this is the function: (x^3+6xy+3y^2-9x) this is the counter plot this is the 3D
ID: 3343265 • Letter: T
Question
this is the function:
(x^3+6xy+3y^2-9x)
this is the counter plot
this is the 3D plot
http://www.wolframalpha.com/input/?i=x%5E%283%29%2B6xy%2B3y%5E%282%29-9x
1)in this picture can you find and show the level curves for this function? and tell me how far they are spaced?
2)Label this diagram with x and y values and the levels corresponding to the curves in the figure and show all of the max, min and saddle point.
3)In one of your labelled diagrams draw in the gradient vector at the origin.
4)Choose a closed level curve. Write an integral to compute the area enclosed by your chosen level curve. Give an estimate for this area.
5)Integrate the function of over the region which is enclosed by the level curve you have considered above. Specify the tools you used and support your answer by explaining how you can estimate the value of this integral.
please use a programe like maple or mathmatica to explin your answers.
Explanation / Answer
Contrast on this page is pretty poor (!) but I think #13 says
f(x,y) = x^3 + 6xy + 3y^2 - 9x
First I'm going to try to figure out the numerical coordinates
of that peak or valley that appears as a small oval
in the right-hand lower part of the figure.
Partial derivative of f w/r/t x = 3x^2 + 6y - 9
Partial derivative of f w/r/t y = 6x + 6y
When are both of these zero?
When y = -x and 3x^2 - 6x = 9,
so 0 = x^2 - 2x - 3, and x is 3 or -1,
so y is -3 or +1.
At these two points, the second partials are
f_xx = 6x and f_yy = 6
Both are positive at (3,-3)
but they have opposite signs at (-1,1) which
then should be a saddle point.
So I deduce that the "cross" near the center of the figure is at (-1,1)
and the center of the small oval lower down is at (3,-3).
This latter point is the local minimum of the function,
where f(x,y) = -27.
The gradient of the function is
(3x^2 + 6y - 9) i + (6x+6y) j
At the origin this should be -9i,
which seems possible, given that the origin
is located 1/4 along the downward diagonal from the saddle point
to the trough at (3,-3).
The crossing contours through the saddle point
must have the value f(x,y) = 5,
and I'm going to guess that the three contours between
the saddle point and the minimum point have contours
-5, -15, and -25.
So I'm saying that the smallest closed curve at lower right
is f(x,y) = -25,
surrounded by the closed curve f(x,y) = -15,
which in turn is surrounded by the closed curve f(x,y) = -5.
But this is only a guess; a contour-plotting tool will get you a better answer.
In the "choose a closed level curve" part, I would choose maybe
f(x,y) < -15, i.e., the interior of the second-smallest closed circuit.
Cut it up into vertical strips.
The limits on x will be at the points where the partial w/r/t y is zero,
i.e., both along the line y = -x.
At these points, you also have
x^3 - 6x^2 + 3x^2 - 9x = -15, or
x^3 - 3x^2 - 9x + 15 = 0
I can use fooplot.com to find out that this is solved when
x = 1.3365 and x = 4.2836.
Hence I would be integrating from x = 1.3365 to x = 4.2836,
and the integrand is found from
3y^2 + 6xy + x^3 - 9x + 15 = 0
The quadratic formula gives:
y = -2x plus or minus (1/6)sqrt(36x^2 - 4x^3 + 36x - 60)
The "plus" gives you the top boundary of the vertical strip and
the "minus" gives you the bottom boundary.
So you'd be integrating
[-2x + (1/3) sqrt(36x^2 - 4x^3 + 36x - 60) ] dx
from x = 1.3365 to x = 4.2836.
It's not obvious that this can be integrated with elementary functions.
The area appears kind of elliptical with a major semi-axis of about 3
and a minor semi-axis of about 1+, so its area should be
something like pi(ab) = about 3pi+ = about 10, but you will do better
if you have access to a contour-plotting tool that gives you
more accurate numbers.
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