# How do you know what variable to replace for a time derivative?

I am a avid reader of Physics articles, and enjoy the mathematics greatly. I have taught myself most of the needed mathematics however one part of it always messes me up. When you take a time derivative, I have read that you have to replace a variable that is dependent to time, with time.

If this is true, then it must apply to other variables as well. Which would make it very difficult to take the time derivative of a equation with 5 different variables dependent to time, so how would you know which variable to replace?

The Article Says:

consider a particle moving in a circular path. Its position is given by the displacement vector $$r=x\hat{i}+y\hat{j}$$, related to the angle, $$θ$$, and radial distance, r, as defined in the figure:

$$x = r \cos\theta$$ $$y = r \sin\theta$$

For purposes of this example, time dependence is introduced by setting θ = t. The displacement (position) at any time t is then

$$f{r}(t) = r\cos(t)\hat{i}+r\sin(t)\hat{j}$$ This form shows the motion described by $$r(t)$$ is in a circle of radius r because the magnitude of r(t) is given by

$$|f{r}(t)| = \sqrt{f{r}(t) \cdot f{r}(t)}=\sqrt {x(t)^2 + y(t)^2 } = r\, \sqrt{\cos^2(t) + \sin^2(t)} = r$$

using the trigonometric identity $$\sin^2(t) + \cos^2(t) = 1$$

If I need to take a time derivative of a function that has multiple variables dependent to time, but the function itself has not time variable, how would I Differentiate it with to respect to time?

• There doesn't seem to be anything in the Wikipedia article you link to that speaks about "you have to replace a variable that is dependant to time, with time". The word "replace" doesn't even occur in the article. Where did you find that strange-sounding claim? – Henning Makholm Jan 4 '16 at 14:43
• "For purposes of this example, time dependence is introduced by setting θ = t. The displacement (position) at any time t is then" – fftk4323 Jan 4 '16 at 15:47

## 2 Answers

That does NOT say anything about "replace a variable that is dependent on time, with time". It says that they are "introducing" time by replacing a variable by time. If you have function of a number of variables that are themselves functions of time, the derivative of that function, with respect to time, is found using the chain rule. If f is a function of x, y, and z, with x, y, and z functions of t then $\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}+ \frac{\partial f}{\partial z}\frac{dz}{dt}$.

If the location $(x,y)$ is defined by a single parameter $\theta$ then

$$\frac{{\rm d}}{{\rm d}t} (x,y) = (\dot{x},\dot{y})= \left( \dot{\theta} \frac{{\rm d}x}{{\rm d} \theta} , \dot{\theta} \frac{{\rm d}y}{{\rm d} \theta} \right)$$

where $\dot{\theta}$ is the time derivative of the parameter. For example with $(x,y) = (r \cos \theta,r \sin \theta)$ you have

$$(\dot{x},\dot{y}) = \left(\dot{\theta} \frac{{\rm d}(r \cos \theta)}{{\rm d}\theta},\dot{\theta} \frac{{\rm d}(r \sin \theta)}{{\rm d}\theta}\right) = \left(-r \dot{\theta} \sin\theta, r \dot{\theta} \cos\theta \right)$$

The above is often denoted with the shorthand that when a variable $x$ depends on a parameter $q$ then $\boxed{\dot{{\bf x}} = {\bf x}'\dot{q}}$ where ${\bf x}'$ denoted derivative in terms of the parameter $q$ and $\dot{q}$ the derivative of the parameter in terms of time.

Here is where it gets exciting. The velocities above are a function of both $\theta$ and $\dot{\theta}$ so to calculate the accelerations the following chain rule is used:

$$(\dot{x},\dot{y}) = (v,u)$$ $$(\ddot{x},\ddot{y}) = \left( \dot{\theta} \frac{{\rm d}v}{{\rm d}\theta} + \ddot{\theta} \frac{{\rm d}v}{{\rm d}\dot{\theta}}, \dot{\theta} \frac{{\rm d}u}{{\rm d}\theta} + \ddot{\theta} \frac{{\rm d}u}{{\rm d}\dot{\theta}}\right)$$

For example when that $(v,u) = (-r \dot{\theta} \sin\theta, r \dot{\theta} \cos\theta)$ the acceleration is

\begin{aligned} \frac{{\rm d}v}{{\rm d}\theta} & = -r \dot{\theta}\cos\theta & \frac{{\rm d}u}{{\rm d}\theta} & = -r \dot{\theta}\sin\theta \\ \frac{{\rm d}v}{{\rm d}\dot{\theta}} & = -r \sin \theta & \frac{{\rm d}u}{{\rm d}\dot{\theta}} & = r \cos \theta \end{aligned}

$$(\ddot{x},\ddot{y}) = \left( -r \dot{\theta}^2 \cos\theta - r \ddot{\theta} \sin\theta, -r \dot{\theta}^2 \sin\theta + r \ddot{\theta} \cos\theta \right)$$

or with the shorthand $\boxed{ \ddot{{\bf x}} = {\bf x}' \ddot{q} +{\bf x}'' \dot{q}^2}$