Find the point on the surface z = 2x2+xy+3y2 where the tangent plane is parallel to
the plane 8x-8y-z=5.
To find the point on the surface where the tangent plane is parallel to the given plane, we need to find the gradient vector of the surface and the normal vector of the given plane.
1. Find the gradient vector of the surface:
The surface equation is z = 2x^2 + xy + 3y^2. To find the gradient vector, we need to find the partial derivatives of z with respect to x and y.
∂z/∂x = 4x + y
∂z/∂y = x + 6y
So, the gradient vector of the surface is ∇z = (4x + y) i + (x + 6y) j + k.
2. Find the normal vector of the given plane:
The given plane equation is 8x - 8y - z = 5. The coefficients of x, y, and z represent the normal vector of the plane.
The normal vector of the plane is n = 8i - 8j - k.
3. Equate the gradient vector and normal vector:
Since the tangent plane is parallel to the given plane, the gradient vector of the surface should be proportional to the normal vector of the plane. This means the two vectors need to be parallel.
Set up the equations:
4x + y = 8k (from equating the x-component)
x + 6y = -8k (from equating the y-component)
1 = -k (from equating the z-component)
From the third equation, we get k = -1.
Substituting this value into the first two equations:
4x + y = 8
x + 6y = 8
Solving these two equations will give us the values of x and y.
4. Solve the equations:
From the second equation, we get x = 8 - 6y. Substituting this value into the first equation:
4(8 - 6y) + y = 8
32 - 24y + y = 8
-23y = -24
y = 24/23
Substituting this value back into x = 8 - 6y:
x = 8 - 6(24/23)
x = 8 - 144/23
x = 176/23
Now we have the values of x and y. To find the value of z, substitute these values into the surface equation:
z = 2(176/23)^2 + (176/23)(24/23) + 3(24/23)^2
Calculating this expression will give us the value of z.
Hence, the point on the surface where the tangent plane is parallel to the plane 8x - 8y - z = 5 is (176/23, 24/23, z), where z is the value calculated from the above expression.