Determine equation of tangent plane?

Question

Determine equation of tangent plane in points $(\frac{1}{2},1,f(\frac{1}{2},1) )$

$f(x,y)=x^{4}-x^{2}+y^{2}$

I know usually how these examples work, but I am confused with these $3$ points. I have only $2$ variables $x$ and $y$, but $3$ points. What to do with third point?

Answer 1

HINT.-The point is $P=(\frac 12, 1,\frac {13}{16})$. Put your equation under the form $$F(x,y,z)=x^4-x^2+y^2-z=0$$ You have to calculate the three partial derivatives (very easy)

$$\frac{\partial F(P)}{\partial x}\\\frac{\partial F(P)}{\partial y} \\\frac{\partial F(P)}{\partial z}$$ and your plan has the equation $$\frac{\partial F(P)}{\partial x}(x-\frac 12)+\frac{\partial F(P)}{\partial y}(y-1)+\frac{\partial F(P)}{\partial z}(z-\frac{13}{16})=0$$

Answer 2

Method 1

The standard equation of the tangent plane on the surface of $z=f(x,y)$ in a point $(x_0,y_0,z_0)$ where $z_0=f(x_0,y_0)$ is given by: $$z-z_0 = \left.\frac{\partial f}{\partial x}\right|_{(x_0,y_0)}(x-x_0)+\left.\frac{\partial f}{\partial y}\right|_{(x_0,y_0)}(y-y_0)$$ where I used the notation $\left.\tfrac{\partial f}{\partial x}\right|_{(x_0,y_0)}$ to indicate that these partial derivatives have to be evaluated in the point $(x_0,y_0)$.

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Method 2

If you rewrite the equation of the surface in the form $F(x,y,z)=0$, so as a level set of a function of three variables, then the equation of the tangent plane at $(x_0,y_0,z_0)$ can be elegantly written as: $$\nabla F(x_0,y_0,z_0) \cdot \left( x-x_0,y-y_0,z-z_0 \right) = 0$$ where $\nabla F(x_0,y_0,z_0)$ is the gradient of $F$, evaluated at $(x_0,y_0,z_0)$. With the notation of above, this boils down to: $$\left.\frac{\partial F}{\partial x}\right|_{(x_0,y_0,z_0)}(x-x_0)+\left.\frac{\partial F}{\partial y}\right|_{(x_0,y_0,z_0)}(y-y_0)+\left.\frac{\partial F}{\partial z}\right|_{(x_0,y_0,z_0)}(z-z_0)=0$$

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Answer

Either way, you should end up with (Wolfram|Alpha) something that can be simplified to:

$$z = -\frac{x}{2}+2y-\frac{15}{16}$$