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); |
![[Maple Plot]](images/m114-ex121.gif)
Here's the actual surface, with the contours on it:
> | plot3d(x*y,x=-4..4,y=-4..4,style=patchcontour); |
![[Maple Plot]](images/m114-ex122.gif)
(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)); |
![1/((x-1)*x)](images/m114-ex123.gif)
Clearly this expression is ok as x approaches 2 and y approaches -4:
![1/2](images/m114-ex124.gif)
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/(x^2+y^2)^(1/2)](images/m114-ex125.gif)
> | limit(limit(f,x=0),y=0); |
![0](images/m114-ex126.gif)
> | limit(limit(f,y=0),x=0); |
![undefined](images/m114-ex127.gif)
> | limit(limit(f,y=0),x=0,right); |
![-1](images/m114-ex128.gif)
(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; |
![eqn := x*y+z(x,y)^3*x-2*y*z(x,y) = 0](images/m114-ex1210.gif)
> | solve(diff(eqn,x),diff(z(x,y),x)); |
![-(y+z(x,y)^3)/(3*z(x,y)^2*x-2*y)](images/m114-ex1211.gif)
and at (1,1,1):
> | subs(x=1,y=1,z(1,1)=1,%); |
![-2](images/m114-ex1212.gif)
Problem 65
![f := exp(-2*y)*cos(2*x)](images/m114-ex1213.gif)
> | simplify(diff(f,x$2)+diff(f,y$2)+diff(f,z$2)); |
![0](images/m114-ex1214.gif)
(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; |
![eqn := 1/R(R1,R2) = 1/R1+1/R2](images/m114-ex1215.gif)
> | Rx:=solve(diff(eqn,R1),diff(R(R1,R2),R1)); |
![Rx := R(R1,R2)^2/R1^2](images/m114-ex1216.gif)
> | Ry:=solve(diff(eqn,R2),diff(R(R1,R2),R2)); |
![Ry := R(R1,R2)^2/R2^2](images/m114-ex1217.gif)
![dR := R(R1,R2)^2/R1^2*dR1+R(R1,R2)^2/R2^2*dR2](images/m114-ex1218.gif)
This is part (a). For part (b), we have numbers:
> | solve(subs(R1=100,R2=400,eqn),R(100,400)); |
![80](images/m114-ex1219.gif)
> | subs(R1=100,R2=400,R(100,400)=80,dR); |
![16/25*dR1+1/25*dR2](images/m114-ex1220.gif)
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); |
![vol := a(t)*b(t)*c(t)](images/m114-ex1221.gif)
![surf := 2*a(t)*b(t)+2*a(t)*c(t)+2*b(t)*c(t)](images/m114-ex1222.gif)
![diag := (a(t)^2+b(t)^2+c(t)^2)^(1/2)](images/m114-ex1223.gif)
> | 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)); |
![3](images/m114-ex1224.gif)
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)); |
![0](images/m114-ex1225.gif)
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)); |
![-3/7*14^(1/2)](images/m114-ex1226.gif)
So the diagonals are getting smaller.
Section 12.7, page 967
Problem 18
> | f:=x^2*y+exp(x*y)*sin(y); |
![f := x^2*y+exp(x*y)*sin(y)](images/m114-ex1227.gif)
> | simplify(subs(x=1,y=0,[diff(f,x),diff(f,y)])); |
![[0, 2]](images/m114-ex1228.gif)
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))); |
![2](images/m114-ex1229.gif)
Problem 31
> | eqn:=cos(Pi*x)-x^2*y+exp(x*z)+y*z=4; |
![eqn := cos(Pi*x)-x^2*y+exp(x*z)+y*z = 4](images/m114-ex1230.gif)
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)])); |
![nvec := [2, 2, 1]](images/m114-ex1231.gif)
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); |
![2*x+2*y+z = 4](images/m114-ex1232.gif)
And parametric equations for the normal line are:
> | x=0+2*t, y=1+2*t, z=2+t; |
![x = 2*t, y = 1+2*t, z = 2+t](images/m114-ex1233.gif)
Section 12.8, page 975
Problem 36
First we look for interior critical points:
> | T:=48*x*y-32*x^3-24*y^2; |
![T := 48*x*y-32*x^3-24*y^2](images/m114-ex1234.gif)
> | crits:=solve({diff(T,x)=0,diff(T,y)=0},{x,y}); |
![crits := {x = 0, y = 0}, {x = 1/2, y = 1/2}](images/m114-ex1235.gif)
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); |
![crits1 := 0, 0](images/m114-ex1236.gif)
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); |
![crits2 := 1](images/m114-ex1237.gif)
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); |
![crits3 := -1/2*2^(1/2), 1/2*2^(1/2)](images/m114-ex1238.gif)
The second of these will have to be checked (i.e., the point
). Finally:
> | crits4:=solve(diff(subs(x=0,T),y)=0,y); |
![crits4 := 0](images/m114-ex1240.gif)
OK - we only have one interior point, one edge point and the four corners to check:
![2](images/m114-ex1241.gif)
> | subs(x=crits3[2],y=1,T); |
![16*2^(1/2)-24](images/m114-ex1242.gif)
![-1.37258301](images/m114-ex1243.gif)
![0](images/m114-ex1244.gif)
![-24](images/m114-ex1245.gif)
![-8](images/m114-ex1246.gif)
![-32](images/m114-ex1247.gif)
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); |
![[Maple Plot]](images/m114-ex1248.gif)
Looks like we got it right.
