contestada

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 6 + ln(t), y = t2 + 2, (6, 3)

Respuesta :

Without elimination:

[tex]\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}[/tex]
[tex]\frac{dy}{dt} = 2t[/tex]

[tex]\frac{dx}{dt} = \frac{1}{t}[/tex]
[tex]\frac{dt}{dx} = t[/tex]

[tex]\therefore \frac{dy}{dx} = \frac{2t}{t} = 2[/tex]

With elimination:

[tex]ln(t) = x - 6; t = e^{x - 6}[/tex]
[tex]y = e^{2(x - 6)} + 2[/tex]
[tex]y = e^{2x - 12} + 2[/tex]
[tex]\frac{dy}{dx} = 2e^{2x - 12}[/tex]

At x = 6:
[tex]\frac{dy}{dx} = 2e^{2\cdot 6 - 12} = 2(1) = 2[/tex]