Question

Find an equation for the plane tangent to the surface given by z = x^2 + y^4 + e^xy

Answer 1

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.