Question

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.)

Answer 1

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 \(\hat\nabla f=\dfrac1{\sqrt6}(1,1,2)=(\frac1{\sqrt6},\frac1{\sqrt6},\frac2{\sqrt6})\)

Answer 2

Sorry, I can't understand what you wrote.

Answer 3

Maybe an example will help, http://mathinsight.org/directional_derivative_gradient_examples

Answer 4

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 \((1,1,2)\).

Answer 5

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 \(1/\sqrt6\).

Answer 6

Thanks.