# Homework Help: Parametric equation for a tangent line (multi vars)

1. Feb 7, 2010

### forest125

1. The problem statement, all variables and given/known data

Find the parametric equation for the line tangent to the curve:

x=t^3-1, y=t^4+1, z=t

at the point (26, 82, 3).

Use the variable t for your parameter.

2. Relevant equations

3. The attempt at a solution

dx/dt=3t^2, dy/dt=4t^3, dz/dt=1

I got then that

x=26+(3*26^2)t
y=82+(4*82^3)t
z=3+t

and only the z component is correct. Where am I going wrong. Any help is really appreciated!

2. Feb 7, 2010

### Dick

Look, at your given point t=3. When you are working out the d/dt parts why are you substituting x and y for t?

3. Feb 7, 2010

### Raziel2701

You should not plug back your given point coordinates into "t" in your derived equations. Your value of t is constant. And technically it's given in your problem.

4. Feb 7, 2010

### forest125

I substituted x and y into my derivatives to find the slope of the that component of the line at that point. I then multiply it by t, and add x and y for intercepts or to start from the origin.

5. Feb 7, 2010

### Dick

t is t. x is x. y is y. z happens to be t. You got lucky on the z component because z=t. Just put t in where your equations say 't'. Don't substitute x for t just because it happens to be the 'x' component.

6. Feb 7, 2010

### forest125

I don't know why I'm so hung up on this question.

For the x component for instance,

it the curve is t^3-1, the slope of the tangent is 3t^2. If we're talking about the slope of the tangent at 26 in the curve, why wouldn't we plug in 26 for t in dx/dt?

What is the next step I should take after finding the derivative?

7. Feb 7, 2010

### forest125

Oh jesus. t=3 right. That can't change.

Wow I feel stupid. Thanks all for the help.