# Recitation 8

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.