# Help understanding derivatives of time; chain rule.

1. Sep 15, 2009

### jimz

I'm having a bit of a hiccup understanding the differentiation that I am doing... I'd like to be clear on the concept rather than just knowing 'apply chain rule'.

So I have a particle with equation:
$$y=a(1+cos\theta)$$

now the derivative with respect to time (the velocity in y) is

$$\frac{dy}{dt}=\dot{y}=\frac{dy}{d\theta} \frac{d\theta}{dt}=a(-sin\theta)\dot{\theta}=a(-\dot{\theta}sin\theta)$$

What I am having the most trouble coming to terms with is the time derivative of theta. How do I even know that theta has a time dependence? Why is theta even a variable when apparently a is not?

2. Sep 15, 2009

### rl.bhat

y = a(1 + cosθ) is a function, in which y changes according to θ.
When θ = 0, y = 2a. When θ = π/2, y = a and so on. It represents a circle with radius a, i.e. the particle is moving in a circular orbit. dθ/dt represents the angular velocity of the particle.