Thursday, February 28 ∗ Solutions ∗ Curves and integration.1. Consider the curve C in R3given byr(t ) =¡etcos t¢i +2j +¡etsin t¢k(a) Draw a sketch of C.Solution. The sketch of C is the following graph.010020001234-150-100-500Figure 1: Sketch of C.(b) Calculate the arc length function s(t ), which gives the length of the segment of C betweenr(0) and r(t) as a function of the time t for all t ≥0. Check your answer with the instructor.Solution. Sincex0(t ) =etcos t −etsin t, y0(t ) =0, z0(t ) =etsin t +etcos t,we have|r0(t )|=p(etcos t −etsin t)2+(etsin t +etcos t)2=p2et.Hence the arc length iss(t ) =Zt0|r0(u)|du =Zt0p2eudu =p2et−p2.(c) Now invert this function to find the inverse function t (s). This gives time as a function ofarclength, that is, tells how long you must travel to go a certain distance.Solution. Solve s =p2et−p2, which gives et=s+p2p2, and sot = t(s) =lnÃs +p2p2!.(d) Suppose h : R →R is a function. We can get another parameterization of C by consideringthe compositionf(s) =r¡h(s)¢This is called a reparametrization. Find a choice of h so thati. f(0) =r(0)ii. The length of the segment of C between f(0) and f(s) is s. (This is called parametrizingby arc length.)Check your answer with the instructor.Solution. From (c) we know t = ln³s+p2p2´. When s = 0, we have t = ln1 = 0. Then wecan chooseh(s) =lnÃs +p2p2!.(e) Without calculating anything, what is |f0(s)| ?Solution. Since s =Rs0|f0(u)|du, then by the fundamental theorem of calculus, we candifferentiate both sides with respect to s and get 1 =|f0(s)| .2. Consider the curve C given by the parametrization r: R →R3where r(t) =(sin t,cos t,sin2t).(a) Show that C is in the intersection of the surfaces z = x2and x2+y2=1.Solution. Since x = sin t , y = cos t, z = sin2t, it is very easy to check that z = x2andx2+ y2= 1. So the curve C lies in both these two surfaces, hence is in the intersectionof them.(b) Use (a) to help you sketch the curve C.Solution. The left graph is the intersection of the two surfaces, while the right one is thecurve.-1.0-0.50.00.51.0-1.0-0.50.00.51.0-2-1012-1.0-0.50.00.51.0-1.0-0.50.00.51.00.00.51.0Figure 2: Two surfaces and the curve C.3. (a) Sketch the top half of the sphere x2+y2+z2=5. Check that P =¡1,1,p3¢is on this sphereand add this point to your picture.Solution. The top half of the sphere is shown in Figure 3 (the black dot is P). Since12+12+(p3)2=5, we know P is on this sphere.-202-2020.00.51.01.52.0Figure 3: Half sphere and the path.(b) Find a function f (x, y) whose graph is the top-half of the sphere. Hint: solve for z.Solution. Since x2+ y2+z2= 5, we have z2= 5 −x2− y2, and so z = ±p5 −x2−y2. Aswe only want the top half of the sphere, we can let f (x, y) =p5 −x2−y2.(c) Imagine an ant walking along the surface of the sphere. It walks down the sphere alongthe path C that passes through the point P in the direction parallel to the yz-plane. Drawthis path in your picture.Solution. The black curve in Figure 3 is the path.(d) Find a parametrization r(t) of the ant’s path along the portion of the sphere shown in yourpicture. Specify the domain for r, i.e. the initial time when the ant is at P and the final timewhen it hits the x y-plane.Solution. x =1 along the path and f (1, y) =p4 −y2, so setting y = t we get the parametriza-tionr(t ) =(1, t,p4 −t2).4. As in 1(d), consider a reparametrizationf(s) =r¡h(s)¢of an arbitrary vector-valued function r: R → R3. Use the chain rule to calculate |f0(s)| in termsof r0and h0.Solution. By the chain rule, f0(s) = r0(h(s)) h0(s). Taking magnitudes of both sides we have|f0(s)|=|r0(h(s))|·|h0(s)|
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