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Thread: Implicit vs. partial

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Implicit vs. partial

    Ok, I know the answer is probably no, and I feel stupid asking it, but say we have an equation (I can't say function for reasons that will become obvious), lets say

    $\displaystyle f(x,y)=g(x,y)$

    Now lets say I differentiate it implicitly and solve for y'. that is the derivative at a point.

    Now bear with me if this is really bad but could we do this

    Let $\displaystyle z=f(x,y)-g(x,y)$

    Then will differentiating implicitly give the same answer as...

    $\displaystyle \frac{\frac{\partial{z}}{\partial{x}}}{\frac{\part ial{z}}{\partial{y}}}$?

    Maybe I am just really lucky, but after pondering this and thinking about it analytically I tried a couple of problems and it seems to pan out, but I do not think it is true, otherwise my book would have it or I would have heard of it.


    Can someone please clarify?

    Mathstud.

    EDIT:

    The more I think about it, the more I don't think so.

    The reason is the difference in the chain rule

    $\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}$

    But I am pretty sure

    $\displaystyle \frac{\frac{\partial{z}}{\partial{x}}}{\frac{\part ial{z}}{\partial{y}}}\ne\frac{\partial{y}}{\partia l{x}}$

    Is that right?

    EDIT EDIT:

    I have done some more thinking and a little experimentation and I think that I was right

    but I think that it is actually

    $\displaystyle y'=\frac{-\frac{\partial{z}}{\partial{x}}}{\frac{\partial{z} }{\partial{y}}}$
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  2. #2
    Super Member Aryth's Avatar
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    According to the Implicit Function Theorem:

    $\displaystyle \frac{dy}{dx} = -\frac{\frac{\partial{F}}{\partial{x}}}{\frac{\part ial{F}}{\partial{y}}} = -\frac{F_x}{F_y}$

    So, technically:

    $\displaystyle -\frac{\frac{\partial{F}}{\partial{x}}}{\frac{\part ial{F}}{\partial{y}}} = -\frac{F_x}{F_y}$

    And therefore:

    $\displaystyle \frac{\frac{\partial{F}}{\partial{x}}}{\frac{\part ial{F}}{\partial{y}}} = \frac{F_x}{F_y}$

    Finally:

    $\displaystyle \frac{\frac{\partial{F}}{\partial{y}}}{\frac{\part ial{F}}{\partial{x}}} = \frac{F_y}{F_x}$

    So you were right in saying that:

    $\displaystyle \frac{\frac{\partial{z}}{\partial{x}}}{\frac{\part ial{z}}{\partial{y}}}\ne\frac{\partial{y}}{\partia l{x}}$

    As a matter of fact... The best way to do this would have to be to use the Chain Rule.
    Last edited by Aryth; Jun 29th 2008 at 12:31 AM.
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  3. #3
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    I am not exactly sure what you are asking but as Aryth said it seems to use the implicit function theorem. In its most basic form it says that if $\displaystyle F:\mathbb{R}^2\mapsto \mathbb{R}$ (this notation it means that $\displaystyle F$ is a function of two variables) is a $\displaystyle \mathcal{C}^1$ function (ignore what this mean - it just means the function is behaved well-enough such as being differenciable). Say that $\displaystyle F(a,b) = 0$. The implicit function theorem says that if $\displaystyle \partial_y F(a,b) \not = 0$ then the equation $\displaystyle F(x,y) = 0$ can be solved uniquely (for $\displaystyle y$) in the "neighborhood" of $\displaystyle (a,b)$. The term "in the neighborhood" means in some small disk around $\displaystyle (a,b)$ we can solve this equation uniquely for $\displaystyle y$ (in terms of $\displaystyle x$).

