Query111mth

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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)

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:

f(0,0,0)=0*0*1+0^2=0

f(1,-1,1)=-20.09+9=-11.09

f(-1,1,-1)=.0498+9=9.0498

d/dx(f(x,x,x))=3e^(3x)(x^2+x^3)+2x

d/dy(f(1,y,1))=d/dy(1*y*e^3 + y^2-4y+4)=e^3+2y-4

d/dz(f(1,1,z^2))=d/dze^3+z^4=4z^3

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Good.

There are at least two ways to find the derivative of f(1, 1, z^2) with respect to z.

One is to substutute and take the derivative. We obtain

f(1, 1, z^2) = e^3 + z^4, and easily find the derivative to be 4 z^3. This is the approach you correctly used, and it is completely appropriate to this question.

Another more generally applicable way (more work that it's worth if all you want to know is the derivative of f(1, 1, z) ) is to find the derivative of f(x, y, z^2) with respect to z. This will involve the chain rule, applied to the 'innermost' function z^2 (the derivative of which is 2 z), and the result is

2 z * 2 (x - y + z)^2.

For x = y = 1, this reduces to the former result 4 z^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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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:

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.

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Given Solution:

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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:

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.

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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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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:

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

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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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&#Good responses. See my notes and let me know if you have questions. &#