# Thread: Parametric tangent line

1. ## Parametric tangent line

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

$x=ln(t), y=2(sqrt(t)), z=t^2; (0,2,1)$

My question is, do I simply solve for t (t=1) and then take a derivative of each? I'm confused on what it is asking for.

2. Originally Posted by Rhode963
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

$x=ln(t), y=2(sqrt(t)), z=t^2; (0,2,1)$

My question is, do I simply solve for t (t=1) and then take a derivative of each? I'm confused on what it is asking for.
Well, it is asking for parametric equations of the tangent line!

You should know that if you are given a curve as (x(t), y(t), z(t)) then the derivative vector <x'(t), y'(t), z'(t)> is tangent to the curve. You should also know that parametric equations for a line through $(x_0, y_0, z_0)$ with "direction vector" <A, B, C> are $x= At+ x_0$, $y= Bt+ y_0$, and $z= Ct+ z_0$.

You are given the point and the "direction vector" of a tangent line is a tangent vector at that point. Differentiate x, y, and z with respect to t, evaluate at t= 1, and use that as the "direction vector" of the line.

3. So $x'=1/t, y'=t^(-1/2), z'=2t$ and evaluated at t=1, $x'=1, y'=1, z'=2$, meaning the parametric tangent line is... $x=1+ln(t), y=1+2(sqrt(t)), z=2+t^2$ ?