course Mth 272
wxx?B????M?·????{···assignment #028028.
Applied Calculus II
04-27-2009
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22:01:49
Query Problem 7.4.8 fy for xy / (x^2+y^2)
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RESPONSE -->
fx= (x^2+y^2)(y)-(xy)(2x) / (x^2+y^2)^2 differentiating
fx= (x^2+y^3 - 2x^2y)/(x^2+y^2)^2
fy= (x^2+y^2)(x)- (xy)(2y)/ (x^2+y^2)^2
fy= (x^3+xy^2- 2xy^2)/ (x^2+y^2)^2
confidence assessment: 3
Your expression simplifies to
[ x^3 - x y^2 ] / (x^2 + y^2) ^ 2 or
x [ x^2 - y^2 ] / (x^2 + y^2) ^ 2 . **
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22:02:00
What is the requested partial derivative?
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RESPONSE -->
answered
confidence assessment: 3
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22:04:44
Query problem 7.4.32 wx, wy, wz at origin for w = 1/sqrt(1-x^2-y^2-z^2)
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RESPONSE -->
wx= 1(1-x^2-y^2-z^2)^-1/2
wx=-1/2(1-x^2-y^2-z^2)^-3/2(-2x)
=(x)(1-x^2-y^2-z^2)^-3/2
for wy= (y)(1-x^2-y^2-z^2)^-3/2
for wz= (z)(1-x^2-y^2-z^2)^-3/2
confidence assessment: 3
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22:04:52
What are the values of the three requested partial derivatives at the specified point?
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RESPONSE -->
answered
confidence assessment: 3
You are correct that the partial derivatives are
wx = -2x * -1/2 ( 1 - z^2 - y^2 - z^2)^(-3/2) = x / (1 - x^2 - y^2 - z^2)^3/2
wy = -2y * -1/2 ( 1 - z^2 - y^2 - z^2)^(-3/2) = y / (1 - x^2 - y^2 - z^2)^3/2
wz = -2z * -1/2 ( 1 - z^2 - y^2 - z^2)^(-3/2) = z / (1 - x^2 - y^2 - z^2)^3/2
At the origin we have x = y = z = 0 so that the three partial derivatives are all of form 0 / 1 = 0.
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22:05:10
What are the three partial derivative functions?
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RESPONSE -->
answered
confidence assessment: 3
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22:13:09
Query problem 7.4.40 (was 7.4.36) fx = fy = 0 for 3x^3-12xy+y^3
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RESPONSE -->
fx= 9x^2-12y= 0
fy= -12x+3y^2= 0
points where it equals zero for fx= (4,12)
for fy= (4,4)
confidence assessment: 3
The partial derivatives are respectively zero at these points, but they are not both zero at either point. There are in fact infinitely many points where fx = 0 (any point where y = 3/4 x^2 makes fx zero, and for any value of x you can easily find the necessary value of y). Similarly there are infinitely may points where fy = 0. You need to find the point(s) where both are true simultaneously.
fx= 9x^2-12y = 0 when y = 3/2 x^2
fy= -12x+3y^2 = 0 when x = 1/4 y^2
Geometrically:
fx = 0 when y = 3/4 x^2, which describes a parabola with vertex at the origin, axis of symmetry conciding with the y axis, opening upward and passing thru the points (1, 3/4) and (-1, 3/4).
fy = 0 when x = 1/4 y^2, which describes a parabola with vertex at the origin, axis of symmetry conciding with the x axis, opening to the right and passing thru the points (0, 1/4) and (0, -1/4).
If you sketch these parabolas it should be clear that they concide at one point in the first quadrant. You should estimate the coordinates of these points.
Then proceed to solve these equations simultaneously to find the accurate coordinates:
**Algebraically:
The partial derivatives are
fx = 9x^2 - 12y
fy = -12x + 3y^2.
If the two partial derivatives are zero we get the equations
9x^2 - 12 y = 0
-12x + 3 y^2 = 0.
Solving the first equation for y we get y = 3x^2 / 4.
Substituting this expression for y in the second we have
-12 x + 3 ( 3x^2 / 4)^2 = 0 so
-12 x + 27 x^4 / 16 = 0. Factoring out -3x:
-3x ( 4 - 9 x^3 / 16) = 0. This is so if
-3x = 0 or 4 - 9 x^3 / 16 = 0.
-3x = 0 gives solution x = 0.
4 - 9 x^3 / 16 = 0 if x^3 = 4 * 16 / 9, which happens when
x = 64^(1/3) / 9^(1/3) = 4 * 3^(-2/3), which is expressed in standard form as 4 * 3 ^(1/3) / 3.
If x = 0 then since 9 x^2 - 12 y = 0 we have y = 0.
If x = 4 * 3^(1/3)/3 then 9 x^2 - 12 y = 0 gives us
9 [ 4 * 3^(1/3)/3 ] ^2 - 12 y = 0 so
y = 9 [ 4 * 3^(1/3)/3 ] ^2 / 12 = 3/4 * 16 * 3^(2/3)/9 = 4 * 3^(2/3) / 3.
So the two partials are both zero at (0,0) and at ( 4 * 3^(1/3)/3, 4 * 3^(2/3)/3.
2-place approximations tell us the point is near (1.9, 8.3); this result should be consistent with your sketch, though your sketch won't be nearly this precise. **
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22:13:18
What are the coordinates of the point or points where the two partial derivatives are both zero?
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RESPONSE -->
answered
confidence assessment: 3
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22:13:58
What system of simultaneous equations did you solve to get your result?
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RESPONSE -->
i just did the partial derivatives, then solved. maybe i did not answer quite correctly, could you explain. Thankyou.
confidence assessment: 3
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&#Your work looks good. See my notes. Let me know if you have any questions. &#