# How do you find parametric equations for the tangent line to the curve with the given parametric equations x=7t^2-4 and y=7t^2+4 and z=6t+5 and (3,11,11)?

##### 1 Answer
Feb 22, 2015

The answer is:

$x = 3 + 14 t$
$y = 11 + 14 t$
$z = 11 + 6 t$

The point $\left(3 , 11 , 11\right)$ is for $t = 1$, as you can see substituting it in the three equations of the curve.

Now let's search the generic vector tangent to the curve:

$x ' = 14 t$
$y ' = 14 t$
$z ' = 6$

So, for $t = 1$ it is: $\vec{v} \left(14 , 14 , 6\right)$.

So, remembering that given a point $P \left({x}_{P} , {y}_{P} , {z}_{P}\right)$ and a direction $\vec{v} \left(a , b , c\right)$ the line that passes from that point with that direction is:

$x = {x}_{P} + a t$
$y = {y}_{P} + b t$
$z = {z}_{P} + c t$

so the tangent is:

$x = 3 + 14 t$
$y = 11 + 14 t$
$z = 11 + 6 t$

N.B. The direction $\left(14 , 14 , 6\right)$ is the same that $\left(7 , 7 , 3\right)$, so the line could be written also:

$x = 3 + 7 t$
$y = 11 + 7 t$
$z = 11 + 3 t$

that's simplier!