Answer to Question #136779 in Calculus for Promise Omiponle

Question

Find the equation for the tangent plane to the surface x-z= 4 arctan(yz) at the point (1 +pi,1,1).

Answer 1

We have $x-4arctan(yz)-z=0$ and point $(1+\pi, 1, 1).$

The equation is $F_{x}^{'}(x_0, y_0, z_0)(x-x_0)+F_{y}^{'}(x_0, y_0, z_0)(y-y_0)+F_{z}^{'}(x_0, y_0, z_0)(z-z_0)=0.$

$F_{x}^{'}=1$, $F_{x}^{'}(1+\pi, 1, 1)=1.$

$F_{y}^{'}=-4\frac{1}{y^2z^2+1}z=\frac{-4z}{y^2z^2+1}$, $F_{y}^{'}(1+\pi, 1, 1)=\frac{-4}{2}=-2$.

$F_{z}^{'}=-4\frac{1}{y^2z^2+1}y-1=\frac{-4y}{y^2z^2+1}-1$, $F_{z}^{'}(1+\pi, 1, 1)=\frac{-4}{2}-1=-3$.

So, $1(x-1-\pi)-2(y-1)-3(z-1)=0$,

$x-1-\pi-2y+2-3z+3=0$,

$x-2y-3z+4-\pi=0$.

Answer: $x-2y-3z+4-\pi=0.$