Answers to the old final exam (spring 2006) in Math 251

Problems with letters whose backgrounds are this color have
answers here. Those with backgrounds in this color do not.
Have you done a problem?

AB 1
Answer by Joseph Buono;
corrected by Samantha Parillo		 CD 1
Answer by Joseph Buono		 2 a,b,
cAB,cCD
3
Answer by Joseph Buono		 4
Answer by Joseph Buono;
a small part of a)'s answer is missing		 5
Answer by Sai Veruva
6 7 8
9 10
Answer by Sai Veruva		 11
Answer by Sai Veruva
12
Answer by Sai Veruva		   

Problem 5
F(x,y)=(2y+1)ex2-y=2yex2-y+ex2-y

5a) ∂f/∂x=(2x)(2y)ex2-y+(2x)ex2-y=ex2-y(4xy+2x)=0. Exponential functions are never =0, so 4xy+2x=0; x(4y+2)=0 → x=0.
∂f/∂y=2ex2-y-2yex2-y-ex2-y=ex2-y(2-2y-1)=0. Exponential functions are not =0, so 2-2y-1=0 → y=1/2.
The critical point is at (0,1/2). No other points satisfy the equations above.

5b) The nature of the critical point is determined by D=D(a,b)=fxx(a,b)fyy(a,b)-fxy(a,b)2.
fxx(x,y)=(4y+2)ex2-y+ex2-y(4xy+2x)2x, where fxx(0,.5)=4e-.5.
fyy(x,y)=-ex2-y(2-2y-1)+(-2)ex2-y, where fyy(0,.5)=-2e-.5.
fxy(x,y)=2xex2-y(2-2y-1), where fxy(0,.5)=0.
Thus, D=(4e-.5)(-2e-.5)-02=-8/e.
Since D<0, the critical point (0,.5) is a saddle point.

Problem 12

(z2)x + 3xy2 + ey2z = 4. Find ∂z/∂x.
∂z/∂x = -Fx/Fz
Fx= z2 + 3y2
Fz=2zx+(y2)ey2z
∂z/∂x=-(z2+3y2)/[2zx+(y2)ey2z]
If the formula wasn't given, then it can be easily derived using implicit diff. ideas from single variable calc. If we assume F(x,y,z)=0 and differentiate both sides with respect to x: 
∂F/∂x(∂x/∂x)+∂F/∂y(∂y/∂x)+∂F/∂z(∂z/∂x)=0
∂F/∂x(1)+∂F/∂y(0)+∂F/∂z(∂z/∂x)=0  
∂F/∂z(∂z/∂x)= -(∂F/∂x)
∂z/∂x= -(∂F/∂x)/(∂F/∂z)= -Fz/Fz

Maintained by greenfie@math.rutgers.edu and last modified 12/11/2008.