1. Find an equation for the horizontal tangent plane to the surface v .) u and d

Question

1. Find an equation for the horizontal tangent plane to the surface v .) u and displaystyle frac{partial F}{partial v} = (Note that your answers should be functions of displaystyle frac{partial F}{partial u} = displaystyle frac{partial F}{partial v} . displaystyle frac{partial F}{partial u} and y=u-v , Use the chain rule to find x=u+v and F(x,y) = x+y^2 where v . Given u and z=F(x(u,v),y(u,v)) depends on y=y(u,v) . Then the function x=x(u,v) and y where x and F(x,y) depends on f(sqrt{pi/2}+.01,sqrt{pi/2}-.01) approx 3.The function x_0=y_0=sqrt{pi/2}, Delta x = .01, Delta y=-.01 ) f(x,y) = sin(xy) . (NOTE: Use f(sqrt{pi/2}+.01,sqrt{pi/2}-.01)) for the function y= Tangent Plane: 2. Use an incremental approximation to estimate x= y coordinates of the point at which the tangent plane is horizontal. x and z=4(x-1)^2+3(y+1)^2 . First, find the

Explanation / Answer

1)F= 4*(x-1)^2 +3*(y+1)^2 - z and Let (a,b,c) be point at which it becomes horizontal

8*(a-1)*(x-a)+6*(b+1)*(y-b)+(-1)*(z-c)=0

Fx=8*(a-1) Fy=6*(b+1) Fz=(-1)

so Fx=Fy=0 at that point its tangent becomes horizontal

a=1, b=-1 c=0

so x-coordinate=1  y-coordinate=-1

3) F(x,y)=x+y^2

dF/du = (df/dx)(dx/du) + (df/dy)(dy/du) = 1+2*y

dF/dv = (df/dx)(dx/dv) + (df/dy)(dy/dv) = 1-2*y

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