Parametric curves

**DarthPipsqueak** · Apr 14th 2010, 12:14 PM

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.

Please help!!!!

**skeeter** · Apr 14th 2010, 01:56 PM

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.

Please help!!!!

eliminating the parameter ...

$x^2 = \tan^2{\theta} = \sec^2{\theta} - 1 = y^2 - 1<br />$

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

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

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

parametrically ...

$x = \tan{\theta}$

$1 = \tan{\theta}$

$y = \sec{\theta}$

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

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

$\frac{dy}{d\theta} = \sec{\theta}\tan{\theta} = \sqrt{2} \cdot 1 = \sqrt{2}<br />$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\sqrt{2}}{2}<br />$

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