A Point on Curve Where Tangent is Parallel How do I find a point on a curve, $r=\langle 2,3t,5t^3\rangle$ in which tangent line is parallel to some plane.
I thought to find the derivative first, but having trouble there as well. Anyway, this would provide the slope. However, I don't know what comes next.
 A: A point on a curve $r(t)=\langle 1,t,t^2\rangle$ in which the tangent line is parallel to plane $x+2y+3z=0$ occurs where the derivative of $r(t)$ is orthogonal to the normal of the plane. That is, when $$r'(t)\cdot \hat{n}=0$$
$$\langle0,1,2t\rangle\cdot\langle1,2,3\rangle=0$$
A: You are correct that you will need to find $r'(t)$, the derivative of your function. This will give you the direction vector of your curve at an arbitrary time $t$.
From there, all we have to do is determine when this direction vector is perpendicular to the normal of the plane and solve for $t$.
Start by finding $r'(t)$. Simply differentiate each component of $r(t)$ and we get:
$$
r'(t) = <0,1,2t>
$$
We can then extract the normal from the plane rather easily, as the normal of a plane in the form $ax + by + cz = 0$ is simply $(a,b,c)$.
Thus our normal vector for our plane is $(1,2,3)$.
To check where this is perpendicular to the direction vector of our curve, we will see where their dot product is $0$:
$$
(0,1,2t)\cdot(1,2,3) = 0 \Rightarrow 0 + 2 + 6t = 0 \Rightarrow 6t = -2 \Rightarrow t = -2/6 = -1/3
$$
Now all we have to do is plug in $t = -1/3$ to the equation of our curve to find the given point: $r(-1/3) = (1, -1/3, 1/9)$.
