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

```(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```