    For example, consider the circle $\displaystyle x^2+y^2 - 1= 0$. Where can we solve for $\displaystyle y$ uniquely (in a neighborhood)? Define the function $\displaystyle F(x,y) = x^2+y^2 - 1$. The point $\displaystyle (1,0)$ lies on this circle because $\displaystyle F(1,0) = 1^2 + 0^2 - 1 = 0$. If we try to solve the equation within the small disk $\displaystyle \sqrt{(x-1)^2+y^2} < \epsilon $ we see that it has two solutions, one in the upper half and one in the lower half (see red circle) - in fact those solutions are $\displaystyle y = \pm \sqrt{ 1-x^2}$. And this happens no matter how small we make the circle. Therefore, the equation $\displaystyle F(x,y) = 0$ cannot be solved uniquely. Let us see what happens when we use the implicit function theorem. The theorem says that if $\displaystyle \partial_y F(1,0) \not = 0$ then we can solve the equation uniquely. Since we cannot solve this equation uniquely it must mean that $\displaystyle \partial_y F(1,0) = 0$. Check this: $\displaystyle \partial_y F(x,y) = 2y$ and so $\displaystyle \partial_y F(1,0) = 0$. However, if $\displaystyle F(a,b) = 0$ and $\displaystyle b\not = 0$ i.e. we stay away from the break in the graph where it lies above and below the x-axis then $\displaystyle \partial_y F(a,b) = 2b \not = 0$ and therefore we can solve this equation uniquely. As in the case of the blue circle, the solution is $\displaystyle y=\sqrt{1-x^2}$.

    Let us say that $\displaystyle F(x,y)$ is function which satisfies the condition of the implicit function theorem as in the first paragraph. Then in the neighborhood of $\displaystyle (a,b)$ (where $\displaystyle (a,b)$ is a solution to $\displaystyle F(x,y)=0$) we can solve for $\displaystyle y$ uniquely in terms of $\displaystyle x$. This means we can define a new function $\displaystyle y = g(x)$ in this neighborhood. Remember a function is a set of pairs so that for each first coordinate there is a unique matching coordinate. Therefore, if we let $\displaystyle x$ be the first coordinate then by the theorem there is a unique matching coordinate so that $\displaystyle F(x,y) = 0$. And this would define a function $\displaystyle y=g(x)$.

    But the implicit function theorem does not stop here, it says more it says that $\displaystyle g(x)$ is itself $\displaystyle \mathcal{C}^1$ (which is differenciable with some more properties). Now since $\displaystyle F(x,y) = 0 \implies F(x,g(x)) = 0$. The functions $\displaystyle F$ and $\displaystyle g$ are differenciable and so by the chain rule for multivariable functions we get that (differenciating both sides of $\displaystyle F(x,g(x)) = 0$) $\displaystyle \partial_x F + \partial_y F \cdot g'(x) = 0$ and this means $\displaystyle g'(x) = - \frac{\partial_x F}{\partial_y F}$ (the partials are evaluated at $\displaystyle (x,y)$). And that gives you a formula.

    If you got that then great. There is just one more point to be addressed. If it confuses you just ignore that. The questionable step is dividing by $\displaystyle \partial_y F$. How do we know it is non-zero? This is where we use two facts. The first one is that $\displaystyle \partial_y F (a,b) \not = 0$. The second one is that $\displaystyle F(x,y)$ is $\displaystyle \mathcal{C}^1$. The first fact is not good enough to say that $\displaystyle \partial_y F\not =0$ for all points around $\displaystyle (a,b)$ because it might be non-zero at $\displaystyle (a,b)$ and yet be zero at some points around $\displaystyle (a,b)$. This is where we need the second fact. The meaning of $\displaystyle \mathcal{C}^1$ is that the function is differenciable and the derivative is continous. Therefore $\displaystyle \partial_y F$ is a continous function. Since $\displaystyle \partial_y (a,b)\not = 0$ it means there is a small enough neighborhood so that $\displaystyle \partial_y (x,y)\not = 0$ for points close to $\displaystyle (x,y)$ by the definition of continuity.
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  4. #4
    Eater of Worlds
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    I think I know what you're saying, and yes, you can differentiate implicitly that way. I have used it before. As a matter of fact, I wrote a program for implicit diff using that method.

