Question

Find an equation for the plane that is tangent to the surface z equals ln (x plus y )at the point Upper P (1 comma 0 comma 0 ).

Answer 1

Let $f(x,y,z)=z-\ln(x+y)$. The gradient of $f$ at the point (1, 0, 0) is the normal vector to the surface, which is also orthogonal to the tangent plane at this point.

So the tangent plane has equation

$\nabla f(1,0,0)\cdot(x-1,y,z)=0$

Compute the gradient:

$\nabla f(x,y,z)=\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)=\left(-\dfrac1{x+y},-\dfrac1{x+y},1\right)$

Evaluate the gradient at the given point:

$\nabla f(1,0,0)=(-1,-1,1)$

Then the equation of the tangent plane is

$(-1,-1,1)\cdot(x-1,y,z)=0\implies-(x-1)-y+z=0\implies\boxed{z=x+y-1}$