I Understanding the Difference Between Partial and Full Derivatives

[member 428835]

1) $f(a,b) = a^2 +3b$.
2)$y(w)$: some function of a single real variable $w$
3)$s(p,q) = f(p,y(q))$, whose definition involves taking a composition of the functions $f$ and $y$.

Ok, so how about this: $s(x,t) = f(x,y(t)) = x^2+3y$. Now $$\frac{\partial s}{\partial t} = \frac{\partial f}{\partial y}\frac{d y}{d t} = 3\frac{d y}{d t}$$ However, if on the other hand $f(x,y(t)) = x^2+3y$ Then $$\frac{\partial f}{\partial t} = 0$$
Do you both agree with this or have I messed stuff up again

 

[member 428835]

Thanks for both your replies by the way!

 

[Stephen Tashi]

Ok, so how about this: $s(x,t) = f(x,y(t)) = x^2+3y$. Now $$\frac{\partial s}{\partial t} = \frac{\partial f}{\partial y}\frac{d y}{d t} = 3\frac{d y}{d t}$$

I agree.

However, if on the other hand $f(x,y(t)) = x^2+3y$
Then $$\frac{\partial f}{\partial t} = 0$$

I agree, if we are thinking of the function as $f(x,y) = x^2 + 3y$. This requires a remark about tradition. In applied math, the derivative of function with respect to a variable that does not represent something in the domain of the function is often assumed to be zero. By that tradition, your answer is correct. However, technically the definition of a derivative does not apply variables not in the domain of a function. If $f$ is not a function of the thing denoted by $t$ then we can't evaluate limits like $lim_{h \rightarrow 0} \frac{ f(...,t + h,...) - f(...,t,...)}{h}$ So, technically, $\frac{\partial f}{\partial t}$ is undefined rather that being a defined function with constant value zero. If we define $f$ to be the function $f(x,y,t) = x + 3y$ then $\frac{\partial f}{\partial t} = 0$ without appealing to tradition.

The ambiguous notation used in applied math is due to the fact that symbols are often used to represent phenomena, not specific functions. For example, if we think of "$h$" as a persons height, then this is a clear idea, but it is ambiguous as an attempt to define a function. We might consider height as the function $h(t)$ of height vs time, or we might consider height as $h(a,b) = a + b$ where $a$ is the distance from a persons belly button to the top of his head and $b$ is the distance from his belly button to the bottom of his feet.

Notions of physics like "velocity" , "distance", "kinetic energy" do not define unique mathematical functions. A discussion of a physics problem may use a single symbol to represent different functions because the focus is associate a phenomena (such as "velocity") with the symbol, not to associate a specific mathematical function with it.

 

[PeroK]

Ok, so how about this: $s(x,t) = f(x,y(t)) = x^2+3y$. Now $$\frac{\partial s}{\partial t} = \frac{\partial f}{\partial y}\frac{d y}{d t} = 3\frac{d y}{d t}$$ However, if on the other hand $f(x,y(t)) = x^2+3y$ Then $$\frac{\partial f}{\partial t} = 0$$
Do you both agree with this or have I messed stuff up again

I would write both of those differently:

$s(x,t) = f(x,y(t)) = x^2+3y(t)$

There's no reason to omit that you are taking $y$ to be a function of $t$ on the RHS. Again, your decision to omit the $t$ can only lead to possible confusion about what is a function of what.

$f(x,y, t) = x^2+3y$ then $\frac{\partial f}{\partial t} = 0$

Here $y$ is not a function of $t$, which is an independent variable.

Let me give a physical example of this idea. Suppose a particle of unit mass has a potential energy function $V(x, t)$. This function depends on position and time and is therefore a function of two variables. Example $V(x, t) = x^2 e^{-t}$, which gives us:

$\frac{\partial V}{\partial x} = 2xe^{-t}$ and $\frac{\partial V}{\partial t} = -x^2e^{-t}$

Now, imagine the particle takes a specific path $x(t)$, for example $x = \sin(t)$. You can now look at the particle's potential along its path, $V_p$ which is solely a function of $t$ given by:

$V_p(t) = V(x(t), t)) = V(\sin(t), t) = \sin^2(t)e^{-t}$

And, we have a normal derivative of $V_p$ given by the one dimensional chain rule:

$\frac{dV_p}{dt} = [2\sin(t) \cos(t) - \sin^2(t)]e^{-t}$

Now, we could verify the multi-dimensional chain rule in this case:

$\frac{dV_p}{dt} = \frac{\partial V}{\partial x} \frac{dx}{dt} + \frac{\partial V}{\partial t} = 2xe^{-t}\cos(t) -x^2e^{-t} = [2\sin(t) \cos(t) - \sin^2(t)]e^{-t}$

And that agrees with the normal derivative of $V_p$ so everything is correct.

My final point is that you should start to use examples like this to check whether you're doing a chain rule correctly.

 

[member 428835]

Cool, thanks for both your help! I really appreciate it!