 , 4t^2 +1>$
Is a curve that lies within a plane. Find the equation of this plane.

3. The attempt at a solution
I am not sure if my approach is correct. These are my results:
$x=2e^t - 5$
$y = e^t +3t^2$
$z = 4t^2 + 1$

$z = 4t^2 + 1 ~~\Rightarrow~~ t = \sqrt{\frac{z-1}{4}}$

$x = 2e^\sqrt{\frac{z-1}{4}} - 5 ~~\Rightarrow~~ \frac{x+5}{2} =e^\sqrt{\frac{z-1}{4}}$
$y = e^\sqrt{\frac{z-1}{4}} + \frac{3z-3}{4}$

$y = \frac{x+5}{2} + \frac{3z-3}{4}$
$= \frac{x}{2} + \frac{5}{2} + \frac{3z}{4} - \frac{3}{4}$

$= \frac{1}{2}x - y + \frac{3}{4}z = -\frac{15}{8}$

2. Oct 15, 2016

pasmith

One way of specifying a plane is as $\{ \lambda \mathbf{a} + \mu\mathbf{b} + \mathbf{c} : (\lambda , \mu) \in \mathbb{R}^2\}$ for given vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ where $\mathbf{a}$ and $\mathbf{b}$ are linearly independent. Then for non-constant functions $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ we have that $$\mathbf{r}(t) = f(t)\mathbf{a} + g(t)\mathbf{b} + \mathbf{c}$$ is a curve which lies on this plane.

3. Oct 15, 2016

Ray Vickson

Easier: we have $2y- x = 6 t^2 + 5$, which has eliminated the $e^t$ terms. Now to eliminate the $t^2$ terms, just add or subtract a suitable multiple of $z$. That will leave you with a constant, having no $t$ in it anywhere.

4. Oct 15, 2016

Staff: Mentor

"I am not sure if my approach is correct."
You can (and really should) check your final equation. To do this choose three t values in your equation for r(t). That will give you three vectors, the endpoints of which lie in the plane. These three points should satisfy the plane equation you ended with. If all three points work, then you can be 99% sure that your work is correct. (Subtracting a tiny amount for arithmetic errors you might make). If the three points don't satisfy the plane equation, that's a sign that you have done something wrong.

5. Oct 15, 2016

LCKurtz

I didn't check all your work, but look at your last 3 lines. You start out with $y =$ and end up with $= \frac{1}{2}x - y + \frac{3}{4}z = -\frac{15}{8}$.

Does $y=$ that or is your final equation supposed to be $\frac{1}{2}x - y + \frac{3}{4}z = -\frac{15}{8}$? If that is your answer then notice that $\vec r(0) = \langle -3,1,1\rangle$, so that point must be on the plane. It doesn't seem to satisfy that last equation. It may be a simple arithmetic mistake.

But one reason I didn't check your work is that I don't think eliminating the parameter is a sensible way to do the problem. I would suggest a different approach. All you need for a plane is a point and a normal vector. It's easy to get a point on the plane. Then notice that given the curve is planar, that means the tangent vectors to the curve must be in the plane. So you could calculate $\vec r'(0)\times \vec r'(1)$ to get a normal.

6. Oct 15, 2016

dlacombe13

Thank you all for your replies. I tried plugging in t=0,1,2 into each parameter, and plugged it into the equation for the plane. For all three points, I get: -1.75. -15/8 = -1.875.
Thank you all for you're replies, I tried checking my values and got close to -15/8, but I don't like the error, since I think my arithmetic is right. Also, my original thinking was exactly as LCKurtz said. However I ran into issues, but I think that is because I did it wrong. I used:
$T = \frac{r'(t)}{|r'(t)|}$
T x r(t)
But it didn't seem to work. So what you're saying is I just need to take find two random tangent points along the curve, which will give me two random vectors on the plane it is within, and cross them which will yield an orthogonal vector (n), and then just form the equation from it?

7. Oct 15, 2016

dlacombe13

Thanks everyone, I did get the equation finally, using r'(0) x r'(1) = <8,-16,12> and got the equation:
8x - 16y +12z +28 = 0
I verified it by plugging in points, and it works out. One last question before I go...
Why didn't my attempt at using T, the unit tangent vector work? I mean it is still a vector that is in the direction of the curve, and thus on the plane, right?

8. Oct 15, 2016

LCKurtz

Yes, it should have worked. But dividing by the magnitude is unnecessary, adds square roots, and makes the calculations more error prone, which apparently got you.