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LeoMessi

  • 4 years ago

Find the equation tangent plane to the graph z = (x-4)^2 + (y-4)^2 at p=(5, 6, 5)...?

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  1. Owlfred
    • 4 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. dpflan
    • 4 years ago
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    All rright, here we go! let f(x,y,z)=(x−4)2+(y−4)2−z we have the following partial derivatives: ∂f/∂x=2x−8 ∂f/∂y=2y−8 ∂f/∂z=−1 gradf(x,y,z)=(2x−8)i+(2y−8)j−k gradf(5,6,5)=2i+4j−k hence an equation of the tangent line at P(5,6,5) is given by 2(x−5)+4(y−6)−(z−5)=0⟹2x+4y−z=29

  3. fauxshaux
    • 4 years ago
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    dw= dz + dx + dy

  4. azumih
    • 4 years ago
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    \[\partial{z}/\partial{x}|_5 = 2(x - 4)|_5 = 2\] \[\partial{z}/\partial{y}|_6 = 2(y - 4)|_6 = 4\] Hence the tangent plane has gradient (2, 4). Its equation is in the form \[z = 2x + 4y + p\] And it passes through (5, 6, 5), so \[5 = 2*5 + 4*6 + p\] \[p = -29\] Result: \[z = 2x + 4y - 28\] How about that...

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