# parametrically defined curve

• Dec 7th 2011, 09:48 PM
icelated
parametrically defined curve
I have a parametrically defined curve

$\displaystyle x(t) = \frac {1}{t}$
and
$\displaystyle y(t) = -2 + lnt$

I need to find the equation for the tangent line to the curve at the point defined by t = 1

Now, i think i need to find $\displaystyle \frac{dy}{dx}$

the dirivitive of x = $\displaystyle -\frac{1}{t^2}$ ?

Would the dirivitive of y = $\displaystyle \frac{1}{t}$ ?

then, i evaluate t= 1?

then, take the dirivitive of dy again?

then use point slope form?
• Dec 7th 2011, 10:13 PM
earboth
Re: parametrically defined curve
Quote:

Originally Posted by icelated
I have a parametrically defined curve

$\displaystyle x(t) = \frac {1}{t}$
and
$\displaystyle y(t) = -2 + lnt$

I need to find the equation for the tangent line to the curve at the point defined by t = 1

Now, i think i need to find $\displaystyle \frac{dy}{dx}$

the dirivitive of x = $\displaystyle -\frac{1}{t^2}$ ?

Would the dirivitive of y = $\displaystyle \frac{1}{t}$ ?

then, i evaluate t= 1?

then, take the dirivitive of dy again?

then use point slope form?

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Plug in the terms you already have determined.

Then use the point-slope-form of the equation of a straight line at t = 1
• Dec 7th 2011, 10:18 PM
princeps
Re: parametrically defined curve
$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Equation of the tangent line is :

$\displaystyle y(t)=\frac{dy}{dx}(t)\cdot x(t)+n$