    An easy-schmeasy example to be sure.

    Say you want to implicitly differentiate $\displaystyle x^{3}+y^{3}=1$

    The derivative wrt x is $\displaystyle \frac{\partial}{\partial{x}}=3x^{2}$

    wrt y it is $\displaystyle -\frac{\partial}{\partial{x}} = -3y^{2}$

    So, by dividing, we get $\displaystyle \frac{-x^{2}}{y^{2}}$

    Is that what you mean?. I think it is?. It makes sense to do it that way I always thought.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I am not exactly sure what you are asking but as Aryth said it seems to use the implicit function theorem. In its most basic form it says that if $\displaystyle F:\mathbb{R}^2\mapsto \mathbb{R}$ (this notation it means that $\displaystyle F$ is a function of two variables) is a $\displaystyle \mathcal{C}^1$ function (ignore what this mean - it just means the function is behaved well-enough such as being differenciable). Say that $\displaystyle F(a,b) = 0$. The implicit function theorem says that if $\displaystyle \partial_y F(a,b) \not = 0$ then the equation $\displaystyle F(x,y) = 0$ can be solved uniquely (for $\displaystyle y$) in the "neighborhood" of $\displaystyle (a,b)$. The term "in the neighborhood" means in some small disk around $\displaystyle (a,b)$ we can solve this equation uniquely for $\displaystyle y$ (in terms of $\displaystyle x$).

    For example, consider the circle $\displaystyle x^2+y^2 - 1= 0$. Where can we solve for $\displaystyle y$ uniquely (in a neighborhood)? Define the function $\displaystyle F(x,y) = x^2+y^2 - 1$. The point $\displaystyle (1,0)$ lies on this circle because $\displaystyle F(1,0) = 1^2 + 0^2 - 1 = 0$. If we try to solve the equation within the small disk $\displaystyle \sqrt{(x-1)^2+y^2} < \epsilon $ we see that it has two solutions, one in the upper half and one in the lower half (see red circle) - in fact those solutions are $\displaystyle y = \pm \sqrt{ 1-x^2}$. And this happens no matter how small we make the circle. Therefore, the equation $\displaystyle F(x,y) = 0$ cannot be solved uniquely. Let us see what happens when we use the implicit function theorem. The theorem says that if $\displaystyle \partial_y F(1,0) \not = 0$ then we can solve the equation uniquely. Since we cannot solve this equation uniquely it must mean that $\displaystyle \partial_y F(1,0) = 0$. Check this: $\displaystyle \partial_y F(x,y) = 2y$ and so $\displaystyle \partial_y F(1,0) = 0$. However, if $\displaystyle F(a,b) = 0$ and $\displaystyle b\not = 0$ i.e. we stay away from the break in the graph where it lies above and below the x-axis then $\displaystyle \partial_y F(a,b) = 2b \not = 0$ and therefore we can solve this equation uniquely. As in the case of the blue circle, the solution is $\displaystyle y=\sqrt{1-x^2}$.

    Let us say that $\displaystyle F(x,y)$ is function which satisfies the condition of the implicit function theorem as in the first paragraph. Then in the neighborhood of $\displaystyle (a,b)$ (where $\displaystyle (a,b)$ is a solution to $\displaystyle F(x,y)=0$) we can solve for $\displaystyle y$ uniquely in terms of $\displaystyle x$. This means we can define a new function $\displaystyle y = g(x)$ in this neighborhood. Remember a function is a set of pairs so that for each first coordinate there is a unique matching coordinate. Therefore, if we let $\displaystyle x$ be the first coordinate then by the theorem there is a unique matching coordinate so that $\displaystyle F(x,y) = 0$. And this would define a function $\displaystyle y=g(x)$.

