## Thomas' Calculus 13th Edition

$2x-z=2$
The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ Given: $\ln(x^2+y^2)-z=0$ The equation of tangent line for $\nabla f(1,0,0)=\lt 2,0,-1 \gt$ is $(2)(x-1)+(0)(y-0)-(1)(z-0)=0 \implies 2x-2-z=0$ Thus, $2x-z=2$