CHAPTER 12 - Multivariable Functions and Partial Derivatives

Section 12.1, page 914

Problem 5

The domain of the function xy is all pairs of real numbers, so we'll just draw the level curves.

>	with(plots,contourplot):

>	contourplot(x*y,x=-4..4,y=-4..4,contours=[-10,-8,-6,-4,-2,0,2,4,6,8,10],color=blue,thickness=2);

Here's the actual surface, with the contours on it:

>	plot3d(x*y,x=-4..4,y=-4..4,style=patchcontour);

(see the Math 104 solved problems for more from this section)

Section 12.2, page 921

Problem 16

We'll do what it says -- simplify first, then take the limit:

>	simplify((y+4)/(x^2*y-x*y+4*x^2-4*x));

Clearly this expression is ok as x approaches 2 and y approaches -4:

>	subs(x=2,y=-4,%);

This is the limit.

Problem 35

We'll show that taking the limit in x first then y is different from y first then x:

>	f:=-x/sqrt(x^2+y^2);

>	limit(limit(f,x=0),y=0);

>	limit(limit(f,y=0),x=0);

>	limit(limit(f,y=0),x=0,right);

(see the Math 104 solved problems for more from this section)

Section 12.3, page 931

Problem 57

If we're going to find  then we must be thinking of z as a function of x and y:

>	eqn:=x*y+z(x,y)^3*x-2*y*z(x,y)=0;

>	solve(diff(eqn,x),diff(z(x,y),x));

and at (1,1,1):

>	subs(x=1,y=1,z(1,1)=1,%);

>

restart;

Problem 65

>	f:=exp(-2*y)*cos(2*x);

>	simplify(diff(f,x$2)+diff(f,y$2)+diff(f,z$2));

(see the Math 104 solved problems for more from this section)

Section 12.4, page 942

Problem 20

We'll be thinking of R as a function of R1 and R2:

>	eqn:=1/R(R1,R2)=1/R1+1/R2;

>	Rx:=solve(diff(eqn,R1),diff(R(R1,R2),R1));

>	Ry:=solve(diff(eqn,R2),diff(R(R1,R2),R2));

>	dR:=Rx*dR1+Ry*dR2;

This is part (a). For part (b), we have numbers:

>	solve(subs(R1=100,R2=400,eqn),R(100,400));

>	subs(R1=100,R2=400,R(100,400)=80,dR);

So the resistance R is more sensitive to variations in R1, because the derivative of R with respect to R1 is bigger.

Section 12.5, page 950

Problem 40

We'll write expressions for volume, surface area and diagonal in terms of a(t), b(t) and c(t):

>	vol:=a(t)*b(t)*c(t); surf:=2*a(t)*b(t)+2*a(t)*c(t)+2*b(t)*c(t); diag:=sqrt(a(t)^2+b(t)^2+c(t)^2);

>	subs(diff(a(t),t)=1,diff(b(t),t)=1,diff(c(t),t)=-3,a(t)=1,b(t)=2,c(t)=3,diff(vol,t));

Volume is increasing at 3 cubic meters per second

>	subs(diff(a(t),t)=1,diff(b(t),t)=1,diff(c(t),t)=-3,a(t)=1,b(t)=2,c(t)=3,diff(surf,t));

Surface area isn't changing at that particular moment.

>	subs(diff(a(t),t)=1,diff(b(t),t)=1,diff(c(t),t)=-3,a(t)=1,b(t)=2,c(t)=3,diff(diag,t));

So the diagonals are getting smaller.

Section 12.7, page 967

Problem 18

>	f:=x^2*y+exp(x*y)*sin(y);

>	simplify(subs(x=1,y=0,[diff(f,x),diff(f,y)]));

OK - so the direction of fastest increase at this point is just in the y-direction. And the (directional) derivative in this direction is just the partial with respect to y:

>	simplify(subs(x=1,y=0,diff(f,y)));

Problem 31

>	eqn:=cos(Pi*x)-x^2*y+exp(x*z)+y*z=4;

The normal vector is just the gradient of the left side:

>	nvec:=simplify(subs(x=0,y=1,z=2,[diff(lhs(eqn),x),diff(lhs(eqn),y),diff(lhs(eqn),z)]));

So the equation of the tangent plane is:

>	2*x+2*y+z=subs(x=0,y=1,z=2,2*x+2*y+z);

And parametric equations for the normal line are:

>	x=0+2*t, y=1+2*t, z=2+t;

Section 12.8, page 975

Problem 36

First we look for interior critical points:

>	T:=48*x*y-32*x^3-24*y^2;

>	crits:=solve({diff(T,x)=0,diff(T,y)=0},{x,y});

Only the second one is in the interior, so we only remember that one for later.

Now for the four edges. First, the y=0 edge:

>	crits1:=solve(diff(subs(y=0,T),x)=0,x);

The origin is not in the interior of the edge, so we move on to the x=1 edge:

>	crits2:=solve(diff(subs(x=1,T),y)=0,y);

The point (1,1) is not in the interior of the edge (it's on a corner), so we move on to the y=1 edge:

>	crits3:=solve(diff(subs(y=1,T),x)=0,x);

The second of these will have to be checked (i.e., the point ). Finally:

>	crits4:=solve(diff(subs(x=0,T),y)=0,y);

OK - we only have one interior point, one edge point and the four corners to check:

>	subs(crits[2],T);

>	subs(x=crits3[2],y=1,T);

>

evalf(%);

>	subs(x=0,y=0,T);

>	subs(x=0,y=1,T);

>	subs(x=1,y=1,T);

>	subs(x=1,y=0,T);

We conclude that the absolute max occurs at the interior critical point [1/2,1/2], and the absolute min occurs at the corner [1,0]. We plot to make sure:

>	plot3d(T,x=0..1,y=0..1,style=patchcontour);

Looks like we got it right.

Problem 42

>	f:=x*y+2*x-ln(x^2*y);

>	crits:=solve({diff(f,x)=0,diff(f,y)=0},{x,y});

>	H:=diff(f,x$2)*diff(f,y$2)-diff(f,x,y)^2;

>	subs(crits,H);

>	subs(crits,diff(x$2));

Since the discriminant is positive and the second derivative with respect to x is positive, the function has a local minimum at (1/2,2). Also note that when x and y approach zero, the log of  goes to , so the function f goes to +. And as x and y grow the function also gets big. So our point must be the global min.

Section 12.9, page 987

Problem 46

>	f:=x*y*z; g1:=x^2+y^2-1; g2:=x-z;

>	F:=f-lambda*g1-mu*g2;

>	crits:=solve({diff(F,x)=0,diff(F,y)=0,diff(F,z)=0,diff(F,lambda)=0,diff(F,mu)=0},{x,y,z,mu,lambda});

>	subs(crits[1],f);

>	subs(crits[2],f);

>	c3:=allvalues(crits[3]);

>	subs(c3[1],[f,g1,g2]);

>	subs(c3[2],[f,g1,g2]);

>	subs(c3[3],[f,g1,g2]);

>	subs(c3[4],[f,g1,g2]);

So the max and min of our function are + and - , and these occur at [,,] and points obtained by negating y or both x and z (independently, so there are 4 choices).