Tangent plane of a surface at points with given gradient

Question

So I'm stuck on the next problem:

I need to find the tangent plane of the surface $$u=\ln\left( x+\frac{1}{y} \right)$$ at all the points where the gradient is equal to $$\nabla u=\hat i-\frac{16}{9}\hat j$$

Please help me. Thank you very much.

Answer 1

First you need to find the point(s) where you have to give the tangent plane.

For the gradient you have: $$\nabla u = \left( \frac{\partial u}{\partial x},\frac{\partial u}{\partial y} \right) = \left( \frac{1}{x+1/y},\frac{-1/y^2}{x+1/y} \right)$$ And you're looking for the points $(x,y)$ where this is equal to $\left( 1,-\tfrac{16}{9}\right)$, so: $$\left\{\begin{array}{rcl} \displaystyle \frac{1}{\color{blue}{x+1/y}} & =& 1 \\[8pt] \displaystyle \frac{-1/y^2}{\color{blue}{x+1/y}} &=& \displaystyle -\frac{16}{9} \end{array}\right.$$ From the first equation it is obvious that $\color{blue}{x+1/y=1}$ , so the second equations yields: $$-\frac{1}{y^2}=-\frac{16}{9} \Leftrightarrow y^2 = \frac{9}{16}$$ This gives $y = \pm \tfrac{3}{4}$ and that leads to the points: $$\left( -\frac{1}{3},\frac{3}{4}\right) \, , \, \left( \frac{7}{3},-\frac{3}{4}\right)$$ Now use the formula for the tangent plane at the surface $u=f(x,y)$ in $(x_0,y_0)$: $$u =f(x_0,y_0) + \frac{\partial u}{\partial x}(x-x_0)+ \frac{\partial u}{\partial y}(y-y_0) $$ Do this for both points with the desired gradient.