# Finding the tangent plane

Find the tangent plane to $$z=4 x^3+3 xy +4 y^3$$ at $(-1,1,-3)$.

Answer on the form $z=Ax+By+C$.

I don't know how to solve these problems, should I find the derivative of x and y, then let those be the vectors on my plane?

If I see how this problem is solved, then I might be able to solve similar ones.

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Your surface is defined by a polynomial equation $f(x,y,z) = 0$. You could interpret this situation geometrically in the following way:

$f:R^3\to R$ gives a color to every point of the 3D space (imagine you have a color palette indexed by real numbers). Assume white is indexed at $0$, then your surface is the locus of points with white color.

The gradient $$\nabla f = ( \partial_x f, \partial_y f, \partial_z f )$$ is orthogonal to the surface when evaluated at points $(x,y,z)$ of the surface. This is because the color stays constant in our white surface.

So it's enough for you to compute $\nabla f$ with partial derivatives, evaluate it at your point $(-1,1,-3)$ to get a vector $V$, and then notice that the set of vectors perpendicular to that one is given by the equation $$V\cdot (X,Y,Z) = 0$$ Doing this you find the equation of a plane parallel to the tangent space, but passing through the origin. To get the tangent space, translate it via $(-1,1,-3)$. Finally, you should get $$Z=15X+9Y+3$$

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Tank you! But solutions says $15,9,9$ is incorrect. – jacob Dec 15 '13 at 13:02
I'm sorry, I was using $(-1,1,3)$ in place of $(-1,1,-3)$. Now it should be correct! – Abramo Dec 15 '13 at 13:05
@jacob: Is it ok now? – Abramo Dec 15 '13 at 15:11
correct, thanks. – jacob Dec 16 '13 at 21:05

$$\nabla z=\left\langle\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}\right\rangle^T=\langle12x^2+3y,12y^2+3x\rangle^T$$ At $(-1,1,-3)$, we have $\nabla z=\begin{pmatrix}15\\9\end{pmatrix}$ so we can conclude that the plane is of the form $z=15x+9y+c$. Solving for $c$, we have \begin{align}-3&=15(-1)+9(1)+c\\3&=c\end{align} So the tangent plane is $z=15x+9y+3$.

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now I'm kind of embarrassed. oh well. I was more after "If I see how this problem is solved, then I might be able to solve similar ones." and hoping OP would get the intuition behind it. – Tim Ratigan Dec 15 '13 at 12:42
In fact your solution is still wrong, simply because you are using $(-1,1,3)$ in place of $(-1,1,-3)$. My solution was wrong as well ;-) – Abramo Dec 15 '13 at 13:06
wow, thanks for the checking. I don't know how I got $z=3$ on top of everything else. – Tim Ratigan Dec 15 '13 at 13:10