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Math Help - Directional derivative

  1. #1
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    Directional derivative

    Having a little trouble with this problem. I keep getting zero as an answer which is incorrect. I know where I'm making the mistake, just don't know how to fix it.

    Problem:
    Compute the directional derivative for f(x,y,z) = zexp(-2xy) at point (1,1,-1), t = (1i,-3j,2k).

    Formula:
    \frac{df}{ds} = t \cdot (\frac{\partial f}{dx}i, \frac{\partial f}{dy}j, \frac{\partial f}{dz}k).

    Basically I'm getting stuck with the partial derivatives. I seems to me that the partial of either x, y, or z will give 0. No?
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  2. #2
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    Quote Originally Posted by Ares_D1 View Post
    Having a little trouble with this problem. I keep getting zero as an answer which is incorrect. I know where I'm making the mistake, just don't know how to fix it.

    Problem:
    Compute the directional derivative for f(x,y,z) = zexp(-2xy) at point (1,1,-1), t = (1i,-3j,2k).

    Formula:
    \frac{df}{ds} = t \cdot (\frac{\partial f}{dx}i, \frac{\partial f}{dy}j, \frac{\partial f}{dz}k).

    Basically I'm getting stuck with the partial derivatives. I seems to me that the partial of either x, y, or z will give 0. No?
    Fx means you consider x a variable and treat all other variables as constants. So if you have f=ae^{-2bx} \Rightarrow f'=(ae^{-2bx})*(-2b) by the rule for differentiating e^x. So now just use z=a and y=b and that's Fx.

    Repeat for Fy and Fz. Also, if this is supposed to be in the direction of vector t, you need to get the corresponding unit vector so you can just apply the direction.
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  3. #3
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    Quote Originally Posted by Jameson View Post
    Fx means you consider x a variable and treat all other variables as constants. So if you have f=ae^{-2bx} \Rightarrow f'=(ae^{-2bx})*(-2b) by the rule for differentiating e^x. So now just use z=a and y=b and that's Fx.

    Repeat for Fy and Fz. Also, if this is supposed to be in the direction of vector t, you need to get the corresponding unit vector so you can just apply the direction.
    Sorry, so e^{-2xy} = exp(-2xy)?

    I understand the rules for partial differentiation, just wasn't sure how to interpret the exp function.
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  4. #4
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    So with e^{-2xy} = exp(-2xy) I have:

    \frac{df}{ds} = t \cdot (\frac{\partial f}{dx}i, \frac{\partial f}{dy}j, \frac{\partial f}{dz}k)

    = (1,-3,-2) \cdot (-2yze^{-2xy}, -2xze^{-2xy}, e^{-2xy})

    = -2yze^{-2xy} + 6xze^{-2xy} + 2e^{-2xy}

    = 2e^{-2} - 6e^{/2} + 2e^{-2}

    = -2exp(-2).

    And with the corresponding unit vector,

    \frac{df}{ds} = \frac{-2exp(-2)}{\sqrt{14}}

    That's the answer, thanks!
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  5. #5
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    Quote Originally Posted by Ares_D1 View Post
    Sorry, so e^{-2xy} = exp(-2xy)?

    I understand the rules for partial differentiation, just wasn't sure how to interpret the exp function.
    Yes, "exp(x)" is defined as exp(x)= e^x.
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