Find the equation of a plane normal to the curve (e^t, t, t^2)

Problem: Find the equation of a plane normal to the curve $(e^t, t, t^2)$ at the point $t=1$. Set the curve to $P(t)=(e^t, t, t^2)$, then taking the derivative twice we get $P''(t)=(e^t, 0, 2)$. Using $(X-P) \cdot N=0$ If $P''(1) = N$ and $P(1)=P$, then I get $ex+2z=e^2+2$. I'm pretty sure $P\cdot N=0$ should be met, but I'm not exactly sure where I went wrong. Is this the correct approach? I have realized with a few counterexamples that this derivation may not yield the correct normal. Any insights?

smokeypeat

asked Sep 28, 2017 at 4:25

smokeypeat smokeypeat

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$\begingroup$ Why are you taking the derivative twice? $\endgroup$

Commented Sep 28, 2017 at 4:27

$\begingroup$ It is asking for the plane normal to the curve, not tangent. The thinking is the first derivative is tangent $\endgroup$

Commented Sep 28, 2017 at 4:34