Question

How do you find the equation of the tangent line to the curve given by parametric equations: `x=1+(1/t^2)`, `y=1-(3/t)` at the point when t=2?

Answer 1

Please see the explanation for the process. `y = -3x - 17/4`

Explanation:

To find the slope of the tangent line, you need `dy/dx` evaluated at a specified point.

You are given `x(t)` and `y(t)`. You can use these to compute `x'(t)` and `y'(t)` which can be written using the notation `dx/dt` and `dy/dt`, respectively.

`dx/dt = -2t^(-3)`

`dy/dt = 3t^(-2)`

From the chain rule, we know that:

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

Solve for `dy/dx`:

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

`dy/dx = (3t^(-2))/(-2t^(-3))`

`dy/dx = (-3/2)t`

To obtain the slope, m, you evaluate the above at `t = 2`:

`m = (-3/2)(2) = -3`

Evaluate x and y at t = 2:

`x = 1 + 1/2^2 = 5/4`

`y = 1 - 3/2 = -1/2`

This is the point `(5/4, -1/2)`

Use the point-slope form of the equation of a line:

`y - y_1 = m(x - x_1)`

`y - -1/2 = -3(x - 5/4)`

`y = -3x - 17/4`