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Math Help - Finding a Critical Point

  1. #1
    Member Ranger SVO's Avatar
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    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?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ranger SVO View Post
    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
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Ranger SVO View Post
    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:

    <br />
\frac{\partial f}{\partial x} = 0<br />
and \frac{\partial f}{\partial y} = 0

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

    <br />
e^x \sin(y) = e^x \cos(y)<br />

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

    RonL
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  4. #4
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    Quote Originally Posted by Ranger SVO View Post
    ...
    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.
    Attached Thumbnails Attached Thumbnails Finding a Critical Point-ex_siny.gif  
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    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.
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  6. #6
    Member Ranger SVO's Avatar
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    Thanks everyone

    Earboth how did you graph that.
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  7. #7
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    Quote Originally Posted by Ranger SVO View Post
    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.
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    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
    Attached Thumbnails Attached Thumbnails Finding a Critical Point-gash.jpg  
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  9. #9
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    Quote Originally Posted by Ranger SVO View Post
    ...
    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
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  10. #10
    Member Ranger SVO's Avatar
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    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
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  11. #11
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    Quote Originally Posted by Ranger SVO View Post

    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.
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