#$&*
course Mth 277
3/21 1
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
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Question: `q001. Let f(x,y,z) = x^2*y*e^3x + (x - y + z)^2. Find the following expressions.
• f(0,0,0)
• f(1,-1,1)
• f(-1,1,-1)
• d/dx(f(x,x,x))
• d/dy(f(1,y,1))
• d/dz(f(1,1,z^2))
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Your solution:
1) f(0,0,0) = 0^2*0*e^3*0 + (0 - 0 + 0)^2 = 0
2) f(1,-1,1) = 1^2*(-1)*e^3 + (1 + 1 + 1)^2
= -e^3 + 3^2 = 9 - e^3 = approx. -11
3) f(-1,1,-1) = (-1)^2*1*e^3*(-1) + ((-1) - 1 + (-1))^2
= e^-3 + -3^2 = 9 + e^3 = approx. 9
4) d/dx(f(x,x,x)) =d/dx[ x^2*x*e^3x + (x - x + x)^2]
x^2*x*e^3x + (x - x + x)^2 = x^3e^3x + x^2
d/dx[x^3e^3x + x^2] = d/dx(x^3e^3x) + d/dx(x^2)
d/dx(x^3e^3x) = (x^3)’*e^(3x) + x^3*(e^(3x))’ = 3x^2e^(3x) + 3x^3e^(3x)
d/dx(x^2) = 2x
d/dx[x^3e^3x + x^2] = 3x^2e^(3x) + 3x^3e^(3x) + 2x
= 3e^(3x)(x^2 + x^3) + 2x
5) d/dy(f(1,y,1)) = 1^2*y*e^3 + (1 - y + 1)^2
1^2*y*e^3 + (1 - y + 1)^2 = ye^3 + (2-y)^2 = ye^3 + (y^2 - 4y +4)
d/dy[ye^3 + (y^2 - 4y +4)] = d/dy(ye^3) + d/dy(y^2 - 4y +4)
d/dy(ye^3) = e^(3)*d/dy(y) = e^3*1 = e^3
d/dy(y^2 - 4y +4) = 2y - 4
d/dy[ye^3 + (y^2 - 4y +4)] = e^3 + 2y -4
= approx. 16 + 2y
6) d/dz(f(1,1,z^2)) = 1^2*1*e^3 + (1 - 1 + z^2)^2
1^2*1*e^3 + (1 - 1 + z^2)^2 = e^3 + z^4
d/dz[e^3 + z^4] = d/dz(e^3) + d/dz(z^4)
d/dx(e^3) = 0
d/dx(z^4) = 4z^3
d/dx[ye^3 + (y^2 - 4y +4)] = 0 + 4z^3
= 4z^3
Confidence rating: 3
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Given Solution:
f(x, x, x) = x^2 * x * e^(3x) + (x - x + x)^2 = x^3 e^(3x) + x^2.
So (d/dx) f(x, x, x) is the x derivative of this expression, equal to
(x^3) ' * e^(3x) + x^3 * e^(3x) ' + (x^2) ' = 3 x^2 e^(3x) + 3 x^3 e^(3x) + 2 x = 3 (x^2 + x^3) e^(3x) + 2x.
f(1, y, 1) = 1 * y^2 * e^(3 * 1) + (1 - y + 1)^2 = y^2 * e^3 + (2-y)^2.
The derivative with respect to y of this expression is 2 y e^3 - 2 ( 2 - y), which simplifies to 2 y ( e^3 + 1) - 4.
f(1, 1, z^2) = e^3 + z^2; the z derivative of this expression is 2 z.
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Self-critique (if necessary):
when you plug in z^2 into equation would it not go in the expression as follows 1^2*1*e^3 + (1 - 1 + z^2)^2 and there for be z^4 that you take derivative of????
@& Right. Should have read
f(1, 1, z^2) = e^3 + z^4; the z derivative of this expression is 4 z^3.*@
Also wouldn’t for f(1, y, 1) and using the equation “Let f(x,y,z) = x^2*y*e^3x + (x - y + z)^2” then when you sub in y it is just y and not y^2 giving you the solution e^3 + 2y -4 because y*e^3 has derivative of e^3 when taken with respect to y.?????????????
@& Again right.
f(1,y,1) = 1^2 * y * e^(3 * 1) + (1 - y + 1)^2 = y e^3 + (2 - y)^2
The y derivative is e^3 - 2 ( 2 - y) = e^3 - 4 + 2 y.*@
Not nit picking just want to make sure that I’m not missing something wrt plugging in and solving functions correctly.
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Question: `q002. Find the domain and range of the function f(u,v) = sqrt(u cos v).
sqrt( u cos(v)) is defined when u cos(v) >= 0.
This occurs when u >= 0 and cos(v) >= 0, or when u <= 0 and cos(v) <= 0.
u >= 0 on the right-hand half of the u-v plane.
cos(v) >= 0 when -pi/2 <= v <= pi/2, or more generally when -pi/2 + 2 n pi <= v <= pi/2 + 2 n pi, for n = ..., -2, -1, 0, 1, 2, ... . The corresponding regions of the u-v plane are alternating infinite horizontal strips of width pi.
