# Partial Derivitive Notation - SIMPLE

1. Jan 5, 2010

### Chewy0087

1. The problem statement, all variables and given/known data
Find;

$$\frac{\partial u}{\partial x} , \frac{\partial u}{\partial y} , \frac{\partial u}{\partial z}$$ for the function

$$u^2 + x^2 + y^2 = a^2$$ given that $$x^2 + y^2 \leq a^2$$ where a is a constant

3. The attempt at a solution

i'm having alot of trouble, and I think it stems from my confusion with the notation;

The textbook i'm using says that the partial derivatives of u = f(x,y,z) are (verbatim);

$$\frac{\partial u}{\partial x} = f_{1}(x,y,z) = \frac{\partial f}{\partial x} =\frac{\partial (f(x,y,z)}{\partial x}$$
$$\frac{\partial u}{\partial y} = f_{2}(x,y,z)$$
$$\frac{\partial u}{\partial z} = f_{3}(x,y,z)$$

Now from this I can only assume that in turn

$$\frac{\partial u}{\partial y} = f_{2}(x,y,z) = \frac{\partial f}{\partial y} = \frac{\partial (f(x,y,z)}{\partial y}$$

and likewise for z, but what exactly does it mean by

$$f_{1}(x,y,z)$$ , is this another type of notation maybe with the 1 pointing to the first variable in (1,2,3) as in (x,y,z)...is that right?

Next, this is an american textbook, however i'm english and i understand the notation for derivitives is different, does;

$$\frac{\partial u}{\partial y} \left|_{x=x_{o}, y=y_{0},z=z_{0}}$$ simply mean the value of the partial derivitive inserting the points x0 y0 and z0?

I KNOW these are very simple questions and i'm probably right but i would really like assurance to try and build confidence a bit...plus that LaTeX took AGES lol.

Finally could someone check my answer;

$$\frac{\partial u}{\partial x} = \frac{-x}{u} , \frac{\partial u}{\partial y} = \frac{-y}{u} , \frac{\partial u}{\partial z} = \frac{-z}{u}$$

sorry for the long post and thank you.

2. Jan 5, 2010

### jambaugh

Remember that the differential is defined in terms of the partial derivatives and you can thence solve for the partial from the form of the differential:
$$d\,u(x,y,z) = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy + \frac{\partial u}{\partial z} dz$$

You should be able to take the differential of the defining equation, solve for du i.t.o. dx, dy, and dz with coefficients which are functions themselves of x,y,z and also of u.
(Note that without z dependence you must add a 0dz term which of course means the partial w.r.t. z is 0.

A longer equivalent method if you are unsure about differentials is to take the partial derivative with respect to each independent variable of your defining equation and solve for the partial derivative of u. E.g. solve:
$$\frac{\partial}{\partial x} (u^2+x^2+y^2) = \frac{\partial}{\partial x} a^2 (=0)$$
for $\partial u / \partial x$.

3. Jan 5, 2010

### vela

Staff Emeritus
The subscript is there because the three partial derivatives are generally different functions of x, y, and z. You can't call them all by the same name, so the book calls them f1, f2, and f3.

Yes.

Looks good except for the last one since u has no z dependence.

4. Jan 5, 2010

### Chewy0087

I'm not entireley sure about your use of notation now lol =P; you say

$$d\,u(x,y,z) = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy + \frac{\partial u}{\partial z} dz$$ i don't understand your use of the dx, dy dz here, what do they represent? surely by having the partial differential there you're indicating that it's with respect to x, or is the partial differential just a symbol to represent it and the dx represents the actual change...? but that wouldn't make sense either

5. Jan 5, 2010

### Chewy0087

Thank you, and yeah >.< i've no idea why I did the z.

6. Jan 5, 2010

### vela

Staff Emeritus
Remember when you used a substitution to do an integral? For example, you might have changed variables by saying

$$u=sin x$$

from which you got

$$du=cos x dx$$.

What you did there is just

$$du = \frac{du}{dx} dx$$

right? What Chewy wrote is just the multivariate generalization of that.

Intuitively, what it says is that for an infinitesimal change from $$(x,y,z)$$ to $$(x+dx,y+dy,z+dz)$$, u changes by the amount

$$du(x,y,z) = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy + \frac{\partial u}{\partial z} dz$$.

