Section 11.6 Problem 25: Suppose that over a certain region of space the electrical potential V is given by V(x, y, z) = 5x^2 + 3xy + xyz. (a) Find the rate of change of the potential at P(3, 4, 5) in the direction of the vector \mathbf{v} =\mathbf{i}+ \mathbf{j} -\mathbf{k}. (b) In which direction does V change most rapidly at P? (c) What is the maximum rate of change at P?
Solution: The gradient of V is \nabla V(x,y,z)=\langle 10x+3y+yz, 3x+xz, xy \rangle. So the gradient at P is \langle 62, 24, 12\rangle. (a) Therefore, in the direction of \mathbf{v} =\mathbf{i}+ \mathbf{j} -\mathbf{k}, the rate of the change is \nabla V(3,4,5)\cdot\frac{\mathbf{v}}{|\mathbf{v}|}=\frac{74}{\sqrt{3}}. (b) In the direction of 62\mathbf{i}+24 \mathbf{j} -12\mathbf{k} (c) the rate of the change reaches its maximum which is |62\mathbf{i}+24 \mathbf{j} -12\mathbf{k}|=2\sqrt{1141}.
Section 11.6 Problem 34: Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point. xy + yz + zx = 5, (1, 2, 1).
Comment: Given a surface F(x,y,z)=C and a point P(a,b,c) on it, the normal vector of the tangent plane at P or the direction of the normal line is given by \nabla F(a,b,c).
Section 11.6 Problem 44: Show that the ellipsoid 3x^2 + 2y^2 + z^2 = 9 and the sphere x^2 +y^2 +z^2 -8x-6y-8z+24=0 are tangent to each other at the point (1, 1, 2). (This means that they have a common tangent plane at the point.)
Comment: It is enough to show the normal vectors of the tangent planes are parallel to each other. Let F(x,y,z)=3x^2 + 2y^2 + z^2 and G(x,y,z)=x^2 +y^2 +z^2 -8x-6y-8z+24. Verify \nabla F(1,1,2) is parallel to \nabla G(1,1,2).
Section 11.6 Problem 48: Show that every normal line to the sphere x^2 + y^2 + z^2 = r^2 passes through the center of the sphere.
Solution: The direction of the normal line that passes through point (a,b,c) is \nabla F(a,b,c)=\langle 2a,2b,2c\rangle where F=x^2+y^2+z^2.
Section 11.7 Problem 4: Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and view- point that reveal all the important aspects of the function. f(x, y)= xy - 2x - 2y - x^2 - y^2
Comment: The critical point (a,b) is given by f_x(a,b)=0, f_y(a,b)=0. Then apply the second derivative test on this critical point to see if it is a local minimum, a local maximum or a saddle point. Let D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-f_{xy}(a,b)^2.
- If D>0 and f_{xx}(a,b)>0,then f(a,b)is a local minimum.
- If D>0 and f_{xx}(a,b)>0, then f(a,b) is a local maximum.
- If D< 0, then f (a, b) is not a local maximum or minimum. The point (a, b) is a saddle point of f.
Note that given a point (a,b) and some perturbation (\Delta x, \Delta y), within a small neighborhood of (a,b), f(a+\Delta x,b+\Delta y)\approx f(a,b)+f_x(a,b)\Delta x+f_y(a,b)\Delta y+q(\Delta x, \Delta y) where q(\Delta x, \Delta y)=f_{xx}(a,b)\Delta x^2+2f_{xy}(a,b)\Delta x\Delta y+f_{yy}(a,b)\Delta y^2.
Suppose (a,b) is a critical point, we have f_x(a,b)=f_y(a,b)=0. Hence f(a+\Delta x,b+\Delta y)\approx f(a,b)+q(\Delta x, \Delta y).
Notice that q(\Delta x, \Delta y) is a quadratic form of \Delta x, \Delta y whose discriminant \Delta=4f_{xy}(a,b)^2-4f_{xx}(a,b)f_{yy}(a,b)=-4D(a,b).
If D(a,b)>0, then the discriminant \Delta < 0. Furthermore if the leading coefficient f_{xx}(a,b)>0, then the quadratic form is positive definite, that is q(\Delta x, \Delta y)\geq 0. Hence f(a+\Delta x, b+\Delta y)\approx f(a,b)+q(\Delta x, \Delta y)\geq f(a,b).
On the other hand, if f_{xx}(a,b)<0, then q(\Delta x, \Delta y)\leq 0. Consequently, f(a,b) is the local maximum.
Section 11.7 Problem 23: Find the absolute maximum and minimum values of f on the set D. f(x, y) = x^2 + y^2 - 2x, D is the closed triangular region with vertices (2, 0), (0, 2), and (0, -2).
Comment: The quickest thing to do is to find the critical point inside the region D and the critical point on each side of the triangle. Evaluate f at those points and three vertices of the triangle and compare all of them.
Section 11.7 Problem 36: Find three positive numbers whose sum is 12 and the sum of whose squares is as small as possible.
Solution (Sketch): Let x,y,z be such that x+y+z=12. We want to minimize x^2+y^2+z^2. Since z=12-x-y, x^2+y^2+z^2=x^2+y^2+(12-x-y)^2. Then find the critical point of this expression of x, y which would lead us to the answer.