Question

How to find the tangent line to a curve given parametric equations?

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.
x = t − t^−1, y = 7 + t^2, t = 1

I am not sure how to go about this. Help is appreciated!

Answer 1

`y =x +8`

Explanation:

You have to determine `dy/dx`.

Recall that

`dy/dx = (dy/(dt))/(dx/(dt))`

We know that `(dx)/(dt) = 1 + 1/t^2` and `dy/(dt) = 2t`. Thus

`dy/dx= (2t)/(1 + 1/t^2)`

The slope of the tangent line is given by evaluating `dy/dx` at the given point.

`m = (2(1))/(1 + 1) = 1`

Now find the values of `x` and `y`.

`x(1) = 1 - 1 = 0`
`y(1) = 1^2 + 7 = 8`

The equation will therefore be

`y - 8 = 1(x -0)`

`y = x + 8`

Hopefully this helps!