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Thread: 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 $\displaystyle \frac{\partial f(x,y)}{\partial x}$ you just differentiate in respect to $\displaystyle x$ holding $\displaystyle y$ constant...same concept for $\displaystyle \frac{\partial f(x,y)}{\partial y}$. 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 $\displaystyle Df,Dg$? My book uses the $\displaystyle Df$ 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 $\displaystyle \phi$ approaches zero from all paths there is no need to consider it

    so

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

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

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

    So we can conclude that

    $\displaystyle \lim_{(x,y)\to(0,0)}f(x,y)=f(0,0)$ thus continuous.
    Last edited by Mathstud28; Jan 14th 2009 at 09:08 PM. Reason: Typo
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  3. #3
    Moo
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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, $\displaystyle \frac{\partial w}{\partial x} = \frac{df}{du}\frac{\partial u}{\partial x}$ 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": $\displaystyle Df= \frac{\partial f}{\partial y}dx+ \frac{\partial f}{\partial y}dy$= $\displaystyle 2xdx- 6ydy$

    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 "$\displaystyle \theta$".
    $\displaystyle f(x,y)= \frac{(r cos(\theta))(r sin(\theta))}{r}= r cos(\theta)sin(\theta)$
    If the limit of that, as r goes to 0, does not depend on $\displaystyle \theta$ then the limit of the function at (0,0) exists and is that value.
    Last edited by mr fantastic; Jan 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 $\displaystyle \frac{\partial f(x,y)}{\partial x}$ you just differentiate in respect to $\displaystyle x$ holding $\displaystyle y$ constant...same concept for $\displaystyle \frac{\partial f(x,y)}{\partial y}$. 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, $\displaystyle \frac{\partial w}{\partial x} = \frac{df}{du}\frac{\partial u}{\partial x}$ Since you are not given f you really just need to find the partial derivatives of u.


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


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