    But the implicit function theorem does not stop here, it says more it says that $\displaystyle g(x)$ is itself $\displaystyle \mathcal{C}^1$ (which is differenciable with some more properties). Now since $\displaystyle F(x,y) = 0 \implies F(x,g(x)) = 0$. The functions $\displaystyle F$ and $\displaystyle g$ are differenciable and so by the chain rule for multivariable functions we get that (differenciating both sides of $\displaystyle F(x,g(x)) = 0$) $\displaystyle \partial_x F + \partial_y F \cdot g'(x) = 0$ and this means $\displaystyle g'(x) = - \frac{\partial_x F}{\partial_y F}$ (the partials are evaluated at $\displaystyle (x,y)$). And that gives you a formula.

    If you got that then great. There is just one more point to be addressed. If it confuses you just ignore that. The questionable step is dividing by $\displaystyle \partial_y F$. How do we know it is non-zero? This is where we use two facts. The first one is that $\displaystyle \partial_y F (a,b) \not = 0$. The second one is that $\displaystyle F(x,y)$ is $\displaystyle \mathcal{C}^1$. The first fact is not good enough to say that $\displaystyle \partial_y F\not =0$ for all points around $\displaystyle (a,b)$ because it might be non-zero at $\displaystyle (a,b)$ and yet be zero at some points around $\displaystyle (a,b)$. This is where we need the second fact. The meaning of $\displaystyle \mathcal{C}^1$ is that the function is differenciable and the derivative is continous. Therefore $\displaystyle \partial_y F$ is a continous function. Since $\displaystyle \partial_y (a,b)\not = 0$ it means there is a small enough neighborhood so that $\displaystyle \partial_y (x,y)\not = 0$ for points close to $\displaystyle (x,y)$ by the definition of continuity.
    That made perfect sense actually! Thank you very much.

    So for the last paragraph since

    $\displaystyle F(x,y)$ being $\displaystyle C^{1}$ (a smooth function) this implies that $\displaystyle F_y$ is continous which implies that

    $\displaystyle \lim_{(x,y)\to(a,b)}F_y=F_y(a,b)$ from all paths, so this implies that there exists a region containing $\displaystyle (a,b)$ such that $\displaystyle F_y\ne{0}$.

    We know this because we have that

    $\displaystyle \lim_{(x,y)\to(a,b)}F_y(a,b)$ represents the area immediately surrounding the point $\displaystyle )(a,b)$, or in other words the aforementioned region.

    And since as was said earlier

    $\displaystyle \lim_{(x,y)\to(a,b)}F_y(x,y)=F_y(a,b)$ (due to its continuity)

    and

    $\displaystyle F(x,y)\ne{0}$

    This implies that

    $\displaystyle \lim_{(x,y)\to(a,b)}F_y(x,y)\overbrace{\ne}^{\text {must}}0$


    Which gives us the guarantee that dividing by $\displaystyle F_y$ will not produce a division by zero.

    Is that right?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    I think I know what you're saying, and yes, you can differentiate implicitly that way. I have used it before. As a matter of fact, I wrote a program for implicit diff using that method.

    An easy-schmeasy example to be sure.

    Say you want to implicitly differentiate $\displaystyle x^{3}+y^{3}=1$

    The derivative wrt x is $\displaystyle \frac{\partial}{\partial{x}}=3x^{2}$

    wrt y it is $\displaystyle -\frac{\partial}{\partial{x}} = -3y^{2}$

    So, by dividing, we get $\displaystyle \frac{-x^{2}}{y^{2}}$

    Is that what you mean?. I think it is?. It makes sense to do it that way I always thought.
    I was asking both actually, the analytic reason behind and the actual application.

