where the velocity field, u = w(, )ˆz, is the fluid density and t is time. 2.Usi
ID: 3141515 • Letter: W
Question
where the velocity field, u = w(, )ˆz, is the fluid density and t is time.
2.Using a computer package, e.g., Matlab, Mathematica, WolframAlpha etc, plot the fluid’s velocity profile for several different values of (0, ]. How does the result for = differ from the result for flow along a cylinder of circular cross-section? Explain your answer.
3.Plot the fluid’s velocity profile for several values of , near = /4. Explain your results and hence comment on the behaviour of the flow at = /4, without solving the problem for = /4.
ou 0tExplanation / Answer
Ans-
. Use the result from class: Area(Dd) = Area(D0) + d · Length(D0)+d2 . This implies Length(D0) = d Area(Dd)/d(d)kd=0. Since any of the domains Dd can be taken on the role of D0, we find Length(Dd) = d Area(Dd)/d(d) = Length(D0) + 2d. Solution 2. Use the method from class. For convex polygons, Length(Dd) = Length(D0) + 2d by direct observation. We obtain the same result for arbitrary convex domains D0 by approximating them with polygons and passing to the limit. Solution 3. Avoid limits and polygons. Let s 7 r(s) be the counterclockwise arc-length parameterization of D0. Then Dd can be parameterized by adding to r(s) the d-multiple of the (unit) vector dr/ds rotated 90 degrees clockwise (we denote the rotation by J: s 7 fd(s) := r(s) + d · Jdr(s)/ds. The velocity vector dfd(s)/ds = dr/ds + d · Jd2 r/ds2 = (1 + d · k(s))dr(s)/ds because the acceleration d 2 r(s)/ds2 of the original curve is proportional to Jdr(s)/ds with the proportionality coefficient k(s) (by the very definition of curvature k). Since |dr/ds| = 1 and k > 0 (convexity), we have Length(Dd) = I |dfd(s)|ds = I ds+d· I k(s)ds = Length(D0)+2d. 2. Verify the invariance of the arc length R b a p x 2 (t) + y 2 (t) dt under reparameterizations t = t( ). By the chain rule, and the variable change law in integrals, we have Z r ( dx d ) 2 + (dy d ) 2 d = Z r x 2 ( dt d ) 2 + y 2 ( dt d ) 2 d = Z p x 2 + y 2 | dt d | d = Z t() t() p x 2 + y 2 dt. 1 2 3. (a) Prove the formula k = (¨xy y¨x)/( x 2+ y 2 ) 3/2 for the curvature of a regular parameterized plane curve t 7 (x(t), y(t)). The determinant (¨xy y¨x) is (up to a sign, which actually should be reversed to agree with our orientation conventions) the area of the parallelogram spanned by the velocity r and the acceleration ¨r and thus equals |r| × |an| (“base times height”). The curvature k = |an|/|r| 2 is therefore obtained from the area by dividing it by |r| 3 = ( x 2 + y 2 ) 3/2 . (b) Compute the curvature of the graph of a smooth function y = f(x). Parameterizing the graph as t 7 (x, y) = ((t, f(t)), we obtain k(x) = f (x)/(1 + f (x))3/2 . (c) Take f = x a/a and find the limit of curvature at x = 0 for a = 5/2, 2, 3/2, 1, 1/2. At the origin, the curve y = x 5/2 has the curvature 0 (since it is best approximated by the parabola y = kx2/2 with k = 0); the curve y = x 2/2 is the parabola with k = 1; the curve y = x 3/2/(3/2) has the curvature (according to part (b)) k(x) = x 1/2/2/(1 + x 1/2 ) 3/2 which tends to as x approaches 0; y = x is a straight line and has k = 0; and y = x 1/2/(1/2) means y 2/4 = x, which is a parabola again with the curvature at the origin equal to 1/2. 4. Draw the typographic symbol (“infinity” or “figure eight”) increased 100 times and then draw an equidistant curve as follows: orient all normal lines to the large figure eight in a continuous fashion, and connect all points removed 1 cm from the large figure eight in the positive normal direction. Which curve is longer — the large figure eight or the curve equidistant to it? Using any of the methods from Problem 1 (e.g. approximating the curve with polygons) one concludes that Length(Dd) = Length(D0)+2d× (rotation index of the tangent line). Since the tangent line to “figure eight” makes 0 number of turnes, the equidistant curve has the same length as the “figure eight
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