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Math Help - Calculus of several variables

  1. #1
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    Post Calculus of several variables

    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)
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by okthenso View Post
    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.
    Last edited by Mathstud28; January 14th 2009 at 09:08 PM. Reason: Typo
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  3. #3
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    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.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    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.
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  5. #5
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    Quote Originally Posted by okthenso View Post
    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.
    Last edited by mr fantastic; January 15th 2009 at 04:36 AM. Reason: This is a recording: Fixed the latex .... This is a recording: ......
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    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...
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    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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