# Parametric curves

• Apr 14th 2010, 12:14 PM
DarthPipsqueak
Parametric curves
The question is:

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter.
x = tan(θ)
y = sec(θ)

(1, sqrt(2))

I am unsure as to what the question is referring to. Maybe I'm being silly and it's actually very easy, I don't know.

• Apr 14th 2010, 01:56 PM
skeeter
Quote:

Originally Posted by DarthPipsqueak
The question is:

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter.
x = tan(θ)
y = sec(θ)

(1, sqrt(2))

I am unsure as to what the question is referring to. Maybe I'm being silly and it's actually very easy, I don't know.

eliminating the parameter ...

$\displaystyle x^2 = \tan^2{\theta} = \sec^2{\theta} - 1 = y^2 - 1$

$\displaystyle \frac{d}{dx}(x^2 = y^2 - 1)$

$\displaystyle 2x = 2y \cdot \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{x}{y} = \frac{1}{\sqrt{2}}$

parametrically ...

$\displaystyle x = \tan{\theta}$

$\displaystyle 1 = \tan{\theta}$

$\displaystyle y = \sec{\theta}$

$\displaystyle \sqrt{2} = \sec{\theta}$

$\displaystyle \frac{dx}{d\theta} = \sec^2{\theta} = 2$

$\displaystyle \frac{dy}{d\theta} = \sec{\theta}\tan{\theta} = \sqrt{2} \cdot 1 = \sqrt{2}$

$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\sqrt{2}}{2}$