The domain corresponding to u >= 0 and cos(v) >= 0 are therefore alternating horizontal strips in the right half-plane.
The domain corresponding to u <= 0 and cos(v) <= 0 are alternating the horizontal strips in the left half-plane corresponding to pi/2 + 2 n pi <= v <= 3 pi/2 + 2 n pi.
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Your solution:
I’m not sure I understand question??????
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& The question was
Question: `q002. Find the domain and range of the function f(u,v) = sqrt(u cos v).
The rest of that stuff is the solution, which I neglected to move to its rightful place.
Basically u cos(v) can't be negative.
For positive u this means cos(v) can't be negative, so v can't be between pi / 2 and 3 pi / 2, or between 5 pi/2 and 7 pi / 2, etc..
For negative u cos(v) can't be positive, so v can't be between -pi/2 and pi/2, or between 3 pi/2 and 5 pi/2, etc..
*@
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Question: `q003. Sketch and describe the level surface f(x,y,z) = 1 when f(x,y,z) = 2x^2 + 2z^2 - y.
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Your solution:
We would have f(x,y,z) = 2x^2 + 2z^2 - y = 1, so I’m kind of confused would this form a level curve that or circular paraboloid that has all the same characteristics except that it extends in the y direction where most are of the form x^2+y^2 = z and extend in the z direction?
I can at least visualize this graph its all starting to come together.
confidence rating #$&*:
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Given Solution:
This corresponds to the equation 2 x^2 + 2 z^2 - y = 1. This is a quadric surface, an elliptic paraboloid. Its intersection with any plane parallel to the x-y plane, and also with any plane parallel to the y-z plane, is a parabola. Its intersection with any surface parallel to the x-z plane is either an ellipse (for y < -1), the point (0, 0, -1) for the plane y = -1, and empty for y > -1.
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Self-critique (if necessary):
Would it not be the x-z parallel planes that formed the parabola???
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Self-critique rating:2
@& In any plane parallel to the x-z plane y would be constant, so f(x, y, z) = 1 would mean that x^2 + y^2 = 1/2 + y/2. That would be a circle of radius sqrt(1/2 + y/2), provided y > = -1. If y < -1, you would not get a point.
I called that an ellipse in the given solution, and since a circle is an ellipse that is technically correct. However it would have been more specific to call it a circle.*@
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Question: `q004.
According to the ideal gas law, PV = kT where P is pressure, V is volume, T is temperature, and k is some constant. Suppose a tank contains 3500in^3 of some gas at a pressure of 24lb/in^2 when the temperature is 270K.
• Determine k for this gas.
• Express T as a function of P and V using the k found in the previous step and describe the isotherms.
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Your solution:
24lb/in^2*3500in^3 = 84000lb/in = k*270K
k = 84000lb/in/270K = approx. 311lbK/in
T(x, y, z) = 373 - x^2 - y^2 - z^2 Kelvin.
Which gives us x^2 + y^2 + z^2 = 373 - k, We found k to be = 311K
We have sphere of radius `sqrt(373 - 311), so as temp increases radius dercreases and the same for if temp decreases radius increases.
confidence rating #$&*:
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Given Solution:
k = P V / T = 24 lb/in^3 * 3500 in^3 / (270 K) = 320 lb / K, approx..
So T = P V / k = P V / (320 lb/K) = .003 K / lb * P V.
An isotherm occurs when T is constant, in which case
P V = constant
and
P = constant / V.
This is a hyperbola in the P V plane, asymptotic to the x and y axes, with the line P = V as the axis of symmetry. (very similar to the graph of y = 1 / x).
The constant is ( .003 K / lb ) / T. The greater the value of T, the greater the constant and the further the hyperbola's closest approach to the origin.
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Self-critique (if necessary):
I thought we didn’t have inches in the units for k, so for T(0, 0, 0) should our center always be boinging temp of whatever temp scale we are using or can it vary to fit the situation as well??
Never mind I just reread the solution only now I don’t understand how you get .003K/lb/T, wouldn’t that be equal to 1/PV, os that P = (.003K/lb*V)/T not = constant/V??????????????????????
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Self-critique rating:
@& Corrected solution:
k = P V / T = 24 lb/in^3 * 3500 in^3 / (270 K) = 320 lb * in / K, approx..
So T = P V / k = P V / (320 lb/K) = .003 K / (lb in) * P V.
An isotherm occurs when T is constant, in which case
P V = constant = k T = 320 lb in / K * T
and
P = constant / V.
This is a hyperbola in the P V plane, asymptotic to the x and y axes, with the line P = V as the axis of symmetry. (very similar to the graph of y = 1 / x).
The constant is 320 lb in / K * T
The greater the value of T, the greater the constant and the more distant the hyperbola's closest approach to the origin.*@
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#*&!
&#Good responses. See my notes and let me know if you have questions. &#