Problem 42
![f := x*y+2*x-ln(x^2*y)](images/m114-ex1249.gif)
> | crits:=solve({diff(f,x)=0,diff(f,y)=0},{x,y}); |
![crits := {x = 1/2, y = 2}](images/m114-ex1250.gif)
> | H:=diff(f,x$2)*diff(f,y$2)-diff(f,x,y)^2; |
![H := 2/x^2/y^2-1](images/m114-ex1251.gif)
![1](images/m114-ex1252.gif)
![1](images/m114-ex1253.gif)
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 := x*y*z](images/m114-ex1257.gif)
![g1 := x^2+y^2-1](images/m114-ex1258.gif)
![g2 := x-z](images/m114-ex1259.gif)
![F := x*y*z-lambda*(x^2+y^2-1)-mu*(x-z)](images/m114-ex1260.gif)
> | 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}); |
![crits := {x = 0, y = 1, lambda = 0, mu = 0, z = 0}, {x = 0, lambda = 0, y = -1, mu = 0, z = 0}, {y = RootOf(3*_Z^2-1), x = RootOf(3*_Z^2-2), lambda = RootOf(3*_Z^2-1), mu = -RootOf(3*_Z^2-2)*RootOf(3*_...](images/m114-ex1261.gif)
![crits := {x = 0, y = 1, lambda = 0, mu = 0, z = 0}, {x = 0, lambda = 0, y = -1, mu = 0, z = 0}, {y = RootOf(3*_Z^2-1), x = RootOf(3*_Z^2-2), lambda = RootOf(3*_Z^2-1), mu = -RootOf(3*_Z^2-2)*RootOf(3*_...](images/m114-ex1262.gif)
![crits := {x = 0, y = 1, lambda = 0, mu = 0, z = 0}, {x = 0, lambda = 0, y = -1, mu = 0, z = 0}, {y = RootOf(3*_Z^2-1), x = RootOf(3*_Z^2-2), lambda = RootOf(3*_Z^2-1), mu = -RootOf(3*_Z^2-2)*RootOf(3*_...](images/m114-ex1263.gif)
![0](images/m114-ex1264.gif)
![0](images/m114-ex1265.gif)
> | c3:=allvalues(crits[3]); |
![c3 := {x = 1/3*6^(1/2), z = 1/3*6^(1/2), y = 1/3*3^(1/2), lambda = 1/3*3^(1/2), mu = -1/9*6^(1/2)*3^(1/2)}, {mu = 1/9*6^(1/2)*3^(1/2), x = 1/3*6^(1/2), z = 1/3*6^(1/2), y = -1/3*3^(1/2), lambda = -1/3*...](images/m114-ex1266.gif)
![c3 := {x = 1/3*6^(1/2), z = 1/3*6^(1/2), y = 1/3*3^(1/2), lambda = 1/3*3^(1/2), mu = -1/9*6^(1/2)*3^(1/2)}, {mu = 1/9*6^(1/2)*3^(1/2), x = 1/3*6^(1/2), z = 1/3*6^(1/2), y = -1/3*3^(1/2), lambda = -1/3*...](images/m114-ex1267.gif)
![c3 := {x = 1/3*6^(1/2), z = 1/3*6^(1/2), y = 1/3*3^(1/2), lambda = 1/3*3^(1/2), mu = -1/9*6^(1/2)*3^(1/2)}, {mu = 1/9*6^(1/2)*3^(1/2), x = 1/3*6^(1/2), z = 1/3*6^(1/2), y = -1/3*3^(1/2), lambda = -1/3*...](images/m114-ex1268.gif)
![c3 := {x = 1/3*6^(1/2), z = 1/3*6^(1/2), y = 1/3*3^(1/2), lambda = 1/3*3^(1/2), mu = -1/9*6^(1/2)*3^(1/2)}, {mu = 1/9*6^(1/2)*3^(1/2), x = 1/3*6^(1/2), z = 1/3*6^(1/2), y = -1/3*3^(1/2), lambda = -1/3*...](images/m114-ex1269.gif)
![[2/9*3^(1/2), 0, 0]](images/m114-ex1270.gif)
![[-2/9*3^(1/2), 0, 0]](images/m114-ex1271.gif)
![[2/9*3^(1/2), 0, 0]](images/m114-ex1272.gif)
![[-2/9*3^(1/2), 0, 0]](images/m114-ex1273.gif)
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).