anonymous
  • anonymous
Let f(x, y, z)=yz+xz-6. At the point (1, 1, 1), find the unit vector that points in the direction for which f is increasing at the fastest rate. (Just find the gradient and I'll solve the rest.)
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
The vector will be the unit vector in the direction of the gradient. The gradient itself is trivial; determine our partial derivatives:$$\frac{\partial f}{\partial x}=z\\\frac{\partial f}{\partial y}=z\\\frac{\partial f}{\partial z}=y+x$$... hence our gradient is just:$$\nabla f=z\mathbf{i}+z\mathbf{j}+(x+y)\mathbf{k}$$Evaluated at our point, we find \(\nabla f(1,1,1)=(1,1,2)\). Now, just normalize this -- so compute its norm and shrink it: \(\|\nabla f(1,1,1)\|=\sqrt{1^2+1^2+2^2}=\sqrt6\) so \(\hat\nabla f=\dfrac1{\sqrt6}(1,1,2)=(\frac1{\sqrt6},\frac1{\sqrt6},\frac2{\sqrt6})\)
anonymous
  • anonymous
Sorry, I can't understand what you wrote.
anonymous
  • anonymous
Maybe an example will help, http://mathinsight.org/directional_derivative_gradient_examples

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anonymous
  • anonymous
The gradient is defined as follows:$$\nabla f=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}+\frac{\partial f}{\partial z}\mathbf{k}$$i.e. the vector whose components correspond to partial derivatives in their direction. It is intuitive that the gradient points in the direction of greatest increase. Do you know how to find partial derivatives?$$\frac{\partial f}{\partial x}=\frac\partial{\partial x}[yz+xz-6]=\frac\partial{\partial x}yz+\frac\partial{\partial x}xz-\frac\partial{\partial x}6=0+z-0=z$$Repeat for \(\dfrac\partial{\partial y},\dfrac\partial{\partial z}\):$$\frac{\partial f}{\partial y}=z+0-0=z\\\frac{\partial f}{\partial z}=y+x-0=x+y$$... hence our gradient function is then$$\nabla f=z\mathbf{i}+z\mathbf{j}+(x+y)\mathbf{k}$$Evaluate at point \((1,1,1)\) by substituting \(x=1,y=1,z=1\):$$\nabla f(1,1,1)=1\mathbf{i}+1\mathbf{j}+(1+1)\mathbf{k}=1\mathbf{i}+1\mathbf{j}+2\mathbf{k}$$\(1\mathbf{i}+1\mathbf{j}+2\mathbf{k}\) is an alternate notation for the vector \((1,1,2)\).
anonymous
  • anonymous
Now, this isn't a unit vector because it has a norm/magnitude/length of \(\|(1,1,2)\|=\sqrt{1^2+1^2+2^2}=\sqrt6\neq1\). To make it a unit vector, we *normalize* by multiplying by \(1/\sqrt6\).
anonymous
  • anonymous
Thanks.

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