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kotykmax [81]
3 years ago
5

Find an equation of the plane tangent to the following surface at the given point.z equals tangent Superscript negative 1 Baseli

ne (xy )​;(0 comma 3 comma 0 )
Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
5 0

Let <em>f(x</em>, <em>y</em>, <em>z)</em> = <em>z</em> - arctan(<em>x y</em>). Compute the gradient of <em>f</em> at the point (0, 3, 0):

∇ <em>f(x</em>, <em>y</em>, <em>z)</em> = (-<em>y</em> / (1 + <em>x</em>²<em>y</em>²), -<em>x</em> / (1 + <em>x</em>²<em>y</em>²), 1)

∇ <em>f </em>(0, 3, 0) = (-3, 0, 1)

This vector is orthogonal to the surface <em>z</em> = <em>f(x</em>, <em>y)</em>. Then the equation of the tangent plane is

∇ <em>f </em>(0, 3, 0) • (<em>x</em>, <em>y</em> - 3, <em>z</em>) = 0

(-3, 0, 1) • (<em>x</em>, <em>y</em> - 3, <em>z</em>) = 0

-3<em>x</em> + <em>z</em> = 0

<em>z</em> = 3<em>x</em>

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