# Thread: Equation of tangent line in 3d

1. ## Equation of tangent line in 3d

[SIZE=+1]The equation of the tangent line to the curve
r(t) = t^2 i + 3t j + t3 k at the point where t = 3

i found the point 9,9,27 by subbing in t=3 into the r(t) position equation, however im unsure how they find the (TANGENT) slope s*(6,3,27).

so im left with this at the moment (9,9,27)+s(?,?,?)

2. Hi sebko

To find the slope, first find $\frac{d}{dt}r(t)$, then subs. t = 3

Note: slope = first derivative

3. Sorry i meant, the tangent slope. "s" is the parameter,

4. sorry was getting confused with tangent and normal...... how would you find the normal curve?

5. Hi sebko

Yes, the tangent slope = first derivative

$\text{slope for normal}=\frac{-1}{\text{tangent slope}}$