Answer:
z = 2x + y - 2
Step-by-step explanation:
To find an equation for a plane tangent to a surface, we need two things:
1) A surface Z = f(x,y)
2) A point P(x0,y0,z0)
The equation for the plane tangent to the surface Z at the point P is given by the equation a)
a) Zx(x0,y0,z0)*(x-x0) + Zy(x0,y0,z0)*(y-y0) -(z-z0) = 0
Where Zx is the partial derivative of Z in relation to x and Zy is the partial derivative of Z in relation to y
In your problem, we have that Z = x^2 + y^4 + e^(xy)
So Zx(x,y,z) = 2x + ye^(xy)
We have that (x0,y0,z0) = (1,0,2), so Zx(1,0,2) = 2
Zy(x,y,z) = 4y^3 + xe^(xy)
Zy(1,0,2) = 1
According to equation a), we have
Zx(1,0,2)*(x-2) + Zy(1,0,2)*(y-0) -(z-2) = 0
2*(x-2) + y - z + 2 = 0
z = 2x + y - 2 is the equation for the plane tangent to this surface at this point.
