# Parametric Equations. Finding dy/dx?

• March 20th 2013, 02:09 PM
Parametric Equations. Finding dy/dx?
I'm unsure on the following C4 Coordinate Geometry Question

A curve has the following parametric equations

http://latex.codecogs.com/gif.latex?x= \frac{t}{2-t} http://latex.codecogs.com/gif.latex?y= \frac{1}{1+t}

a) show that

http://latex.codecogs.com/gif.latex?...c{2-t}{1+t})^2, would this be done using dy/dx = (dy/dt)/(dx/dt)?

b) find an equation for the normal to the curve at point where t = 1

this would just be substituting t=1 into the x, y and dy/dx equations and using (y-y1) = m(x-x1), yeah?

c) show that the cartesian equation of the curve can be written in the form

http://latex.codecogs.com/gif.latex?y= \frac{1+x}{1+3x}

unsure on this
• March 20th 2013, 02:22 PM
HallsofIvy
Re: Parametric Equations. Finding dy/dx?
Quote:

I'm unsure on the following C4 Coordinate Geometry Question

A curve has the following parametric equations

http://latex.codecogs.com/gif.latex?x= \frac{t}{2-t} http://latex.codecogs.com/gif.latex?y= \frac{1}{1+t}

a) show that

http://latex.codecogs.com/gif.latex?...c{2-t}{1+t})^2, would this be done using dy/dx = (dy/dt)/(dx/dt)?

Yes, that's the idea.

Quote:

b) find an equation for the normal to the curve at point where t = 1

this would just be substituting t=1 into the x, y and dy/dx equations and using (y-y1) = m(x-x1), yeah?
No, that would give you the tangent to the curve. The normal is perpendicular to the tangent line.

Quote:

c) show that the cartesian equation of the curve can be written in the form

http://latex.codecogs.com/gif.latex?y= \frac{1+x}{1+3x}

unsure on this
Your parametric equations are $x= \frac{t}{2- t}$ and $y= \frac{1}{1+ t}$.
Solve each of those equations for t, set them equal, and solve for y.
• March 20th 2013, 02:35 PM