Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 2 + ln(t), y = t2 + 2, (2, 3)

Respuesta :

Answer:

Without parameter : [tex]y = 2x -1[/tex]

Step-by-step explanation:

from x = 2 + int => t = [tex]e^{x-2}[/tex]

substitute into y equation

y = [tex]e^{2x -4} + 2[/tex]

[tex]\frac{dy}{dx} = 2e^{2x - 4}[/tex] at(x= 2)

[tex]\frac{dy}{dx} = m = 2e^{4-4} = 2e^{0} = 2 x1 = 2\\[/tex]

equation  of the tangent is

[tex]y - 3 = m(x-2)\\y-3 = 2(x-2)\\y = 2x -1[/tex]