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.

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

Problem 5

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
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/∂z(∂z/∂x)= -(∂F/∂x)
∂z/∂x= -(∂F/∂x)/(∂F/∂z)= -Fz/Fz

Maintained by and last modified 12/11/2008.