# Parametric tangent line

• Feb 2nd 2010, 07:43 AM
Rhode963
Parametric tangent line
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

\$\displaystyle 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.
• Feb 2nd 2010, 08:55 AM
HallsofIvy
Quote:

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

\$\displaystyle 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 \$\displaystyle (x_0, y_0, z_0)\$ with "direction vector" <A, B, C> are \$\displaystyle x= At+ x_0\$, \$\displaystyle y= Bt+ y_0\$, and \$\displaystyle 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.
• Feb 2nd 2010, 09:15 AM
Rhode963
So \$\displaystyle x'=1/t, y'=t^(-1/2), z'=2t\$ and evaluated at t=1, \$\displaystyle x'=1, y'=1, z'=2\$, meaning the parametric tangent line is... \$\displaystyle x=1+ln(t), y=1+2(sqrt(t)), z=2+t^2\$ ?