#$&* course Mth 277 Question: `q001. Let f(x,y,z) = x^2*y*e^3x + (x - y + z)^2. Find the following expressions.f(0,0,0)
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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: since the expression is inside a square root it has to be positive to result in a real number. Therefore, ucos(v) has to be greater than or equal to 0. Therefore, we need to look when each, u and cos(v), is negative and positive. The cos of a number is positive in the 1 and 4 quadrant. Therefore, the number has to be between any multiples of (pi/2,0) and (3pi/4,1 pi). So when v is in one of these ranges, u has to be positive. When v is outside that range, u has to be negative. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 2x^2 + 2z^2 - y=1 XYplane y=2x^2-1 this is obviously a parabola that starts from (-1,0). YZ plane y=2z^2-1 This is the same parabola except that it is now in a different plane. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This corresponds to the equation 2 x^2 + 2 z^2 - y = 1. This is a quadric surface, an elliptic paraboloid. Its intersection with any plane parallel to the x-y plane, and also with any plane parallel to the y-z plane, is a parabola. Its intersection with any surface parallel to the x-z plane is either an ellipse (for y < -1), the point (0, 0, -1) for the plane y = -1, and empty for y > -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. According to the ideal gas law, PV = kT where P is pressure, V is volume, T is temperature, and k is some constant. Suppose a tank contains 3500in^3 of some gas at a pressure of 24lb/in^2 when the temperature is 270K. Determine k for this gas. Express T as a function of P and V using the k found in the previous step and describe the isotherms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: PV=kT k=PV/T=24lb/in^3*3500in^3/(270K)=310lb*in/K Since T is equal on an isotherm we can remove that variable from the equation. k=PV V=k/p P=k/V T=.00323*V*T confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Self-critique (if OK ------------------------------------------------ Self-critique rating:"