Calculus of several variables

**okthenso** · January 14th 2009, 08:35 PM

2 questions that aren't familiar to me...

1.

w(x,y) = f(u), where u = xy/(x^2 + y^2)

evaluate and simplify x dw/dx + y dw/dy

2.

Let f:R2 ->R be defined by f(x,y) = x^2 - 3y^2 and G:R2 ->R2 be defined by G(s,t) = {st,s+(t^2)}. Calculate Df,Dg and eventually D(f o G)(1,2).

and one that I can't remember the method for (not as important)...
3.

f(x,y) = xy / sqrt(x^2 + y^2), if (x,y) does not = (0,0), f(0,0) = 0

Prove the f is continuous at (0,0)

**Mathstud28** · January 14th 2009, 08:54 PM

Originally Posted by okthenso

2 questions that aren't familiar to me...

1.

w(x,y) = f(u), where u = xy/(x^2 + y^2)

evaluate and simplify x dw/dx + y dw/dy

What are you having troubles with? Do you know how to differentiate partially? If it is $\frac{\partial f(x,y)}{\partial x}$ you just differentiate in respect to $x$ holding $y$ constant...same concept for $\frac{\partial f(x,y)}{\partial y}$ . Try it out and report back with problems?

2.

Let f:R2 ->R be defined by f(x,y) = x^2 - 3y^2 and G:R2 ->R2 be defined by G(s,t) = {st,s+(t^2)}. Calculate Df,Dg and eventually D(f o G)(1,2).

and one that I can't remember the method for (not as important)...

What is meant by $Df,Dg$ ? My book uses the $Df$ notation to denote the directional derivative...which doesnt appear to be the case since you did not supply a unit position vector...so is this the so called "total derivative"?

3.

f(x,y) = xy / sqrt(x^2 + y^2), if (x,y) does not = (0,0), f(0,0) = 0

Prove the f is continuous at (0,0)

Convert to polar coordinates. Note that when you convert since $\phi$ approaches zero from all paths there is no need to consider it

so

$\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}\stac krel{{\color{red}\star}}{\longmapsto}\lim_{r\to 0}\frac{\cos(\phi)r\cdot\sin(\phi) r}{\sqrt{r^2}}=0$

Which clearly is independent upon $\phi$
${\color{red}\star}:\left(\begin{array}{rcl} x\\ y \end{array}\right)\longmapsto\left (\begin{array}{rcl} r\cos(\phi)\\ r\sin(\phi)\end{array}\right)$
To illustrate this point further note that

$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leqslant\le ft|\frac{xy}{\sqrt{x^2}}\right|=\left|\frac{xy}{|x |}\right|$ which clearly tends to zero as $(x,y)\to(0,0)$

So we can conclude that

$\lim_{(x,y)\to(0,0)}f(x,y)=f(0,0)$ thus continuous.

**Moo** · January 15th 2009, 12:10 AM

What is meant by Df,Dg? My book uses the Df notation to denote the directional derivative...which doesnt appear to be the case since you did not supply a unit position vector...so is this the so called "total derivative"?

Differential (infinitesimal) - Wikipedia, the free encyclopedia

I think we can sometimes present Df as a matrix, like the gradient.

**Mathstud28** · January 15th 2009, 12:16 AM

Originally Posted by Moo

Differential (infinitesimal) - Wikipedia, the free encyclopedia

I think we can sometimes present Df as a matrix, like the gradient.

Thank you Moo, I was not aware of this notation.

**HallsofIvy** · January 15th 2009, 04:31 AM

Originally Posted by okthenso

2 questions that aren't familiar to me...

I was sorely tempted to remark on your counting ability!

1.

w(x,y) = f(u), where u = xy/(x^2 + y^2)

evaluate and simplify x dw/dx + y dw/dy

By the chain rule, $\frac{\partial w}{\partial x} = \frac{df}{du}\frac{\partial u}{\partial x}$ Since you are not given f you really just need to find the partial derivatives of u.

2.

Let f:R2 ->R be defined by f(x,y) = x^2 - 3y^2 and G:R2 ->R2 be defined by G(s,t) = {st,s+(t^2)}. Calculate Df,Dg and eventually D(f o G)(1,2).

and one that I can't remember the method for (not as important)...

You want the "total differential": $Df= \frac{\partial f}{\partial y}dx+ \frac{\partial f}{\partial y}dy$ = $2xdx- 6ydy$

3.

f(x,y) = xy / sqrt(x^2 + y^2), if (x,y) does not = (0,0), f(0,0) = 0

Prove the f is continuous at (0,0)

I recommend changing to polar coordinates. That way "closeness" to (0,0) is determined entirely by "r" and does not depend on " $\theta$ ".
$f(x,y)= \frac{(r cos(\theta))(r sin(\theta))}{r}= r cos(\theta)sin(\theta)$
If the limit of that, as r goes to 0, does not depend on $\theta$ then the limit of the function at (0,0) exists and is that value.

**okthenso** · January 17th 2009, 05:23 AM

Originally Posted by Mathstud28

What are you having troubles with? Do you know how to differentiate partially? If it is $\frac{\partial f(x,y)}{\partial x}$ you just differentiate in respect to $x$ holding $y$ constant...same concept for $\frac{\partial f(x,y)}{\partial y}$ . Try it out and report back with problems?

Hi.. my problem wasn't with that part. Take dw/dx for example. I split that into (dw/du)(du/dx), or as the other fella says (df/du)(du/dx).

And then obviously du/dx is just u differentiated with respect to x. But I don't know how to differentiate f (or w) with respect to u in this case...

**okthenso** · January 17th 2009, 05:24 AM

Originally Posted by HallsofIvy

I was sorely tempted to remark on your counting ability!

By the chain rule, $\frac{\partial w}{\partial x} = \frac{df}{du}\frac{\partial u}{\partial x}$ Since you are not given f you really just need to find the partial derivatives of u.

You want the "total differential": $Df= \frac{\partial f}{\partial y}dx+ \frac{\partial f}{\partial y}dy$ = $2xdx- 6ydy$

I recommend changing to polar coordinates. That way "closeness" to (0,0) is determined entirely by "r" and does not depend on " $\theta$ ".
$f(x,y)= \frac{(r cos(\theta))(r sin(\theta))}{r}= r cos(\theta)sin(\theta)$
If the limit of that, as r goes to 0, does not depend on $\theta$ then the limit of the function at (0,0) exists and is that value.

Haha! I meant 2 that weren't familiar and then a third that I didn't know for a different reason. Thanks for the help.

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