Asked by Anonymous
Find an equation of the tangent plane to the surface given by z = f (x,y )=π₯^2 y + xπ¦^3 at the point (x,y)=(2,1).
Answers
Answered by
oobleck
βz/βx = 2xy + y^3
βz/βy = x^2 + 3xy^2
z(2,1) = 4+2 = 6
So the tangent plane at (1,2) is
z-6 = 5(x-2) + 10(y-1)
βz/βy = x^2 + 3xy^2
z(2,1) = 4+2 = 6
So the tangent plane at (1,2) is
z-6 = 5(x-2) + 10(y-1)
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