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dpflan

  • 5 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. AnwarA
    • 5 years ago
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    let \[f(x,y,z)=(x-4)^2+(y-4)^2-z\] we have the following partial derivatives: \[\partial f/\partial x=2x-8 \] \[\partial f/\partial y=2y-8\] \[\partial f/\partial z=-1\] \[grad f(x,y,z)= (2x-8)i+(2y-8)j-k \] \[grad f(5,6,5)=2i+4j-k\] hence an equation of the tangent line at P(5,6,5) is given by \[2(x-5)+4(y-6)-(z-5)=0 \implies 2x+4y-z=29\] hope it's right :)

  2. dpflan
    • 4 years ago
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    Thanks

  3. AnwarA
    • 4 years ago
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    Wow, that was two months ago :D .. You're welcome anyways!!

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