# Homework Help: Finding an equation of the tangent plane -- need help with steps

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1. Aug 8, 2016

### RoboNerd

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Hi everyone.

I am told to find the equation of the tangent plane to the surface x^2 + 2xy^2 -3z^3 = 6.
I do not know how to approach this problem, and I was wondering if anyone would be kind enough to help.

I know that for example if I had an equation z = x^2 + y^2, with a point P(x0,y0) and I was asked to find a tangent plane at that point, I would just do z = z0 + partialX(x0,y0) * (x-x0) + partialY(x0,y0) * (y-y0).
I can solve these problems quickly. Piece of cake.

However, I have the equation x^2 + 2xy^2 -3z^3 = 6, with the equation not being equal to a variable (like 'z' in the previously mentioned example in the preceding paragraph) but rather equal to a constant.
I have been searching for some time on approaches on how to solve such a problem with an equation equaling a constant (6 in this case) and I have not been successful.

Could anyone be kind enough to help me out? I would be very grateful to anyone who would.

Thanks, and have a good day!

2. Aug 8, 2016

### Staff: Mentor

So usually you do $0 = (z-z_0) \frac{\partial f}{\partial z} + (x-x_0) \frac{\partial f}{\partial x}+ (y-y_0)\frac{\partial f}{\partial y}$?
Where's the difference?
And btw. you really should use the homework section on PF for this.

3. Aug 9, 2016

### RoboNerd

Hi. My book has the answer as being 10x + 2y + 3z = 17 being the answer as the equation for the tangent plane.
This book has a lot of typos so I doubt that their answer is valid.

I got the answer as 17 = -9z + 10x + 8y. I do not know if it is correct. Could anyone please check it?

What I do not understand is why you would do
Suppose I have f(x,y). We then have dimensions x, y, and z (for the output of the function f(x,y) ).
We thus can do what you mentioned here
without any problem whatsoever.

OK. That was a basic case. I am now returning to my problem. I have f(x,y,z) = x^2 + 2xy^2 -3z^3 = 6.
I have three dimensions for my inputs (x,y,z) and then a fourth dimension for my output (let's call it "w" for the sake of naming it) which is equal to 6 all the time.

I thought that we would do the following:
w - w0 = (x - x0) partialF/partialx + (y - y0) partialF/partialy + (z - z0) partialF/partialy

However I do not have the output of my given function being a variable as it is equal to 6. Maybe we then have the left side of the equation equal zero because we have w - w0 = 6 - 6 = 0?

Thanks again for the help and I am looking forward to hearing from you.

4. Aug 9, 2016

### Staff: Mentor

If you imagine $\{(x,y,z)\;|\;0 = x^2 + 2xy^2 -3z^3 - 6\}$ then it will be a mountain like picture in $\mathbb{R}^3$.
Where do you want to find the tangent plane? At different points you get different slopes and therefore different planes, as with a real mountain.
In the simple case of $y=x^2$ you get as many tangents as the parabola has points.

5. Aug 9, 2016

### Ray Vickson

Your question cannot be answered because you do not specify a point at which you want to take the tangent plane. Of course, you could leave it general by writing it as $(x_0,y_0,z_0)$, and at least that would lead to a problem that makes sense.

6. Aug 10, 2016

### RoboNerd

Hi. I forgot to type the point. It is point P (1,2,1). My apologies.

How would I be able to approach it this time?

Thanks!

7. Aug 10, 2016

### Staff: Mentor

How did you get there and what could be done to decide whether your answer or the textbook's answer is the correct one?

8. Aug 11, 2016

### RoboNerd

I started off here. I then took the partials of the expression for each variable: x, y, and z. I set the left side of my expression with the part w-w0 = 0 because the output of the function is constant and so the change in the function's values is zero.

I thus have the following:

0 = df/dx (x-x0) + df/dy (y-y0) + df/dz (z - z0).

df/dx = 2x + 2y^2
df/dy = 4y
df/dz = -9z^2

I plug in the results into the general expression:

0 = (2*1 + 2*2^2) * (x-1) + (4*2) * (y -2) + (-9 * 1^2) * (z -1).

Cleaning up the terms and moving the constants to the left side gave me
17 = 10x + 8y - 9z

As for checking whether this attempt at a solution is right or if the book's answer is right: I have no idea. This is why I appealed here for help

9. Aug 11, 2016

### Staff: Mentor

Your calculation is correct (and you see, that you actually didn't need to introduce $w$ and $w_0$).
What you have said to do usually, the equation in post #2, and your equation above are all the same.

So the question is: How can we distinguish between two planes that both contain the point $(1,2,1)$?
We need another point on the plane to make a decision.

What would happen, if we regarded a cut through our "mountain"? Say, at constant $x=1$?
This would get us a two dimensional situation in which tangents are more easy to calculate. But those tangents will still have to be within our original tangent plane in three dimensions and therefore give us a second point to check (which is on our plane but not on our "mountain").

10. Aug 11, 2016

### RoboNerd

Hi fresh_42,

I am trying to understand somethings that you have said.
So I understand that your phrased question is how can we see if these two planes that contain the point (1,2,1) are the same?

What do you mean a "cut through our mountain"?
I am a bit confused as to what you said here. Could you please try to reword your explanation so I can understand what you said, if possible?

What two dimensional situation results here? What tangents result here?

I apologize for this lack of understanding. I am currently learning vector analysis, and it might be that I do not know of a certain concept(s)

11. Aug 11, 2016

### Staff: Mentor

Ok, the mathematical correct wording would have been: a projection of our algebraic variety. I assumed that wouldn't have helped much.
Therefore I said "mountain" instead of algebraic variety, which both is simply $\{(x,y,z) \in \mathbb{R}^3\;|\;x^2+2xy^2-3z^3-6=0\}$
A plot of this with coordinates $(x,y,z)$ would look like a mountain.

The projection or cut means, that if we fix $x=1$, then we have only the plane $\mathbb{R}^2 = \{(1,y,z) \;|\; y,z \in \mathbb{R}\}$ left.
It is like cutting the plot at $x=1$ with a knife.
But now you can calculate a tangent straight $z = my+b$ at $(1,2,1)$ in $(1,\mathbb{R}^2)$ and find another point by moving, e.g. one step starting at $(1,2,1)$ along that straight.

Edit: And this new point has still to satisfy either $17=10x +8y-9z$ or if the textbook was right $17=10x+2y+3z$.

Last edited: Aug 11, 2016
12. Aug 11, 2016

If I may make a quick input here: I didn't read the entire posting, but you need to find the gradient of the function (which defines the surface) which you correctly did and put in the values (1,2,1). The equation of the tangent plane is found by simply taking a dot product (of the gradient with the vector originating from (1,2,1) ) and setting it equal to zero: (rather than using a formula that you then need to memorize.): $((x-1) \hat{i}+(y-2) \hat{j} +(z-1)\hat{k} ) \cdot (\nabla{f})=0$. And I agree with the answer that you (the OP) got.

Last edited: Aug 11, 2016
13. Aug 12, 2016

### RoboNerd

Thank you everyone for your kind help!

It is really appreciated.