ANSWER ALL QUESTIONS. BE SURE TO ANSWER 7. A sphere of radius a is… ANSWER ALL QUESTIONS

ANSWER ALL QUESTIONS. BE SURE TO ANSWER 7. A sphere of radius a is… ANSWER ALL QUESTIONS. BE SURE TO ANSWER 7. A sphere of radius a is moving in a fluid with the constant velocity u. Show that the fluid satisfies the boundary condition (v ? u) · (r ? ut) = 0 at the surface of the sphere, if the center of the sphere coincides with the origin at t = 0 and v denotes the velocity of the fluid.  Solve the differential equations, Equation 4.1.13 and Equation 4.1.14 with the initial conditions x(0) = y(0) = y ? (0) = 0, and x ? (0) = A0 p g/L assuming that  g/L. 26. If a fluid is bounded by a fixed surface f(x, y, z) = c, show that the fluid must satisfy the boundary condition v · ?f = 0, where v is the velocity of the fluid s, if the divergence is positive, either the fluid is expanding and its density at the point is falling with time, or the point is a source at which fluid is entering the field. When the divergence is negative, either the fluid is contracting and its density is rising at the point, or the point is a negative source or sink at which fluid is leaving the field. If the divergence of a vector field is zero everywhere within a domain, then the flux entering any element of space exactly balances that leaving it and the vector field is called nondivergent or solenoidal (from a Greek word meaning a tube). For a fluid, if there are no sources or sinks, then its density cannot change Let the vectors that are illustrated represent the motion of fluid particles. In the case of divergence only, fluid is streaming from the point, at which the density is falling. Alternatively the point could be a source. In the case where there is only curl, the fluid rotates about the point and the fluid is incompressible. Finally, the point that possesses both divergence and curl is a compressible fluid with rotation 16. If f and g are continuously differentiable scalar fields, show that ?f × ?g is solenoidal. Hint: Show that ?f × ?g = ? × (f?g). 17. An inviscid (frictionless) fluid in equilibrium obeys the relationship ?p = ?F, where ? denotes the density of the fluid, p denotes the pressure, and F denotes the body forces (such as gravity). Show that F · ? × F = 0. Let v(x, y, z) denote the velocity at the point (x, y, z) in a moving fluid. If it varies with time, this is the velocity at a particular instant of time. The integral H C v·dr around a closed path C is called the circulation around that path. The average component of velocity along the path is vs = H C vs ds s = H C v · dr s , (4.3.14) where s is the total length of the path. The circulation is thus H C v · dr = vss, the product of the length of the path and the average velocity along the path. When the circulation is positive, the flow is more in the direction of integration than opposite to it. Circulation is thus an indication and to some extent a measure of motion around the path 10. F = x 2 i + y 2 j + (z 2 + 2xy)k and C consists of the edges of the triangle with vertices at (0, 0, 0), (1, 1, 0), and (0, 1, 0). Proceed from (0, 0, 0) to (1, 1, 0) to (0, 1, 0) and back to (0, 0, 0). Use MATLAB to illustrate the parametric curves Integral around every simply connected circuit is zero, (2) its curl equals zero, and (3) it is the gradient of a scalar function. For continuously differentiable vectors, these properties are equivalent. For vectors that are only piece-wise differentiable, this is not true. Generally the first property is the most fundamental and is taken as the definition of irrotationality  Surface integrals appear in such diverse fields as electromagnetism and fluid mechanics. For example, if we were oceanographers we might be interested in the rate of volume of seawater through an instrument that has the curved surface S. The volume rate equals RR S v · n d?, where v is the velocity and n d? is an infinitesimally small element on the surface of the instrument. The surface element n d? must have an orientation (given by n) because it makes a considerable difference whether the flow is directly through the surface Let us review our analysis to date. We found that each harmonic in the forcing function yields a corresponding harmonic in the particular solution, Equation 5.6.8. The only difficulty arises with the harmonic n = 2. Although our particular solution is not correct because it contains cos(3t), we suspect that if we remove that term then the remaining harmonic solutions are correct. The problem is linear, and difficulties with one harmonic term should not affect other harmonics. But how shall we deal with the cos(3t) term in the forcing function? The first two terms on the right side of Equation 5.6.25 represent the complementary solution. Although this expansion is equivalent to Equation 5.6.16, we have all of the advantages of dealing with exponentials rather than sines and cosines. These advantages include ease of differentiation and integration, and writing the series in terms of amplitude and phase Assuming that these series are useful, the next question is how do we find the Fourier coefficients? We could compute them by numerically integrating Equation 5.1.6. However, the results would suffer from the truncation errors that afflict all numerical schemes. On the other hand, we can avoid this problem if we again employ the orthogonality properties of sines and cosines, now in their discrete form. Just as in the case of conventional Fourier series, we can use these properties to derive formulas for computing the Fourier coefficients. These results will be exact except for roundoff errorsImage transcription textEvaluate [F . dr for the followingvector fields and curves: 1. F =ysin(nz)i + re”j + 3c… Show more… Show moreImage transcription textUsing the potential found in Example4.4.1, let us find the value of the lineintegral [F . dr from the po… Show more… Show moreImage transcription textSolve the following ordinary differentialequations by Fourier series if the forcingis given by the periodic func… Show more… Show more  Math MATH 133 Share QuestionEmailCopy link