Question

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5) Find the equation for the tangent plane passing through the point (3,1,0) for the function: z = ln(x – 2y) Please put your final answer in the form z = ax + by + c, where a, b, and c are real valued constants.

Answer 1

Solution Given surface: z = ln (x-2y) at P(3.1,0) An equation of the tangent plane to the surface z=. = f (x, y) at the point P(Xq, Yo, 2o)is: 2-2n = : (4,7)(x-x)+1,(29,7)(y-y) For surface: z = ln (x-2y) ܚ ܕ 1 1 Differentiate writx: 2x = X -2y 1 At point (3,1,0): 2x = 3-2 -2 Differentiate w.r.ty: 2 (-2) 2y x-2y -2 A point (3,1,0): zy -2 3-2 :: Equation of tangent plane is: 2-0 =1(x-3)+(-2)(x - 1) z= x-2y-1 is the equation of the tangent plane to the given surface 1.