Find an equation of the tangent line to the curve defined parametrically by x=t³−t and y=t⁴−3t²+3 at t=2

Question

asked May 18, 2024 182k views

1 Answer

Kwaku · Answer 1 · 2024-05-23T09:04:41+0000

Explanation:

Firstx let find the coordinate point, plug in 2 for t into each functions

$x(2) = 2 {}^(3) - 2 = 6$

$y(2) = 2 {}^(4) - 3(2) {}^(2) + 3 = 7$

So the coordinate pair is (6,7)

Given the functions

x(t)

and

y(t)

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

Take the derivative of the y function first

$(dy)/(dt) = 4t {}^(3) - 6t$

$(dx)/(dt) = 3 {t}^(2) - 1$

So

$(dy)/(dx) = \frac{2t( 2{t {}^(2) } - 3)}{3 {t}^(2) - 1}$

Plug in x=2 into the day/dx to find the tangential slope

$(dy)/(dx) (2) = \frac{2(2)((2 * {2}^(2) - 3 )}{3(2) {}^(2) - 1} = (4(5)/(11) = (20)/(11)$

We know it passes through (6,7) so

Or simplified