# Parametric equations

1. Apr 1, 2005

### ILoveBaseball

find $$\frac{d^2y}{dx^2}$$ as a function of $$t$$, for the given the parametric equations:

x = 2-4*cos(t)
y= 4+cos(t)^2

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

dy/dt = $$-2*cos(t)*sin(t)$$
second derv. $$2*sin(t)^2-2*cos(t)^2$$

dx/dt = $$4*sin(t)$$
second derv. $$4*cos(t)$$

$$\frac{d^2y}{dx^2}$$ = $$\frac{2*sin(t)^2-2*cos(t)^2}{4*cos(t)}$$

what did i do wrong?

2. Apr 1, 2005

### Oxymoron

$$\frac{d^2y}{dx^2} \neq \frac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}}$$

Here is how I would do it.

$$x = 2 - 4\cos(t)$$
$$4\cos(t) = 2-x$$
$$t = \arccos\left(\frac{2-x}{4}\right)$$

Now we have $$t$$ explicitly.

Substituite this back in for $$t$$ and we have

$$y = 4 + \cos\left(\arccos\left(\frac{2-x}{4}\right)\right)^2$$
$$\sqrt{y} = 2 + \cos\left(\arccos\left(\frac{2-x}{4}\right)\right)$$
$$\sqrt{y} = 2 + \left(\frac{2-x}{4}\right)$$
$$\sqrt{y} = \frac{3}{2} - \frac{x}{4}$$
$$y = \frac{9}{4} - \frac{x^2}{16}$$

Now differentiate.

$$\frac{dy}{dx} = -\frac{x}{8}$$
$$\frac{d^2y}{dx^2} = -\frac{1}{8}$$

3. Apr 1, 2005

### whozum

Also in general, for y(t) and x(t),

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

$$\frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2}$$

Simplifying the RHS of the second equation you get:

$$\frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2} = \frac{d^2y}{dt^2} \frac{dt^2}{dx^2}$$ and by cancelling you get the desired result $$\frac{d^2y}{dx^2}$$

4. Apr 1, 2005

### ILoveBaseball

awesome thanks

5. Apr 2, 2005

### xanthym

whozum:
Highlighted equation above for (d2y/dx2) is NOT correct. For parametric equations {y=y(t) & x=x(t)}, the second derivative is given by:

$$:(2): \ \ \ \ \ \color{blue} \frac{d^2y}{dx^2} \ = \ \left( \frac{d}{dt} \left( \frac{dy}{dx} \right) \right) / \left( \frac{dx}{dt} \right)$$

where (dy/dx) is function of "t" obtained from the correct Equation #1 above.

~~

Last edited: Apr 2, 2005