Question

I have this question for MultiA: f(x,y,z)=sin(xy)cos(z)+xyz,
a)FInd the rate of change of f at point P=(pi,0,pi) in the direction of the vector v=i +j - k
b) In which direction does f increase more rapidly at the point P
c) What is the max rate of change of f at P
d)Determine the equation of the tangent plane to the level surface of f at P?

Answer 1

The directional derivative at P_0 is given by:
\[\vec{\nabla}f(x,y,z)|_{P_0} \cdot \vec{u}_{\vec{v}}\]
Where u_v denotes a unit vector in the v direction. So first:
\[| \vec{v}|=\sqrt{1^2 + 1^2 + 1^2}=\sqrt{3} \implies \vec{u}_{\vec{v}}=\frac{1}{\sqrt{3}}<1,1,-1>\]
Then:
\[\vec{\nabla}f(x,y,z)=<\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}>=<y \cos(xy)\cos(z)+yz,\]
\[x \cos(xy)\cos(z) + xz, -\sin(xy) \sin(z) + xy>\]
Evaluate the gradient at P.
Form the dot product and you're done :P
The direction is of most increase is the gradient.
Most rapid increase is the magnitude of the gradient: \[||\vec{\nabla}f(x,y,z)|_P||\]
And to find the plane you have the perpendicular vector (most increase) then:
\[\hat{n} \cdot (x-P_1,y-P_2,z-P_3)=0; P=(P_1,P_2,P_3)\]
Using normal tangent vector definition.