Question

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)?

Answer 1

The answer is:

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

The point `(3,11,11)` 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'=14t`
`y'=14t`
`z'=6`

So, for `t=1` it is: `vecv(14,14,6)`.

So, remembering that given a point `P(x_P,y_P,z_P)` and a direction `vecv(a,b,c)` the line that passes from that point with that direction is:

`x=x_P+at`
`y=y_P+bt`
`z=z_P+ct`

so the tangent is:

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

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

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

that's simplier!