The partial derivative $$\partial u/\partial x$$ is the rate of change of u with respect to x, so if you increase x by $$\Delta x$$ and hold the other variables constant, u will change by approximately $$(\partial u}/{\partial x})\Delta x$$. The approximation becomes better as $$\Delta x$$ goes to zero. You do the same thing for y and z, and if you vary all three variables, to first order, the change in u is just the sum of the individual differences. Higher-order terms don't matter because they'll be much, much smaller.

7. Jan 5, 2010

### Chewy0087

Thank you, that's a great explanation!

8. Jan 5, 2010

### yungman

I thought if $$u^2 + x^2 + y^2 = a^2$$

$$\Rightarrow u=\sqrt{a^2 -x^2-y^2}$$

$$\frac{du}{dx}=\frac{-2x}{\sqrt{a^2 -x^2-y^2}}$$

$$\frac{du}{dy}=\frac{-2y}{\sqrt{a^2 -x^2-y^2}}$$

$$\frac{du}{dz}=0$$

9. Jan 5, 2010

### Chewy0087

you mean

$$\frac{du}{dx}=\frac{-x}{\sqrt{a^2 -x^2-y^2}}$$ , right?

Yeah i think this is a really easy question but i've only started partial differentiation =P

10. Jan 5, 2010

### yungman

A Chinese 2 is equal to a 1!!!!!:shy::tongue2:

You are right, I forgot the 1/2 from the differentiation of u that cancel the differentiation of x^2!!!!

11. Jan 6, 2010

### jambaugh

Firstly as I said, if you are not up to speed on differentials (dx dy et al) you can just repeatedly take partial derivatives of the whole equation w.r.t. each independent variable and solve for the partials you want.

Differentials and derivatives are related but distinct types of mathematical objects. A derivative is a function while a differential is a variable.
Recall the 'dx' appearing in the integrand of an integral $$\int f(x)dx$$
That is a differential.

You can think of this dx as an infinitesmial change in x and likewise with the others. However that's a bit old school. In a more rigorous setting, if you look at a curve in some variable space (u,v,w,x,y,z) defined by some equation or set of equations,
then you define the hyper-plane tangent to the curve at some point (u,v,w,x,y,z) and then parametrize any point on this tangent hyper-plane via (u+du,v+dv,w+dw,x+dx,y+dy,z+dz).

The differentials are like local coordinates for the tangent hyper-plane with origin at the point of tangency. The virtue of differentials is that there is no need to specify which variables are dependent vs independent. It is working with variable calculus rather than functional calculus (where we use derivatives).

Generally the advantage of differentials is that we don't have to require relations be functions. Given the circle:
$$x^2 + y^2 = 1$$
we have a differential relationship everywhere of:
$$2x\,dx + 2y\,dy = 0$$
then where you can define y as a function of x you can define a derivative dy/dx or likewise where you can define x as a function of y you have the derivative dx/dy. But you don't have to specify who is a function of what to work with dy and dx as (differential) variables in the equation above. It is the better way to deal with implicit forms.

Anyway in the notation I used in the prior post I promulgated a bad use of notation for brevity's sake. The equation:
$$d\,u(x,y,z) = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy + \frac{\partial u}{\partial z} dz$$

should rather have been written with variable and function names distinguished as below.

Given: $$u=f(x,y,z)$$
$$du = \frac{\partial}{\partial x}f(x,y,z) \cdot dx + \frac{\partial}{\partial y}f(x,y,z) \cdot dy + \frac{\partial}{\partial z}f(x,y,z) \cdot dz$$

One should always distinguish the variables from the functions when the reader is not used to the bad habit of giving both the same name.

In summary you should think of differentials of variables as new variables related by the derivatives of the original variables functional derivatives. If you want to skip this for now then as I mentioned just work with partial derivatives of the equations. However when you have time do some studying up on them they are and essential component of higher calculus.

12. Jan 11, 2010

### Chewy0087

much obliged sir, you've cleared many things up for me, but I still think i'm struggling with many of the issues, having only just met mathematical analysis in any sort of formal sense, hopefully I get the hang of it soon.

13. Jan 11, 2010

### jambaugh

No problem. You may think you are making little progress later but looking back you'll realize how quickly you've picked up the material. There is so much in the way of tricks techniques and notations to learn that one is always struggling with the material at hand and can forget how far they progress in a short time. Don't be discouraged and pay attention to the notation it has an internal consistency which makes it much easier to learn the concepts and catch errors in execution provided you learn to carefully follow the notation consistently and rigorously.

If you get stuck on anything else, don't hesitate to PM me.