Approximating equation for tangent plane at a point

[toforfiltum]

Homework Statement

Suppose that you have the following information concerning a differentiable function $f$:

$f(2,3)=12$, $\space$ $f(1.98,3)=12.1$, $\space$ $f(2,3.01)=12.2$

a) Give an approximate equation for the plane tangent to the graph of $f$ at $(2,3,12)$.

b) Use the result of part (a) to estimate $f(1.98,2.98)$

Homework Equations

The Attempt at a Solution

I'm guessing that the reason they gave me $f(1.98,3)$ and $f(2,3.01)$ is so that I could compute the partial derivatives wrt $x$ and $y$ at the point $(2,3)$. I'm guessing it's such because the values of $x$ and $y$ differ very little from the original point.

So, for a)

$\frac{\partial f}{\partial x}= \frac{0.1}{-0.02}$
$\space\space\space\space\space$ $=-5$

$\frac{\partial f}{\partial y}= \frac{0.2}{0.01}$
$\space\space\space\space\space$ $=20$

$z=12-5(x-2)+20(y-3)$

Then for b),

$f(1.98,2.98)=12-5(1.98-2)+20(2.98-3)$
$=11.7$

Is my approach correct?

Thanks.

 

[andrewkirk]

Yes, I think so.
Note that the plane given by that equation is the unique plane that passes through those three points, and that hence the answer to part (a) would have been the same if they had asked you for the equation to the tangent at (1.98,3,12.1) or (2,3.01,12.2).

 

[toforfiltum]

Yes, I think so.
Note that the plane given by that equation is the unique plane that passes through those three points, and that hence the answer to part (a) would have been the same if they had asked you for the equation to the tangent at (1.98,3,12.1) or (2,3.01,12.2).

Hmm...so to clarify, there were 3 sets of points in the function that lie on the tangent plane? If so, that means a tangent plane isn't really the same as a tangent line, since a tangent line touches the graph of a function at a point, while for a tangent plane, there can be more than a point. If so, it's confusing to me because I thought the definition of a tangent plane means the plane touches the graph at point only?

 

[Mark44]

Hmm...so to clarify, there were 3 sets of points in the function that lie on the tangent plane?

No, just one point -- (2, 3, 12).

If so, that means a tangent plane isn't really the same as a tangent line, since a tangent line touches the graph of a function at a point, while for a tangent plane, there can be more than a point. If so, it's confusing to me because I thought the definition of a tangent plane means the plane touches the graph at point only?

Right. Also, earlier in this thread you said this:

so that I could compute the partial derivatives wrt x and y at the point (2,3).

You're not actually computing those partials -- you are approximating them with $\frac{\Delta f}{\Delta x}$ and $\frac{\Delta f}{\Delta y}$, respectively.

 

[toforfiltum]

No, just one point -- (2, 3, 12).

Wait, I'm confused now. So, if the 3 sets of points $(2,3,12), (1.98,3,12.1), (2,3.01,12.2)$ do not all lie on the tangent plane, why did @andrewkirk say that the answer to (a) would have been the same for the other 2 points?

Right. Also, earlier in this thread you said this:

I'm so sorry, but when you said right, what did you really mean?

 

[Mark44]

No, just one point -- (2, 3, 12).

Wait, I'm confused now. So, if the 3 sets of points $(2,3,12), (1.98,3,12.1), (2,3.01,12.2)$ do not all lie on the tangent plane, why did @andrewkirk say that the answer to (a) would have been the same for the other 2 points?

These three points all lie on the tangent plane, but only (2, 3, 12) is a point on the surface z = f(x, y). The other two points, (1.98, 3, 12.1) and (2, 3.01, 12.2) don't lie on this surface. Do you understand the distinction?

Right. Also, earlier in this thread you said this:

I'm so sorry, but when you said right, what did you really mean?

Right, as in "correct."

 

[Mark44]

You're not actually computing those partials -- you are approximating them with $\frac{\Delta f}{\Delta x}$ and $\frac{\Delta f}{\Delta y}$, respectively.

To elaborate on what I said earlier, here's a correction to what you wrote in post #1:
$\frac{\partial f}{\partial x} \approx \frac{\Delta f}{\Delta x} = \frac{0.1}{-0.02} = .5$
The partial above is really evaluated at (2, 3); i.e., $\left.\frac{\partial f}{\partial x}\right|_{(2, 3)}$
And similar for the other partial.

 

[toforfiltum]

These three points all lie on the tangent plane, but only (2, 3, 12) is a point on the surface z = f(x, y). The other two points, (1.98, 3, 12.1) and (2, 3.01, 12.2) don't lie on this surface. Do you understand the distinction?

How so? The other 2 points also came from the function $f(x,y)$. Shouldn't they lie on the surface as well?

 

[Mark44]

How so? The other 2 points also came from the function $f(x,y)$.

Yes, that's correct. I was mistaken in thinking that these points weren't on the surface. However, the three points aren't on the tangent plane, as andrewkirk seemed to believe. The tangent plane should have a slightly different orientation (and normal) from the plane that the three given points define.

 

[toforfiltum]

Yes, that's correct. I was mistaken in thinking that these points weren't on the surface. However, the three points aren't on the tangent plane, as andrewkirk seemed to believe. The tangent plane should have a slightly different orientation (and normal) from the plane that the three given points define.

Ah, that makes much more sense. Thanks for clearing the confusion!

 

[Mark44]

Another point is that using the three given points, you could use vectors to write the equation of the plane through those points -- no calculus needed. What they're asking though is that you approximate the two partial derivatives so that you can get an estimate of the tangent plane at (2, 3, 12).

 

[andrewkirk]

However, the three points aren't on the tangent plane, as andrewkirk seemed to believe.

That's not what I wrote. What I wrote was

the answer to part (a) would have been the same if they had asked you for the equation to the tangent at (1.98,3,12.1) or (2,3.01,12.2).

The actual tangent planes at the three points are different, but the best estimates that can be made of the parameters of those three tangent planes, given only the info in the OP, are all the same, and they are the parameters of the plane that passes through the three points. So if they had replaced (2,3,12) by (1.98,3,12.1) or (2,3.01,12.2) in the question, the answer (which is about an estimate) would not have changed.