Math264-1
Find critical points and 2nd derivative test. f(x,y)=1+(2x+1)^2+y2
(-1/2,0) point of local minimum
critical points for f(x,y)=2x4+y4
(0,0) Inconclusive
Critical points for f(x,y)=4+2x2+3y2
(0,0) local minimum
Critical points of f(x,y)=x2+2y2-4x+4y+6
(2,-1) local minimum
Find an equation for the plane that passes through Po(2,-3,4) with the normal n=<-1,2,3>
-x+2y+3z=4
Linear approximation o f(x,y)=5/(x2+y2) (-1,2,1)
0.4x-0.8y+3
lim (x,y)-->(1,0) (x+y)
1+0=1
Find an equation of the plane that passes through P(2,-1,3), Q(1,4,0), R(0,-1,5)
10x+8y+10z=42
lim(x,y)-->(4,1) (xy-4y2)/sqrt(x)-(2sqrt(y))
4
Find linear approximation and use to approximate f(1.2,1.3) f(x,y)=4+x6y+y6x (1,1)
7x+7y-8 9.5
Find the gradient f(x,y)=e^(-x2-2y2) at P(1,1)
<-2e^-3, -4e^-3>
Gradient of f(x,y)=x^2 +2xy-y^3 P(3,2)
<10.-6>
Domain of sin(x/y)
All pairs such that y doesnt =0
Find the domain of f(x,y)=e^(x/y+1)
All the pairs (x,y) such that y+1 doesnt equal 0, y doesnt equal -1
Directional derivative of f(x,y)=1/4(x2+2y2+2 P(3,2) u=<1/sqrt2,1/sqrt2> v=<1/2,-sqrt3/2>
Duf=7/2 * 1/sqrt2 Duv=3/4 - sqrt3
How to find directional derivatives
Get the gradient and multiply by u
How to find linear approximation
L(x,y)=fx(a,b)(x-a)+fy(a,b)(y-b)+f(a,b) Plug points into original equation. get fx and fy and plug in points.
Linear approximation f(x,y)=ln(1+x+y) (0,0) compute at (0.1,-0.2)
L(x,y)=x+y -0.1
Find the domain of f(x,y)=x+y
R2
Find the domain of f(x,y)=y/x2+y2
R2 is all the pairs (x,y) where x2+y2 does not equal 0
Find domain of f(x,y)=sqrt(4-x2-y2)
R2 such that 4-x2-y2>0 x2+y2<4
Partial derivative of f(x,y)=e^(x2+y2)
fx=(e^y2)2xe^x2 fy=2ye^y2 ex^2
Find partial derivative f(x,y)=(y2+1)e^x
fx=(y2+1)e^x fy=2ye^x
Partial derivative of f(x,y)=cos(xy)
fx=-ysinxy fy=-xsinxy fxx=-y^2cosxy fyy=-x^2cosxy fyx=-sinxy-xycosxy
Partial derivative f(x,y)=3x4y-2xy+5xy3
fx=12x3y-2y+5y3 fy=3x4-2x+15xy2 fxx=36x2y fyy=30xy fyx=12x3-2+15y2
Find partial derivatives of f(x,y)=x^2 y
fx=2xy fy=x2 fxx=2y fyy=0 fxy=2x fyx=2x
Partial derivative f(x,y)=e^(x+y)
fx=e^y e^x fy+e^x e^y fxx=e^y e^x fyy=e^x e^y fxy=e^x e^y
Partial derivative of f(x,y)= xe^y
fx=e^y fy=xe^y fxy=e^y
Partial derivative f(x,y)=e^(x2+y2)
fx=ey^2 2xe^x2 fy=2yey^2 e^x2
Find partial derivatives of f(x,y)=xy
fx=y fy=x fxx=0 fyy=0 fxy=1 fyx=1
Partial derivative of f(x,y)=sin(xy)
fx=ycosxy fy=xcosxy fxx=-y2sinxy fyy=-x2sinxy fxy=cosxy-xysinxy
Find partial derivatives of f(x,y)=e^(xy)
fx=ye^xy fy=xe^xy fxx=y^2 e^xy hyy=x^2 e^xy fxy= e^(xy) +xye^xy fyx=e^(xy)+xye^xy
How to find tangent plane equation
get fx and fy plug in the points, then add the last number to the end of the equation z=fx(a,b)(x-a)+fy(x,b)(y-b)+f(a,b)
How to find gradient
get fx and fy then plug in the P. then <>
How to find critical points
get fx and fy. set fx=0 and fy=0 to get points. get fxx and fyy and fxy D=fxx*fyy -fxy2 compare D and fxx to 0
Critical Points D=0
inconclusive
lim (x,y)-->(1,-1) (x2+xy/x+y)
limX=1
Critical Points D>0 fxx<0
local maximum
Critical Points: D>0 fxx>0
point of local minimum
Critical Points D<0
saddle point
Find directional derivative f(x,y)=10+x2+y2 u=<sqrt3/(2),1/2> P(1,0)
sqrt3
Critical points of f(x,y)=xy(x-2)(y+3)
y=0, y=-3, x=1
Tangent plane of z=32-3x2-4y2 (2,1,16)
z=-12x-8y+48
Find an equation of the tangent plane z=4+x3-y2 at (1,1,4)
z=3x-2y+3