Find the equation of the tangent to the curve at x=3 for parametric equations

x = t + 1/t

y = t^2 + 1 / (t^2) when t is greater than 0

Question

Raven Salas

Mathematics

27 September, 13:46

Find the equation of the tangent to the curve at x=3 for parametric equations

x = t + 1/t

y = t^2 + 1 / (t^2) when t is greater than 0

+5

Answers (1)

Know the Answer?

Aidan Wells · Answer 1

y = 6x - 11

Step-by-step explanation:

x = t + (1/t), y = t² + (1/t²)

If we square x:

x² = t² + 2 + (1/t²)

x² = y + 2

When x = 3, y = 7.

Taking derivative with respect to time:

2x = dy/dx

dy/dx = 6

So the equation of the tangent line is:

y - 7 = 6 (x - 3)

y - 7 = 6x - 18

y = 6x - 11

Find the equation of the tangent to the curve at x=3 for parametric equationsx = t + 1/ty = t^2 + 1 / (t^2) when t is greater than 0

Find the equation of the tangent to the curve at x=3 for parametric equations

x = t + 1/t

y = t^2 + 1 / (t^2) when t is greater than 0