#$&*
course Mth 277
320pm 11/14/2011
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
*********************************************
Question: `q001. Let f(x,y,z) = x^2*y*e^3x + (x - y + z)^2. Find the following expressions.
f(0,0,0)
f(1,-1,1)
f(-1,1,-1)
d/dx(f(x,x,x))
d/dy(f(1,y,1))
d/dz(f(1,1,z^2))
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
f(x,y,z) = x^2*y*e^3x + (x - y + z)^2
f(0,0,0) = 0^2*0*e^30 + (0 - 0 + 0)^2
= 0
f(1,-1,1) = x^2*y*e^3x + (x - y + z)^2
= (1)^2*(-1)*e^3(1) + [(1) - (-1) + (1)]^2
= -e^3 + 9
=approx -11.1 units
f(-1,1,-1) = x^2*y*e^3x + (x - y + z)^2
= (-1)^2*(1)*e^3(-1) + [(-1) - (1) + (-1)]^2
= (-1)^2*(-1)*e^3(1) + [(1) - (-1) + (1)]^2
= [1*1*e^-3] + 9
= e^-3 + 9 =approx 9.1 units
f(x,y,z) = x^2*y*e^3x + (x - y + z)^2
(f(x,x,x)) = x^2*x*e^3x + (x - x + x)^2
= x^3*e^3x + x^2
d/dx(f(x,x,x)) = x^3*3e^3x + e^3x*(2x^2) + 2x
= 3x^3*e^3x + 2x^2*e^3x + 2x
f(x,y,z) = x^2*y*e^3x + (x - y + z)^2
(f(1,y,1)) = 1^2*y*e^3(1) + (1 - y + 1)^2
= ye^3 + (2 - y)^2
d/dy(f(1,y,1)) = y*3e^3 + e^3 - (2-y)/2
= e^3*(3y + 1) - (2-y)/2
f(x,y,z) = x^2*y*e^3x + (x - y + z)^2
(f(1,1,z^2)) = (1)^2*(1)*e^3(1) + (1 - 1 + z^2)^2
= e^3 + z^4
d/dz(f(1,1,z^2)) = 4z^3
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
2 y ( e^3 + 1) - 4. I don't understand how you got this answer. I plugged in the values for f(1,y,1) and then found the derivative. My answer came out to e^3*(3y + 1) - (2-y)/2
I also got a different answer for the derivative of z. I think the given solution is missing (z^2)^2 before the expression was derived. ?????????????????
------------------------------------------------
Self-critique rating: 3
@&
f(1, y, 1) = 1 * y * e^(3 * 1) + (1 - y + 1)^2 = y * e^3 + (2-y)^2.
The y derivative of this is
e^3 - 2 ( 2 - y).
f(1, 1, z) = e^3 + z^4, with z derivative 4 z^3.
*@
*********************************************
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
f(u,v) = sqrt(u cos v).
ucos(v) > 0
sometimes cos(v) can be negative as long as u is also negative.
the doman of f(u,v) is all real numbers, so long as the u and cos(v) are the same sign(+ or negative).
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question: `q003. Sketch and describe the level surface f(x,y,z) = 1 when f(x,y,z) = 2x^2 + 2z^2 - y.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
looking at the x-y plane: z= 2x^2 - y, looks like a parabola
looking at the x-z plane, y= 2x^2 + 2z^2, looks like an ellipse, unless x and z are the same values its a circle
looking at the y-z plane, x= 2z^2 - y, looks like a parabola
It is an elliptic paraboloid
I don't know where the 1 comes in, though???????????
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I still don't understand the value for 1.?????????????
------------------------------------------------
Self-critique rating:
@&
The level curve is f(x, y, z) = 1, giving you 2 x^2 + 2 z^2 - y = 1.
This is a circular paraboloid. Technically a circle is an ellipse, but it still should have been described as circular.
*@
*********************************************
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
PV = kT
k = (PV)/T
= (24lb/in^2 * 3500 in^3) / 270K
= 311.1 lb-in/K
T(P,V,k) = PV/k
T(24, 3500, 311.1) = PV/k
Isotherm is when temperature is constant.
I don't understand what else you are asking for.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I got 311 lb-in/K. I took a class of thermodynamics, so I understand the meaning behind PV=constant. I am assuming the reasoning behind the question is to understand a function with multiple variables.
------------------------------------------------
Self-critique rating: 3"
@&
The constant should have been 311 in * lb / K.
This doesn't change the nature of the level curves, but it certainly does change the values of the levels.
In general
P V = n k T
describes a level hypercurve (of 3 dimension) P V / (n T) = k of the 4-dimensional function
f(P, v, n, T) = P V / (n T)
If n is constant then
P V / T = n k
is a 2-dimensional level surface for the function g(P, V, T).
The techniques of this chapter provide the basis for a lot of insight into the nature of these functions, and a deeper insight into the thermodynamics of an ideal gas.
*@
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
&#Your work looks good. See my notes. Let me know if you have any questions. &#