Thursday October 11 Solutions Curves and integration 1 Consider the curve C in R3 given by r t e t cos t i 2j e t sin t k a Draw a sketch of C Solution The sketch of C is the following graph 01234 0 100 200 0 50 100 150 Figure 1 Sketch of C b Calculate the arc length function s t which gives the length of the segment of C between r 0 and r t as a function of the time t for all t 0 Check your answer with the instructor Solution Since x 0 t e t cos t e t sin t y 0 t 0 z 0 t e t sin t e t cos t we have r0 t p e t cos t e t sin t 2 e t sin t e t cos t 2 p t 2e Hence the arc length is t Z s t 0 Z tp p p r u d u 2e u d u 2e t 2 0 0 c Now invert this function to find the inverse function t s This gives time as a function of arclength that is tells how long you must travel to go a certain distance Solution Solve s p t p 2e 2 which gives e t p s p 2 and so 2 p s 2 t t s ln p 2 d Suppose h R R is a function We can get another parameterization of C by considering the composition f s r h s This is called a reparametrization Find a choice of h so that i f 0 r 0 ii The length of the segment of C between f 0 and f s is s This is called parametrizing by arc length Check your answer with the instructor Solution From c we know t ln can choose p s p 2 2 When s 0 we have t ln 1 0 Then we p s 2 h s ln p 2 e Without calculating anything what is f0 s Rs Solution Since s 0 f0 u d u then by the fundamental theorem of calculus we can differentiate both sides with respect to s and get 1 f0 s 2 Consider the curve C given by the parametrization r R R3 where r t sin t cos t sin2 t a Show that C is in the intersection of the surfaces z x 2 and x 2 y 2 1 Solution Since x sin t y cos t z sin2 t it is very easy to check that z x 2 and x 2 y 2 1 So the curve C lies in both these two surfaces hence is in the intersection of 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 the curve 2 1 0 1 0 1 0 0 5 1 0 5 0 0 0 0 1 0 1 0 0 5 0 5 2 0 0 0 0 0 5 1 0 0 5 0 5 0 5 0 0 0 5 1 0 1 0 1 0 1 0 Figure 2 Two surfaces and the curve C 3 p a Sketch the top half of the sphere x 2 y 2 z 2 5 Check that P 1 1 3 is on this sphere and add this point to your picture Solution The top half of the sphere is shown in Figure 3 the black dot is P Since p 12 12 3 2 5 we know P is on this sphere 2 0 2 2 0 1 5 1 0 0 5 0 0 2 0 2 Figure 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 p Solution Since x 2 y 2 z 2 5 we have z 2 5 x 2 y 2 and so z 5 x 2 y 2 As p we only want the top half of the sphere we can let f x y 5 x 2 y 2 c Imagine an ant walking along the surface of the sphere It walks down the sphere along the path C that passes through the point P in the direction parallel to the y z plane Draw this 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 your picture Specify the domain for r i e the initial time when the ant is at P and the final time when it hits the x y plane p Solution x 1 along the path and f 1 y 4 y 2 so setting y t we get the parametrization p r t 1 t 4 t 2 4 As in 1 d consider a reparametrization f s r h s of an arbitrary vector valued function r R R3 Use the chain rule to calculate f0 s in terms of r0 and h 0 Solution By the chain rule f0 s r0 h s h 0 s Taking magnitudes of both sides we have f0 s r0 h s h 0 s
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