#$&* course Mth 277 10/29 2 *********************************************
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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: domain is the set of all ordered pairs (x,y) where sqrt(ucos(v)) is defined where ucos(v) >= 0 or <=0. range is z=sqrt(ucos(v)) when again ucos(v) >= 0 or <=0. being that the sqrt cannot be defined when it is negative. thus two negatives result in a positive, which is why you can have u and v < 0. the range for f is the set of all numbers z = sqrt(ucos(v)) for (x,y) in the domain ucos(v) >=0. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The solution given was actually in the question, but i'm fuzzy on how you chose pi/2 +2n for the domain values. ------------------------------------------------ Self-critique 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): How do you know it just corresponds to that equation? I dont see this equation anywhere in the book. how many other equations similar to this are there or do we need to know? ------------------------------------------------ Self-critique 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see that you have in^3 where in the question was in^2 but nonetheless its just algebra. I messed up on the isotherm part, I said T AND K should be constant, but now that i think about it, temperature changes as pressure and volume changes, so if temperature was constant, nothing else would change, so If only T is constant, you are left with like you said P V = constant. ------------------------------------------------ Self-critique rating:"