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 | 8 | |
9 | 10 Answer by Sai Veruva |
11 Answer by Sai Veruva |
12
Answer by Sai Veruva |
Problem 5
F(x,y)=(2y+1)e^{x2-y}=2ye^{x2-y}+e^{x2-y}
5a)
∂f/∂x=(2x)(2y)e^{x2-y}+(2x)e^{x2-y}=e^{x2-y}(4xy+2x)=0.
Exponential functions are never =0, so 4xy+2x=0; x(4y+2)=0 →
x=0.
∂f/∂y=2e^{x2-y}-2ye^{x2-y}-e^{x2-y}=e^{x2-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)=f_{xx}(a,b)f_{yy}(a,b)-f_{xy}(a,b)^{2}.
f_{xx}(x,y)=(4y+2)e^{x2-y}+e^{x2-y}(4xy+2x)2x,
where f_{xx}(0,.5)=4e^{-.5}.
f_{yy}(x,y)=-e^{x2-y}(2-2y-1)+(-2)e^{x2-y},
where f_{yy}(0,.5)=-2e^{-.5}.
f_{xy}(x,y)=2xe^{x2-y}(2-2y-1), where
f_{xy}(0,.5)=0.
Thus, D=(4e^{-.5})(-2e^{-.5})-0^{2}=-8/e.
Since D<0, the critical point (0,.5) is a saddle point.
(z^{2})x + 3xy^{2} + e^{y2}z = 4. Find ∂z/∂x. ∂z/∂x = -F_{x}/F_{z} F_{x}= z^{2} + 3y^{2} F_{z}=2zx+(y^{2})e^{y2z} ∂z/∂x=-(z^{2}+3y^{2})/[2zx+(y^{2})e^{y2z}]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)= -F_{z}/F_{z}
Maintained by greenfie@math.rutgers.edu and last modified 12/11/2008.