we have shown that the expression ρ(x,t) = [1/(1 + t^2)] ρ0 [(x/(1 + t^2)) - t] satisfies the advection equation ∂ρ/∂t + v ∂ρ/∂x = -ρ ∂v/∂x.
The density of radioactive material, denoted by ρ(x,t), evolves according to the equation:
∂ρ/∂t + v ∂ρ/∂x = -ρ ∂v/∂x
This equation describes the transport of a substance by a moving medium, where the rate of movement of the radioactive material is influenced by the velocity of the wind, determined by the function v(x,t).
To solve the equation, we use the method of characteristics. We define the characteristic equation as:
x = ξ(t)
and
ρ(x,t) = f(ξ)
where f is a function of ξ.
Using the method of characteristics, we find that:
∂ρ/∂t = (∂f/∂t)ξ'
∂ρ/∂x = (∂f/∂ξ)ξ'
where ξ' = dξ/dt.
Substituting these derivatives into the original equation, we have:
(∂f/∂t)ξ' + v(∂f/∂ξ)ξ' = -ρ ∂v/∂x
Dividing by ξ', we get:
(∂f/∂t)/(∂f/∂ξ) = -ρ ∂v/∂x / v
Letting k(x,t) = -ρ ∂v/∂x / v, we can integrate the above equation to obtain f(ξ,t). Since f(ξ,t) = ρ(x,t), we can express the solution ρ(x,t) in terms of the initial value of ρ and the function k(x,t).
Now, let's solve the advection equation using the method of characteristics. We define the characteristic equation as:
x = x(t)
Then, we have:
dx/dt = v(x,t)
ρ(x,t) = f(x,t)
We need to find the function k(x,t) such that:
(∂f/∂t)/(∂f/∂x) = k(x,t)
Differentiating dx/dt = v(x,t) with respect to t, we have:
dx/dt = (2xt)/(1 + t^2) + 1 + t^2
Integrating this equation with respect to t, we obtain:
x = (x(0) + 1)t + x(0)t^2 + (1/3)t^3
where x(0) is the initial value of x at t = 0.
To determine the function C(x), we use the initial condition ρ(x,0) = ρ0(x).
Then, we have:
ρ(x,0) = f(x,0) = F[x - C(x), 0]
where F(ξ,0) = ρ0(ξ).
Integrating dx/dt = (2xt)/(1 + t^2) + 1 + t^2 with respect to x, we get:
t = (2/3) ln|2xt + (1 + t^2)x| + C(x)
where C(x) is the constant of integration.
Using the initial condition, we can express the solution f(x,t) as:
f(x,t) = F[x - C(x),t] = ρ0 [(x - C(x))/(1 + t^2)]
To simplify this expression, we introduce A(x,t) = (2/3) ln|2xt + (1 + t^2)x|/(1 + t^2). Then, we have:
f(x,t) = [1/(1 +
t^2)] ρ0 [(x - C(x))/(1 + t^2)] = [1/(1 + t^2)] ρ0 [(x/(1 + t^2)) - A(x,t)]
Finally, we can write the solution to the advection equation as:
ρ(x,t) = [1/(1 + t^2)] ρ0 [(x/(1 + t^2)) - A(x,t)]
where A(x,t) = (2/3) ln|2xt + (1 + t^2)x|/(1 + t^2).
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In the figure, the square ABCD and the AABE are standing on the same base AB and between the same parallel lines AB and DE. If BD = 6 cm, find the area of AEB.
To find the area of triangle AEB, we use base AB (6 cm) and height 6 cm. Applying the formula (1/2) * base * height, the area is 18 cm².
To find the area of triangle AEB, we need to determine the length of the base AB and the height of the triangle. Since both square ABCD and triangle AABE is standing on the same base AB, the length of AB remains the same for both.
We are given that BD = 6 cm, which means that the length of AB is also 6 cm. Now, to find the height of the triangle, we can consider the height of the square. Since AB is the base of both the square and the triangle, the height of the square is equal to AB.
Therefore, the height of triangle AEB is also 6 cm. Now we can calculate the area of the triangle using the formula: Area = (1/2) * base * height. Plugging in the values, we get Area = (1/2) * 6 cm * 6 cm = 18 cm².
Thus, the area of triangle AEB is 18 square centimeters.
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In triangle ABC the angle bisectors drawn from vertices A and B intersect at point D. Find m
m
The measure of angle ADB is equal to the square root of ([tex]AB \times BA[/tex]).
