Question

Find an equation of the tangent plane to the given surface at the specified point. z=2(x-1)^2+6(y+3)^2+4,\ (3,-2,18)

Negative numbers and coordinate plane
ANSWERED
asked 2021-05-18
Find an equation of the tangent plane to the given surface at the specified point.
\(z=2(x-1)^2+6(y+3)^2+4,\ (3,-2,18)\)

Answers (1)

2021-05-19
Find an equation of the tangent plane to the given surface at the specified point.
\(z=2(x-1)^2+6(y+3)^2+4,(3,-2,18)\)
\(\Rightarrow z_x=4(x-1)\)
\(=4x-4\),
\(z_y=12(y+3)\)
\(=12y+36\)
At the point (3,-2,18):
\(z_x=4x-4\)
\(=4(3)-4\)
\(=12-4\)
\(=8\)
\(z_y=12y+36\)
\(=-12(-2)+36\)
\(=-24+36\)
\(=12\)
Equation of the tangent plane is,
\(z-z_1=z_x(x-x_1)+z_y(y-y_1)\)
\(z-18=8(x-3)+12(y-(-2))\)
\(z-18=8x+12y\)
\(z=8x+12y+18\)
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