So I have this function that looks something like $2x^2-x-5y=17$ and I need to find out the tangent plane for this surface on the point $P=(x_1,y_1,z_1)$. I understand that the gradient for a function on a point is the normal vector to the level surface on that point. Given that normal-vector I can define a plane that will be tangent to the level surface.
Now what I do not understand is that this function does not contain a z-component. And to find out the level surface for a point of a surface I basically need to take the funtion $f(x,y)$ such that $f(x,y)=C$. Am I supposed to bring the whole expression to one side, as in my example such that $2x^2-x-5y-17=0$ and then define $f(x,y)$ such that $f(x,y)=2x^2-x-5y-17$ and furthermore $f(x,y)=z$? Is this $C$ that is supposed to be $C=z_1$?
Of course the functiongraph is not equal to the function but when I plot both $2x^2-x-5y=17$ and $z=2x^2-x-5y-17$ I (as I expect) get different results in geogebra.
So my question is, how do I calculate the tangentplane on the point $(x_1,y_1,z_1)$ of a surface, but the function is unbound by one component?
Best Answer
Consider the curve $2x^2-x-5y=17$ on the plane $z=0$. Now compute the tangent line to that curve passing through $(x_1,y_1,0)$. You will get a line $ax+by=c$ on the plane $z=0$. Then your tangent plane will be the plane $ax+by=c$, which is the vertical plane containing the line that you got.