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Thread: Evaluating a First Partial at a Point

  1. #1
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    Evaluating a First Partial at a Point

    Evaluate $\displaystyle z_x$ at $\displaystyle (1, 2, 3)$ if $\displaystyle x^3+y^3+z^3+6xyz=72$

    Part of me says I must solve this in terms of z in order to find $\displaystyle z_x$ but I can't see how.

    The other part thinks I can just find $\displaystyle z_x$ implicitly and plug in the point but I am getting an answer that doesn't make any sense.

    What should I do?
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  2. #2
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    You need to find:
    $\displaystyle \frac{\partial z}{\partial x}$

    So differentiate implicitly with respect to x, assuming that y is constant and you will get $\displaystyle z_x$
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  3. #3
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    You need to find:
    $\displaystyle \frac{\partial z}{\partial x}$

    So differentiate implicitly with respect to x, assuming that y is constant and you will get $\displaystyle z_x$
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  4. #4
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    Quote Originally Posted by Shananay View Post
    Evaluate $\displaystyle z_x$ at $\displaystyle (1, 2, 3)$ if $\displaystyle x^3+y^3+z^3+6xyz=72$

    Part of me says I must solve this in terms of z in order to find $\displaystyle z_x$ but I can't see how.

    The other part thinks I can just find $\displaystyle z_x$ implicitly and plug in the point but I am getting an answer that doesn't make any sense.

    What should I do?
    You're not going to be able to define $\displaystyle z$ explicitly in terms of $\displaystyle x$ and $\displaystyle y$.

    So you will need to differentiate both sides implicitly, treating $\displaystyle y$ as constant.
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  5. #5
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    How do I treat z?

    I have something like $\displaystyle 3x^2+3z^2(z_x)+6y(z_x)=0$ then factor and solve for $\displaystyle z_x$?

    I don't feel like that's right.
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  6. #6
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    Quote Originally Posted by Shananay View Post
    Evaluate $\displaystyle z_x$ at $\displaystyle (1, 2, 3)$ if $\displaystyle x^3+y^3+z^3+6xyz=72$

    Part of me says I must solve this in terms of z in order to find $\displaystyle z_x$ but I can't see how.

    The other part thinks I can just find $\displaystyle z_x$ implicitly and plug in the point but I am getting an answer that doesn't make any sense.

    What should I do?
    We assume that $\displaystyle z$ is a function of $\displaystyle x$, so you need to use the Chain Rule...


    $\displaystyle x^3 + y^3 + z^3 + 6xyz = 72$

    $\displaystyle \frac{\partial}{\partial x}(x^3 + y^3 + z^3 + 6xyz) = \frac{\partial}{\partial x}(72)$

    $\displaystyle \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial x}(y^3) + \frac{\partial}{\partial x}(z^3) + \frac{\partial}{\partial x}(6xyz) = 0$

    $\displaystyle 3x^2 + 0 + \frac{\partial z}{\partial x}\,\frac{\partial}{\partial z}(z^3) + 6y\,\frac{\partial}{\partial x}(xz) = 0$ (Chain Rule)

    $\displaystyle 3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6y\left[x\,\frac{\partial}{\partial x}(z) + z\,\frac{\partial}{\partial x}(x)\right] = 0$ (Product Rule)

    $\displaystyle 3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6y\left[x\,\frac{\partial z}{\partial x} + z(1)\right] = 0$

    $\displaystyle 3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6xy\,\frac{\partial z}{\partial x} + 6yz = 0$

    $\displaystyle 3z^2\,\frac{\partial z}{\partial x} + 6xy\,\frac{\partial z}{\partial x} = -(3x^2 + 6yz)$

    $\displaystyle (3z^2 + 6xy)\frac{\partial z}{\partial x} = -(3x^2 + 6yz)$

    $\displaystyle \frac{\partial z}{\partial x} = -\frac{3x^2 + 6yz}{3z^2 + 6xy}$

    $\displaystyle \frac{\partial z}{\partial x} = -\frac{x^2 + 2yz}{z^2 + 2xy}$.


    Now substitute $\displaystyle (x, y, z) = (1, 2, 3)$.
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  7. #7
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    Quote Originally Posted by Shananay View Post
    How do I treat z?

    I have something like $\displaystyle 3x^2+3z^2(z_x)+6y(z_x)=0$ then factor and solve for $\displaystyle z_x$?

    I don't feel like that's right.
    Treat z as a composition which contains variables x,y. So at compositions you must apply the chain rule.

    Now you did well except for one mistake:

    $\displaystyle 3x^2+3z^2z'+(6xyz)'=0$

    $\displaystyle 3x^2+3z^2z'+6y(xz)'=0$

    $\displaystyle 3x^2+3z^2z'+6y(x'z+xz')=0$

    Now factor for z' and substitute (1,2,3) for x,y,z or first substitute and then factor z'.

    Regards.
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  8. #8
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    Thank you both very much. I understand now.
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