# Simultaneous Equations with partial derivatives

The function $$x^3 - 3x^2 - 6xy + 2y^2+ 8y + 8$$ yields

$$f_x = 3x^2 - 6x - 6y$$

$$f_y = 4y + 8 - 6x$$

When we equate both of these to $$0, y = \frac{x^2}{2} - x$$ and $$y = \frac{3}{2}x - 2$$

Plotting these on a graph we can see the points of interaction are $$(1, -0.5)$$ and $$(4, 4)$$ and although that is the answer, I was wondering how to get those results using simultaneous equations, as the $$x^2$$ means that Gaussian elimination is inadequate, but I can't seem to figure it out using simultaneous equations.

Also, is the equation of a tangent plane the same formula as linear approximation? The tangent plane equation I found when at $$(1, -2)$$ is 13 + 9x - 6y

• $13+9x-6y$ is not an equation Jan 9, 2021 at 5:20
• For both equations, you have $y$ on the left-hand side. So $\frac{x^2}{2} - x = \frac{3}{2}x - 2$: is this what you are asking? Jan 9, 2021 at 5:20
• I was trying to understand how (1, -0.5), (4, 4) arise from those equations Jan 10, 2021 at 3:52

Just equate them. We need $$y=\frac{x^2}2-x=\frac{3x}2-2$$ giving $$x^2-5x+4=0\iff x=1,4$$ and corresponding $$y=-1/2,4$$.

$$13+9x-6y$$ is not an equation. For the functional equation $$f(x,y,z)=0$$, the equation of the tangent plane at $$(x_0,y_0,z_0)$$ is $$[(x,y,z)-(x_0,y_0,z_0)]\cdot\vec\nabla f(x_0,y_0,z_0)=0$$ where $$\vec\nabla f=(f_x,f_y,f_z)$$ represents the gradient or normal vector to the surface at a given point. Simplifying, you get$$(x-x_0)f_x(x_0,y_0,z_0)+(y-y_0)f_y(x_0,y_0,z_0)+(z-z_0)f_z(x_0,y_0,z_0)=0$$

We also have$$f(x-x_0,y-y_0,z-z_0)=\underbrace{\require{cancel}\cancelto{0}{f(x_0,y_0,z_0)}}_{(x_0,y_0,z_0)\text{ lies on the surface }f(x,y,z)=0}\\+(x-x_0)f_x(x_0,y_0,z_0)+(y-y_0)f_y(x_0,y_0,z_0)+(z-z_0)f_z(x_0,y_0,z_0)+\text{higher order terms}$$so indeed the equation of the tangent plane of $$f(x,y,z)=0$$ resembles the linear approximation of $$f(x,y,z)$$.

• I used the equation $f(a, b) + f_x (a, b)(x - a) + f_y (a,b) (y-b)$ At $(1, 2)$ The value of $a = 9$ and $b = -6$ which led me to $10 + 9(x - 1) + (-6)(y - 2)$ which led me to $-11 + 9x - 6y$ Jan 10, 2021 at 3:57