Differentiable Function second derivative.

[PsychonautQQ]

Homework Statement

Let f(x,y) be a differentiable function with x = rcosθ and y = rsinθ. find the df(x,y)^2/d^2θ (second derivative with respect the theta)

Homework Equations

The Attempt at a Solution

Don't exactly know what I'm doing here.. The notes from class give me this equation (btw f_xx will mean the second derivative with respect to x etc)

f_r = f_xcosθ + f_ysinθ
f_rr = f_xxcos^2θ + 2f_xycosθsinθ + f_yysin^2θ
f_xx + f_yy = f_rr + f_r / r + f_θθ / r^2

So I'm assuming I have to make use of these or something... Anyone want to help me get started here?

 

[brmath]

The formula you need to work with is df/dθ = (∂f/∂x)(dx/dθ) + (∂f/∂y)(dy/dθ). $\hspace{50px}$ (1)

So what are x(θ) and y(θ)? And what are dx/dθ and dy/dθ? You have enough information to answer those questions. What you don't have is any idea what f is. So you can't find the first and second derivatives of f explicitly. You will have to express them as terms involving ∂f/∂x and ∂f/∂y, or the second derivatives.

So get the first derivative in place, using (1). Now you want to differentiate with respect to θ again, again using (1). You should see terms like $f_{xx}, f_{yy} \text { and } f_{xy}$.

 

[PsychonautQQ]

so in (df/dx)(dx/dθ) is df/dx = 1? if not how do I find that?

 

[brmath]

Let's back up a step and do this in one dimension. And if you don't mind, I'm going to switch the $\theta$ to a t, since that's easier to type.

Step 1. Suppose f(x) = $x^2$ and x(t) = cos t. What is df/dt?

Step 2. Suppose I don't tell you what f(x) is, only that df/dx exists; and x(t) = cos t. What is df/dt? You are going to have to express the answer in terms of df/dx times some other stuff.

Get this much sorted out, and we'll go back to the 2 dim problem.

 

[PsychonautQQ]

How can I express it in terms of df/dx times some other stuff if I don't know what df/dx is??
Is the answer to the one dimensional problem just df/dx * dx/dt?

 

[brmath]

Whenever someone asks you to do that you just write df/dx and then show what it is multipllied by (or added to or divided by or whatever). Everyone understands that df/dx is the derivative of f with respect to x. We often do not know exactly what f(x) is because we are trying to show something for a class of functions.

For example, I might say "let f(x) be a differentiable function"; or let f(x) be twice differentiable on [a,b]. The reason would be that we're trying to prove a result about all differentiable functions; or about twice differentiable functions on [a,b]. So we don't know exactly what f is -- we just know some of its properties.

You are right that df/dt = (df/dx)(dx/dt). Now can you write out the answers to step 1 and step 2?

Please don't get impatient. This stuff does make sense, but it can take time to wrap your brain around it.

 

[Ray Vickson]

How can I express it in terms of df/dx times some other stuff if I don't know what df/dx is??
Is the answer to the one dimensional problem just df/dx * dx/dt?

I strongly urge you to avoid the df/dx notation sometimes: use f'(x) instead of df(x)/dx. The advantage of doing this is that you no longer need to be confused by not knowing what f(x) is; you just call its derivative f'(x) and just treat that as another function of x.

In fact, the very best way to do such problem is to "mix up" the notation, switching from one to the other when needed or convenient. The chain rule is most easily remembered in the form
\frac{df}{dt} = \frac{df}{dx} \cdot \frac{dx}{dt},
but as soon as you write that you should switch to
\frac{df(x(t))}{dt} = f'(x(t)) \cdot x'(t).
This last form is harder to remember but easier to use---guaranteed to cause less confusion. Remember also that $f'(x(t)) = f'(y)|_{y = x(t)}$.

 

[brmath]

I strongly urge you to avoid the df/dx notation sometimes: use f'(x) instead of df(x)/dx. The advantage of doing this is that you no longer need to be confused by not knowing what f(x) is; you just call its derivative f'(x) and just treat that as another function of x.

Ray, I agree that different notation works best in different situations, and do that myself as convenient. I have found that the df/dx notation is very helpful for people who are just starting out. The reason is that the chain rule involves several different variables (expecially in several dimensions), and it's important to keep track of just what is being differentiated with respect to what.

df/dx and dx/dt are very explicit. To you f'(x(t)) is very clear, but I've had many students who simply couldn't grasp it.

This is just a stylistic difference. Probably whatever way you teach things works well with the way you notate them.