# Parametric equation

1. Jun 20, 2004

### naav

Hi...i was just wondering if anyone gets the same answer to what i get for the following question...thanks...

find $$\frac{dy^2}{dx^2}$$ in terms of t for...

x = 2cost - cos2t, y = 2sint + sin2t...

i got my answer to be $$\frac{1 + cost}{2sin^3t(1 -2cost)}$$

the answer is given as $$\frac{-1}{sin^3t(2cost -1)}$$

have i gone wrong somewhere...do i need to simplify further...the answer i got and the one that is given do have some similarities so i'm just wondering...???...

2. Jun 21, 2004

### arildno

You have:
$$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\to\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
Also:
$$\frac{d^{2}y}{dt^{2}}=\frac{d^{2}y}{dx^{2}}(\frac{dx}{dt})^{2}+\frac{dy}{dx}\frac{d^{2}x}{dt^{2}}$$
Rearranging, we get:
$$\frac{d^{2}y}{dx^{2}}=\frac{\frac{d^{2}y}{dt^{2}}\frac{dx}{dt}-\frac{dy}{dt}\frac{d^{2}x}{dt^{2}}}{(\frac{dx}{dt})^{3}}$$

Last edited: Jun 21, 2004
3. Jun 21, 2004

### naav

Hi...thank you...

i found:

$$\frac{d^2y}{dx^2}$$

= (d/dt)(dy/dx) * (dt/dx)

did you get the same answer as me...???...

4. Jun 21, 2004

### arildno

Do not use that form; instead let:
y(t)=y(x(t))
Follow the derivation in post 3 to find the correct expression.
(The notation used here is rather sloppy, but it shouldn't be too difficult to follow)

5. Jun 21, 2004

### naav

Hi...thank you...

the method i mentioned is the way we've been shown in the textbook and in the notes...and that's the way i've tackled other questions...it's better for me to stick to the method shown in the classes...

i got dx/dt = -2sint + 2sin2t and dy/dt = 2cost + 2cos2t...

then i got dy/dx = (cost + 1)/sint

then (d^2t)/(dx^2) = (d/dt)(dy/dx) * dt/dx...

i got (d/dt)(dy/dx) = (-1 - cost)/(sin^2t)

and dt/dx = 1/(-2sint + 2cost)

then (d^2t)/(dx^2) = (1 + cost)/[(2sin^3(t)(1 - cost)]...

but the answer is given as something else...???...

6. Jun 21, 2004

### e(ho0n3

Personally, I would stick with the more efficient/clever method regardless of where I've learned it from, but anywho...

Correct.
Where are you getting this from? You might one to check this again.

7. Jun 21, 2004

### naav

Hi...thank you...

i got dx/dy = (dy/dt)/(dx/dt)...

according to the answer i've got that bit right...

8. Jun 22, 2004

### HallsofIvy

Staff Emeritus
No, dx/dy= (dx/dt)/(dy/dt)

9. Jun 22, 2004

### naav

Hi...sorry, i was meaning the first derivative - dy/dx...

dy/dx = (dy/dt)/(dx/dt)