    Thanks Galactus!
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  7. #7
    Eater of Worlds
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    I think PH gave you the in depth of why it works.
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    Is that right?
    Here is what I was thinking when I wrote it. Since $\displaystyle \partial_y F$ is continous at $\displaystyle (a,b)$ it means $\displaystyle \lim_{(x,y)\to (a,b)}\partial_y F(x,y) = \partial_y F(a,b)$. But this means for any $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ such that if $\displaystyle \sqrt{(x-a)^2+(y-b)^2} <\delta \implies |\partial_y F(x,y) - \partial_y F(a,b) | < \epsilon$. Rewrite $\displaystyle |\partial_y F(x,y) - \partial_y F(a,b) | < \epsilon$ as $\displaystyle - \epsilon < \partial_y F(x,y) - \partial_y F(a,b) < \epsilon $ and so $\displaystyle \partial_y F(a,b) - \epsilon < \partial_y F(x,y) < \partial_y F(a,b) + \epsilon$ for all $\displaystyle \sqrt{(x-a)^2+(y-b)^2} < \delta$. Now $\displaystyle \partial_y F(a,b) \not = 0$. Thus, either $\displaystyle \partial_y F(a,b) > 0$ or $\displaystyle \partial_y F(a,b) < 0$. If $\displaystyle \partial_y F(a,b) > 0$ choose an $\displaystyle \epsilon$ small enough so that $\displaystyle \partial_y F(a,b) - \epsilon > 0$. And then $\displaystyle \partial_y F(x,y) > 0$ (because $\displaystyle \partial_y F(a,b) - \epsilon < \partial_y F(x,y)$ ) for $\displaystyle \sqrt{(x-a)^2+(y-b)^2}< \delta$ (and this is a neighborhood around $\displaystyle (a,b)$) thus $\displaystyle \partial_y F(x,y) \not = 0$. Now if $\displaystyle \partial_y F(a,b) < 0$ choose $\displaystyle \epsilon >0$ small enough so that $\displaystyle \partial_y F(a,b) + \epsilon < 0$. It will follow from $\displaystyle \partial_y F(x,y) < \partial_y F(a,b) + \epsilon $ that $\displaystyle \partial_y F(x,y) < 0$ in this neighborhood and so $\displaystyle \partial_y F(x,y) \not = 0$.
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Here is what I was thinking when I wrote it. Since $\displaystyle \partial_y F$ is continous at $\displaystyle (a,b)$ it means $\displaystyle \lim_{(x,y)\to (a,b)}\partial_y F(x,y) = \partial_y F(a,b)$. But this means for any $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ such that if $\displaystyle \sqrt{(x-a)^2+(y-b)^2} <\delta \implies |\partial_y F(x,y) - \partial_y F(a,b) | < \epsilon$. Rewrite $\displaystyle |\partial_y F(x,y) - \partial_y F(a,b) | < \epsilon$ as $\displaystyle - \epsilon < \partial_y F(x,y) - \partial_y F(a,b) < \epsilon $ and so $\displaystyle \partial_y F(a,b) - \epsilon < \partial_y F(x,y) < \partial_y F(a,b) + \epsilon$ for all $\displaystyle \sqrt{(x-a)^2+(y-b)^2} < \delta$. Now $\displaystyle \partial_y F(a,b) \not = 0$. Thus, either $\displaystyle \partial_y F(a,b) > 0$ or $\displaystyle \partial_y F(a,b) < 0$. If $\displaystyle \partial_y F(a,b) > 0$ choose an $\displaystyle \epsilon$ small enough so that $\displaystyle \partial_y F(a,b) - \epsilon > 0$. And then $\displaystyle \partial_y F(x,y) > 0$ (because $\displaystyle \partial_y F(a,b) - \epsilon < \partial_y F(x,y)$ ) for $\displaystyle \sqrt{(x-a)^2+(y-b)^2}< \delta$ (and this is a neighborhood around $\displaystyle (a,b)$) thus $\displaystyle \partial_y F(x,y) \not = 0$. Now if $\displaystyle \partial_y F(a,b) < 0$ choose $\displaystyle \epsilon >0$ small enough so that $\displaystyle \partial_y F(a,b) + \epsilon < 0$. It will follow from $\displaystyle \partial_y F(x,y) < \partial_y F(a,b) + \epsilon $ that $\displaystyle \partial_y F(x,y) < 0$ in this neighborhood and so $\displaystyle \partial_y F(x,y) \not = 0$.
    Ok got it! Thanks.
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