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find an equation of the tangent plane to the surface at the given point. h(x, y) = ln x2 y2 , (3, 4, ln 5)

User GianFS
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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.

User Ligerdave
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