A student was asked to find the equation of the tangent plane to the surface determined by z = x3 - y4 at the point (x, y) = (1, 3).

Solution:

Given z = x3 - y4 at the point (x, y) = (1, 3).

F(x, y,z) = x3 - y4 -z = 0

The equation of  the tangent line passing through a point is given as

f’x(x0, y0, z0)(x - x0) + f’y(x0, y0, z0)(y - y0) + f’z(x0, y0, z0)(z - z0) = 0

z0 = x03 - y04

= 13 - 34

= 1 - 81

= -80

(x0, y0, z0) = (1, 3, -80)

f’x(x, y, z) = 3x2: f’x(x0, y0, z0) = 3(1)2 = 3

f’y(x, y, z) = -4y3: f’y(x0, y0, z0) = -4(3)3 = -108

f’z(x0, y0, z0) = -1

Equation = 3(x - 1) -108(y - 3) -(z +80) = 0

3x -3 -108y+ 324-z-80= 0

3x -108y-z+241 = 0

Required equation of tangent is 3x -108y-z+241 = 0

A student was asked to find the equation of the tangent plane to the surface determined by z = x3 - y4 at the point (x, y) = (1, 3).

Summary:

A student was asked to find the equation of the tangent plane to the surface determined by z = x3 - y4 at the point (x, y) = (1, 3), the equation of the tangent is 3x -108y-z+241 = 0

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