Question

find the equation of the tangent plane and the normal line given the surface xy^2z^3=8 at (2,2,1)

Answer 1

Given, equation of the surface is $f(x,y,z)=xy^2z^3-8$ .

And ,we need to find the equation of tangent plane and normal line at (x₀,y₀,z₀)=(2,2,1).

$\frac{\partial f }{\partial x}=y^2z^3,\frac{\partial f }{\partial y}=2xyz^3,\frac{\partial f }{\partial z}=3xy^2z^2\\ \text{At point (2,2,1),} \frac{\partial f }{\partial x}=4, \frac{\partial f }{\partial y}=8, \frac{\partial f }{\partial z}=24\\ \text{Equation of tangent plane,}\\ \frac{\partial f }{\partial x}(x-x_0)+\frac{\partial f }{\partial y} (y-y_0)+\frac{\partial f }{\partial z} (z-z_0)=0\\ 4(x-2)+8(y-2)+24(z-1)=0\\ 4x+8y+24z=48\\ \text{Equation of normal line,}\\ \frac{x-x_0}{\frac{\partial f }{\partial x}}=\frac{y-y_0}{\frac{\partial f }{\partial y}}=\frac{z-z_0}{\frac{\partial f }{\partial z}}\\ \frac{x-2}{4}=\frac{y-2}{8}=\frac{z-1}{24}$