The Ideal Gas Law The pressure, temperature, and volume of an ideal gas are related by PV = kT, where k > 0 is a constant. Any two of the variables may be considered independent, which determines the dependent variable. a. Use implicit differentiation to compute the partial derivatives ∂ P ∂ V , ∂ T ∂ P , and ∂ V ∂ T . b. Show that ∂ P ∂ V ∂ T ∂ P ∂ V ∂ T = −1 (See Exercise 75 for a generalization.)
The Ideal Gas Law The pressure, temperature, and volume of an ideal gas are related by PV = kT, where k > 0 is a constant. Any two of the variables may be considered independent, which determines the dependent variable. a. Use implicit differentiation to compute the partial derivatives ∂ P ∂ V , ∂ T ∂ P , and ∂ V ∂ T . b. Show that ∂ P ∂ V ∂ T ∂ P ∂ V ∂ T = −1 (See Exercise 75 for a generalization.)
The Ideal Gas Law The pressure, temperature, and volume of an ideal gas are related by PV = kT, where k > 0 is a constant. Any two of the variables may be considered independent, which determines the dependent variable.
a. Use implicit differentiation to compute the partial derivatives
∂
P
∂
V
,
∂
T
∂
P
,and
∂
V
∂
T
.
b. Show that
∂
P
∂
V
∂
T
∂
P
∂
V
∂
T
= −1 (See Exercise 75 for a generalization.)
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
Write the given third order linear equation as an equivalent system of first order equations with initial values.
Use
Y1 = Y, Y2 = y', and y3 = y".
-
-
√ (3t¹ + 3 − t³)y" — y" + (3t² + 3)y' + (3t — 3t¹) y = 1 − 3t²
\y(3) = 1, y′(3) = −2, y″(3) = −3
(8) - (888) -
with initial values
Y
=
If you don't get this in 3 tries, you can get a hint.
Question 2
1 pts
Let A be the value of the triple integral
SSS.
(x³ y² z) dV where D is the region
D
bounded by the planes 3z + 5y = 15, 4z — 5y = 20, x = 0, x = 1, and z = 0.
Then the value of sin(3A) is
-0.003
0.496
-0.408
-0.420
0.384
-0.162
0.367
0.364
Question 1
Let A be the value of the triple integral SSS₂ (x + 22)
=
1 pts
dV where D is the
region in
0, y = 2, y = 2x, z = 0, and
the first octant bounded by the planes x
z = 1 + 2x + y. Then the value of cos(A/4) is
-0.411
0.709
0.067
-0.841
0.578
-0.913
-0.908
-0.120
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