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My entire Calculus III high school class (including the teacher) have been struggling on this problem and we really need some help.

Find all points of the surface given by

$x^3 + 3yx^2 + 6yz - 3z^2 = 7$

where the tangent plane has an equation of the form $Ax + Ay = B$, for some constants $A$ and $B$

Currently, we've tried to work through the problem like so:

Implicitly derive $x^3 + 3yx^2 + 6yz - 3z^2 = 7$ to recieve:

$\displaystyle\frac{\partial z}{\partial x} = \frac{3x^2+6xy}{6z-6y}$

$\displaystyle\frac{\partial z}{\partial y} = \frac{3x^2+6z}{6z-6y}$

We were under the assumption that in the tangent plane formula, $z - c$ would be $0$, as there is no $z$ after the $B$ coefficient in the given equation

This gave us a final tangent plane equation of:

$\displaystyle 0 = \frac{3a^2x+6abx-3a^3-6a^2b+3a^2y+6cy-3a^2b-6cb}{6c-6b}$

$\displaystyle \frac{3a^3+9a^2b+6cb}{6c-6b} = \left(\frac{3a^2+6ab}{6c-6b}\right)x + \left(\frac{3a^2+6c}{6c-6b}\right)y$

(where $a$, $b$, and $c$ are the $x$, $y$, and $z$ values at the points we're solving for)

Since the coefficients for $x$ and $y$ are the same, we then set

$\displaystyle\frac{3a^2+6ab}{6c-6b} = \frac{3a^2+6c}{6c-6b}$

This told us $ab = c$, which we used to reformat the original equation given into

$a^3 + 3a^2b+6b^2a-3a^2b^2=7$

Which we thought would be almost there.

Our class was under the belief that another equation would be needed to solve for $a$ and $b$ in a system, which would then give us the values of $A$ and $B$ and allow us to solve the question.

Some of the class is still thinking along this line of reasoning, and others think we've gone down a completely incorrect path towards the solution, but none of us have solved it and just a point in the right direction would be great.

Math Attack
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aheiman
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    Yes, edit your question to include your work. I can’t guess where you ran into difficulties. I find three points. – Ted Shifrin Nov 10 '23 at 18:39
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    What a way to make it too hard! This is a level surface of $f(x,y,z)$. What should be the automatic approach to find a normal vector to the surface? (Note that your approach deals at best only with points where the surface is a graph over the $xy$-plane,) – Ted Shifrin Nov 10 '23 at 19:33
  • Oh, silly of me not to have made a stronger comment. The equation you were given of the tangent plane says specifically that the tangent plane is vertical, and so, in particular, $z$ is not locally given as a function of $x$ and $y$. – Ted Shifrin Nov 10 '23 at 19:45
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    You are right, although your derivation is flawed, that $c=ab$. But what you’re missing is that $b=c$ — i.e., the denominators of your fractions must be $0$. You almost always want to (must) use the gradient vector of $f$. – Ted Shifrin Nov 10 '23 at 19:57
  • Thank you. I used the gradient of $f$ this time and got two working points (1,-2,-2) and (1,1,1)

    Can you confirm if these are the right points? I'm also wondering if I'm missing any points as well.

     – aheiman Nov 10 '23 at 20:37
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    You're missing one point, I believe. There's another solution of the equations $b=c$ and $c=ab$. Students often make a mistake of missing solutions when they cancel without considering other possibilities. – Ted Shifrin Nov 10 '23 at 20:47

1 Answers1

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I approached this question the wrong way.

Instead, start with the partial derivatives of $f(x,y,z) = x^3+3yx^2+6yz−3z^2-7$

$\frac{\partial f}{\partial x} = 3x^2 + 6xy$

$\frac{\partial f}{\partial y} = 3x^2 + 6z$

$\frac{\partial f}{\partial z} = 6y - 6z$

Note that $Ax + Ay = B$, which means that the tangent plane is vertical and the $z$ component of the gradient vector must equal $0$ meaning

$0 = 6y - 6z$

$6z = 6y$

$z = y$

Also, note that because $Ax + Ay = B$, the $x$ and $y$ components of the gradient vector must be equal to each other

$3x^2 + 6xy = 3x^2 + 6z$

$6xy = 6z$

$xy = z$ which means $y$ and $z$ could both equal $0$ or $x = 1$

From here, we can begin solving for some points, beginning with the one where $y$ and $z$ are zero.

$x^3 + 3(0)x^2 + 6(0)(0) - 3(0)^2 = 7$

$x^3 = 7$

$x = \sqrt[3]{7}$ and we've found one point $(\sqrt[3]{7}, 0, 0)$

Now, to solve for the points where $x = 1$

$(1)^3 + 3y(1)^2 + 6yz − 3y^2 = 7$

$3y^2 + 3y - 6 = 0$

$(y-1)(y+2) = 0$

$y = 1$ or $y = -2$ so we've found the final points $(1,1,1)$ and $(1,-2,-2)$

Final points are $(\sqrt[3]{7}, 0, 0)$, $(1,1,1)$, and $(1,-2,-2)$

aheiman
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