Parametric Equation of Tangent Line for f(t)= (t2,1/t) at t=2

[Loppyfoot]

Homework Statement

The function f(t)= (t²,1/t) represents a curve in the plane parametrically.
Write an equation in parametric form for the tangent line to this curve at the point where t=2

The Attempt at a Solution

I can solve the gradient from an implicit equation, but solving from a parametric equation confuses me.

Would the gradient of f(t)= <2t, -1/t²>?

Then plug in the value t=2 to get the point?

Would that be correct?

Thanks for the input..

 

[tiny-tim]

Hi Loppyfoot!

Would the gradient of f(t)= <2t, -1/t²>?

It's parallel to that …

the gradient itself is a multiple of that, and the exact multiple depends on the choice of parameter (t in this case) …

fortunately the multiple does't matter in this case.

Then plug in the value t=2 to get the point?

Yes, that'll give you the slope of the tangent line, from which you can get its equation in parametric form, as asked for (with a different parameter, of course!)

 

[Loppyfoot]

Alright!

So the final answer after plugging in t=2 to get the slope, and f(2) to get the points, the parametric equation would be:

f(t)= (4,1/4) + t<4,-1/4>

Thanks a lot TIny TIm!

 

[tiny-tim]

f(t)= (4,1/4) + t<4,-1/4>

Neater would be to simplify it … f(t)= (4,1/4) + t<1,-1>

(and possibly to use a different parameter)

 

[Loppyfoot]

Oh good point. thanks!