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!

1 Answer
Apr 29, 2018

#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!