Find an equation of the tangent plane to the given surface at the specified poin

Question

Find an equation of the tangent plane to the given surface at the specified point. z = 5(x - l)2 + 4(y + 3)2 + 1, (2, -1, 22) z = z = y ln(x), (1, 2, 0) z = z = ex2 - y2, (3, -3, 1) Z = Find the linear approximation of given function at (0, 0). f(x, y) = y + (cos(x))2 f(x, y) Find the linear approximation of given function at (0, 0). f(x, y) = 2x + 5/4y + 1 f(x, y) Find the linear approximation of the function below at the indicated point. f(x, y) = 29 - x2 - 4y2 at (4, 1) f(x, y) Find the linear approximations of the function below and at the indicated point. f(x, y) = ln (x - 4y) at(13, 3) f(x, y) Use the approximation to find f(12.93, 3.04). (Round the answer to two decimal places.) f(12.93, 3.04)

Explanation / Answer

1. gradient gives normal vector= -10i^ -16j^ +1k^

tangent plane=:-

10(x-2) +16(y+1)-(z-22) =0

2.

gradient gives normal vector= -2i^ +1k^

tangent plane=:-

2(x-1) -z =0

3.

gradient gives normal vector= -6i^ -6j^ +1k^

tangent plane=:-

6(x-3) +6(y+3)-(z-1) =0

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