# Thread: Finding a Critical Point

1. ## Finding a Critical Point

Lets make sure I'm doing this right

I need to find the critical points for the function
f (x, y) = e^x * sin(y)

I need to find where the slopes are horizontal or 0, so I differentiate for both x and y

I get df(x, y) for x = e^x * sin(y) and df(x, y) for y = e^x * cos(y)

For the derrivative with respect to x the only time the slope is 0 is when y=pi and for the derivative with respect to y the only time the slope is 0 is when y=pi/2. So my critical points are (pi, pi/2)

Am I right or wrong?

Also how do I tell if the critical point is a local max, local min or a saddle point?

2. Originally Posted by Ranger SVO
Lets make sure I'm doing this right

I need to find the critical points for the function
f (x, y) = e^x * sin(y)

I need to find where the slopes are horizontal or 0, so I differentiate for both x and y

I get df(x, y) for x = e^x * sin(y) and df(x, y) for y = e^x * cos(y)

For the derrivative with respect to x the only time the slope is 0 is when y=pi and for the derivative with respect to y the only time the slope is 0 is when y=pi/2. So my critical points are (pi, pi/2)

Am I right or wrong?

Also how do I tell if the critical point is a local max, local min or a saddle point?
You are doing the derivatives right, but you have chosen an unfortunate notation. Might I suggest you call the "x" partial derivative something like fx and the "y" partial derivative fy, or some such. The problem is that df is already defined as:
$df(x,y) = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$

As far as local max's, etc. A function f(x,y) only has a local maximum or local minimum when the partial derivative in all directions is zero. A saddle point is where the function has a local maximum in one direction (say the x) and a local minimum in another (say the y), where the derivatives are taken at the same point.

Also, your critical points are solved correctly (though for fx = 0 the solution is more generally y = n $\pi$, where n is an integer and a similar comment holds for your fy solution.) And again, being a bit picky but with reason, we should probably label ALL the coordinates to label the critical point, just to be clear. So your first critical "point" would be (x, 0, 0) (so it's really the whole x-axis) and the second would be (x, $\pi /2$, $e^x$).

Since the fx critical "point" is not a location where fy is zero the x-axis is neither a local max nor min for the function. The same comment holds for the second critical point. (And thus neither are they saddle points.)

-Dan

3. Originally Posted by Ranger SVO
Lets make sure I'm doing this right

I need to find the critical points for the function
f (x, y) = e^x * sin(y)

I need to find where the slopes are horizontal or 0, so I differentiate for both x and y

I get df(x, y) for x = e^x * sin(y) and df(x, y) for y = e^x * cos(y)

For the derrivative with respect to x the only time the slope is 0 is when y=pi and for the derivative with respect to y the only time the slope is 0 is when y=pi/2. So my critical points are (pi, pi/2)

Am I right or wrong?

Also how do I tell if the critical point is a local max, local min or a saddle point?
Now I thought the critical points of a function of two variables are
the points where the tangent plane is horizontal, which implies that
the partial derivatives with respect to both the variables will simultaneously
be zero, that is:

$
\frac{\partial f}{\partial x} = 0
$
and $\frac{\partial f}{\partial y} = 0$

Which in this case would be at a point $(x,y) \in \mathbb{R}^2$ where:

$
e^x \sin(y) = e^x \cos(y)
$

but there is no such point, so there are no critical points.

RonL

4. Originally Posted by Ranger SVO
...
I need to find the critical points for the function
f (x, y) = e^x * sin(y)...
Hi,

have a look at the graph of your function. You'll notice that there are no such points. CaptBlack has already demonstrated how to get this result.

5. Originally Posted by CaptainBlack
Now I thought the critical points of a function of two variables are
the points where the tangent plane is horizontal,
Also where the the function is not differenciable on some open continous disk.
----
Alternatively, you show that the function is differenciable (show it is countinous and the partials are countinous) then you seach for points where,
$\nabla f=\bold{0}$
(Which is exactly what CaptainBlank) said.

6. Thanks everyone

Earboth how did you graph that.

7. Originally Posted by Ranger SVO
Earboth how did you graph that.
I know, I wish I had that too.
I asked him one time, he has Maple.
My College uses MATLAB, but people say it is the worst.
These programs are really expensive if you want to buy them honestly.

8. Originally Posted by ThePerfectHacker
I know, I wish I had that too.
I asked him one time, he has Maple.
My College uses MATLAB, but people say it is the worst.
These programs are really expensive if you want to buy them honestly.
There are a number of FREE substitutes for MATLAB (and Maple for that
matter). My favourite is Euler but there are also SciLab and Octave
to name two.

An example of this function plotted using Euler is attached (axes and surface
grid lines can be added if desired - admittedly not as naturally as in Maple
and possibly MATLAB but that could be automated if desired).

RonL

9. Originally Posted by Ranger SVO
...
Earboth how did you graph that.
Hi,

I use DERIVE 6.0, which now is distributed by TI.

(As you may have noticed, the TPH didn't pay attention when I told him, which program I use . I'm used to it - it's the fate of a teacher)

EB

10. I'm wondering if there is another way to tell if I have a local max, local min or local, or a saddle point. If I put the second derivatives in a Matrix

and look at the first element and the determinant, I should be able to tell.

If the first element is greater than 0 and the determinant is greater than 0, I should have a local minimum.
If the first element is less than 0 and the determinant is greater than 0, I should have a local maximum.

Is this right so far?

Can I use the same method for a saddle points?

I have got to get Derive. Thats awesome

11. Originally Posted by Ranger SVO

Is this right so far?

Can I use the same method for a saddle points?
Consider,
$D=f_{xx}f_{yy}-f_{xy}^2$
If, $D>0$ then you check the sign of the function at that point. If it is negative then you have a maximum if it is negative you have a minimum.

If $D<0$ a saddle point.

If $D=0$ it is inconclusive.