# Thread: Parametric Equations. Finding dy/dx?

1. ## Parametric Equations. Finding dy/dx?

I'm unsure on the following C4 Coordinate Geometry Question

A curve has the following parametric equations

$x= \frac{t}{2-t}$ $y= \frac{1}{1+t}$

a) show that

$\frac{dy}{dx}= -\frac{1}{2}(\frac{2-t}{1+t})^2$, 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

$y= \frac{1+x}{1+3x}$

unsure on this

2. ## Re: Parametric Equations. Finding dy/dx?

I'm unsure on the following C4 Coordinate Geometry Question

A curve has the following parametric equations

$x= \frac{t}{2-t}$ $y= \frac{1}{1+t}$

a) show that

$\frac{dy}{dx}= -\frac{1}{2}(\frac{2-t}{1+t})^2$, would this be done using dy/dx = (dy/dt)/(dx/dt)?
Yes, that's the idea.

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.

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

$y= \frac{1+x}{1+3x}$

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.

3. ## Re: Parametric Equations. Finding dy/dx?

I now understand the methods but seem to be getting different answers to those that the answer wants. I must be making mistakes :/ Could you go through them...pretty please :P

4. ## Re: Parametric Equations. Finding dy/dx?

If you have gotten some answers, please post what you have done so we can check them and give you some guidance where you have made your mistakes...