In triangle ABC, let the angle bisectors drawn from vertices A and B intersect at point D. To find the measure of angle ADB, we can use the angle bisector theorem. According to this theorem, the angle bisector divides the opposite side in the ratio of the adjacent sides.
Let AD and BD intersect side BC at points E and F, respectively. Now, we have triangle ADE and triangle BDF.
Using the angle bisector theorem in triangle ADE, we can write:
AE/ED = AB/BD
Similarly, in triangle BDF, we have:
BF/FD = BA/AD
Since both angles ADB and ADF share the same side AD, we can combine the above equations to obtain:
(AE/ED) * (FD/BF) = (AB/BD) * (BA/AD)
By substituting the given angle bisector ratios and rearranging, we get:
(AD/BD) * (AD/BD) = (AB/BD) * (BA/AD)
AD^2 = AB * BA
Note: The solution provided assumes that points A, B, and C are non-collinear and that the triangle is non-degenerate.
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Bearing used in an automotive application is supposed to have a nominal inside diameter 1.5 inches. A random sample of 25 bearings is selected, and the average inside diameter of these bearings is 1.4975 inches. Bearing diameter is known to be normally distributed with standard deviation σ=0.1 inch. We want to test the following hypothesis at α=0.01. H0:μ=1.5,H1:μ=1.5 (a) Calculate the type II error if the true mean diameter is 1.55 inches. (b) What sample size would be required to detect a true mean diameter as low as 1.55 inches if you wanted the power of the test to be at least 0.9 ?
(a) Without knowing the effect size, it is not possible to calculate the type II error for the given hypothesis test. (b) To detect a true mean diameter of 1.55 inches with a power of at least 0.9, approximately 65 bearings would be needed.
(a) If the true mean diameter is 1.55 inches, the probability of not rejecting the null hypothesis when it is false (i.e., the type II error) depends on the chosen significance level, sample size, and effect size. Without knowing the effect size, it is not possible to calculate the type II error.
(b) To calculate the required sample size to detect a true mean diameter of 1.55 inches with a power of at least 0.9, we need to know the chosen significance level, the standard deviation of the population, and the effect size.
Using a statistical power calculator or a sample size formula, we can determine that a sample size of approximately 65 bearings is needed.
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Jocelyn estimates that a piece of wood measures 5.5 cm. If it actually measures 5.62 cm, what is the percent error of Jocelyn’s estimate?
Answer:
The percent error is -2.1352% of Jocelyn's estimate.
(a) Find the work done by a force 5 i^ +3 j^ +2 k^ acting on a body which moves from the origin to the point (3,−1,2). (b) Given u =− i^ +2 j^ −1 k^and v = 2l −1 j^ +3 k^ . Determine a vector which is perpendicular to both u and v .
a) The work done by the force F = 5i + 3j + 2k on a body moving from the origin to the point (3, -1, 2) is 13 units.
b) A vector that is perpendicular to both u = -i + 2j - k and v = 2i - j + 3k is -6i - 7j - 3k.
a) The work done by a force F = 5i + 3j + 2k acting on a body that moves from the origin to the point (3, -1, 2) can be determined using the formula:
Work done = ∫F · ds
Where F is the force and ds is the displacement of the body. Displacement is defined as the change in the position vector of the body, which is given by the difference in the position vectors of the final point and the initial point:
s = rf - ri
In this case, s = (3i - j + 2k) - (0i + 0j + 0k) = 3i - j + 2k
Therefore, the work done is:
Work done = ∫F · ds = ∫₀ˢ (5i + 3j + 2k) · (ds)
Simplifying further:
Work done = ∫₀ˢ (5dx + 3dy + 2dz)
Evaluating the integral:
Work done = [5x + 3y + 2z]₀ˢ
Substituting the values:
Work done = [5(3) + 3(-1) + 2(2)] - [5(0) + 3(0) + 2(0)]
Therefore, the work done = 13 units.
b) To find a vector that is perpendicular to both u = -i + 2j - k and v = 2i - j + 3k, we can use the cross product of the two vectors:
u × v = |i j k|
|-1 2 -1|
|2 -1 3|
Expanding the determinant:
u × v = (-6)i - 7j - 3k
Therefore, a vector that is perpendicular to both u and v is given by:
u × v = -6i - 7j - 3k.
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