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anonymous

  • 5 years ago

Find the sum direction and magnitude of the greatest rate of change of the function z+ln(x+y^2) at (1,1)

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  1. anonymous
    • 5 years ago
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    it should not say sum... it should just say find the direction and magnitude my bad

  2. anonymous
    • 5 years ago
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    I know you need to take the directional derivative and turn (1,1) into a unit vector and I did that, but i'm stuck at this point. I have \[DuF(x,y) = (1+2y)\div(\sqrt{x}(x+y ^{2}))\]

  3. anonymous
    • 5 years ago
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    I’m not very knowledgeable in vector calc, but this comes from http://tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx#Gradient_Defn There’s a theorem in vector calculus that says that the gradient of a function (grad(f(x)) points in the direction of maximal change, and that the magnitude of this change is ||grad(f(x))|| at that point x. If f(x, y, z) = z + ln(x + y^2), then grad(f) = <1/(x + y^2), 2y/(x + y^2), 1>. Plugging in our point (1, 1), we see that the direction of maximal change is <1/2, 1, 1>, and the magnitude of that change is ||<1/2, 1, 1>|| = sqrt(9/4) = 3/2. Again, if you want, you can check out that link above for more explanation and the proof for the theorem.

  4. anonymous
    • 5 years ago
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    thank you thank you thank you

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