Explanation:
To find the equation of the tangent plane to the surface at (3, 4, ln 5), we need to first find the partial derivatives of h with respect to x and y:
hx(x, y) = 2y^2/x
hy(x, y) = 2x^2/y
Then, we can plug in the point (3, 4) to get the partial derivatives evaluated at that point:
hx(3, 4) = 2(4^2)/3 = 32/3
hy(3, 4) = 2(3^2)/4 = 9/2
Now we can use the point-normal form of the equation of a plane, where the normal vector is given by ⟨hx(3, 4), hy(3, 4), -1⟩:
hx(3, 4)(x - 3) + hy(3, 4)(y - 4) - (z - ln 5) = 0
Substituting in the values for hx(3, 4) and hy(3, 4), we get:
(32/3)(x - 3) + (9/2)(y - 4) - (z - ln 5) = 0
Simplifying, we get:
(32/3)x + (9/2)y - z = 55/6
So the equation of the tangent plane to the surface at (3, 4, ln 5) is (32/3)x + (9/2)y - z = 55/6.