Exam practice question to which I don't have an answer. Would be greatful for any help.
Consider the function defined by
and let P=(2,1,3).
(i) Find the rate of change of f(x,y,z) at the point P in the direction .
(ii) Write down the direction of maximum increase of f at P, and the rate of maximum increase. Show that P is on the surface and write down the equation of the tangent plane to the surface at P.
(iii) There is another point Q on the surface where the tangent plane is parallel to the tangent plane at P. Find Q.
(i) first, I find partial derivatives of f at P
Let be the angle between the above vector (6,-10,2)^T and the unit vector v=(1,-1,1)^T. Then I can find alpha:
Then the rate of change of in the direction of v=(1,-1,0)^T is just the length of the projection of this vector on the direction of v, which can be calculated as
(ii) the maximum rate of change of f at P is just the length of its gradient vector at this point, ie as per (i).
Since coordinates of P satisfy the equation , it lies on the surface .
The tangent plane to the surface at P (2,1,3) is
is the tangent plane to the surface f(x,y,z)=3 at P(2,1,3).
For a parallel tangent plane, I understand that I need to find a point that satisfies the equation f(x,y,z)=3 AND an equation of a plane parallel to 6x-10y+z=5. However, I don't know yet how to find it.