Help understanding derivatives of time; chain rule.

[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?

 

[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.

 

[tomcue]

The concept of differentiation and the chain rule can be challenging to grasp at first, but with practice and understanding, it becomes easier. Let's break down the problem to help clarify the concept.

First, it is important to understand that the derivative is a measure of how much a quantity changes with respect to another quantity. In this case, we are looking at how the position of the particle, represented by "y", changes with respect to time, "t". This is why we use the notation \frac{dy}{dt} to represent the derivative.

Now, let's look at the chain rule. This rule is used when we have a function of a function, which is the case in your problem. The function we are looking at is y=a(1+cos\theta), but this function is also dependent on another variable, \theta. So, we can rewrite the function as y(a,\theta)=a(1+cos\theta). The chain rule tells us that when we take the derivative of this function with respect to time, we need to take into account the derivative of the inner function, in this case, \theta, with respect to time as well. This is why we have the term \frac{d\theta}{dt} in our final expression.

Now, to address your question about why \theta has a time dependence, it is important to understand the context of the problem. In this case, \theta represents the angle of rotation of the particle. As time passes, the particle will continue to rotate, and therefore, the angle \theta will change with respect to time. This is why \theta has a time dependence.

Finally, it is worth noting that the variable "a" does not have a time dependence because it is a constant in this problem. It does not change with respect to time, and therefore, it does not need to be included in the derivative.

I hope this explanation has helped clarify the concept of differentiation and the chain rule for you. Remember, practice makes perfect, and with more exposure and understanding, you will become more comfortable with these concepts.