{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

4.2 pt 2 (1)

# 4.2 pt 2 (1) - 240 4 Vector-Valued Functions 3(sin 3t cos...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 240 4. Vector-Valued Functions 3.. (sin 3t, cos 3t,2t3f2)-,0 g t g 1 4.(i+1,[email protected]+7,§t2);1gig 2; 5. (anti). 1 s t 52-, 6. (t,tsint,tcost);0 g t 5 11' deﬁned by c(t) = {2 cost,2sint.t.. ,if2a' 3: g 4n. t): (1,3t2,t3),if0§tg 1 a 7. Find the length of the path eft), D g t 5 2a and C(t) == (2,t — 211,16) 8. Find the length of the path C(t), where cf C(t) = team], if 1 g t g 2. _ ath C(t), deﬁned by s(t- s the distance a particle traversing the trajectory- 5 out at time a; that is, it gives -' d the arc length functions for - = (cos t, sint,t), with a : f}. I; UleNldT. mpresent c will have traveled by time t if it start length of c between C(91) and C(t). Fin curves aft] = [cosht, sinht,t) and ﬁt) 10. Let c be the path C(t) = (2t,t2, logt), deﬁned for t > 0. Find the length of c between the points (2, 1,0} and [4,4, log2). 11. Find the arc length of the path C(t) : (t,tsint,tcos t) between (0.0. and (17,0, —1r]. 12. Let C(t) be a given path, a S t g b. Lets = aft) be a new variable, whe is a strictly increasing C:l function given on [cu b]. For each s in [0501), of there is a unique is with Mt) = 3. Define the function (1 : [15):(tt)I cc[b)1 —~ by d(s] = C(t). The path (1 is said to be a reparametﬁzation of c. (a) Argue that the image curves of c and d are the same. that c and d have the same arc length. (b) Show (c) Let s = aft} = I: Hc’[r)lldr. Define d as above by d[s) : C(t). Sh that d ‘ d is called the arc length reparametﬁzation of c. —» R3 be a smooth path. Assume c"(t) 75 0 for any t. 15. Let c : [ ,b] (1 since RTE)“ = 1- vector (2’ (t) / Hc’ (t) 11 = T(t) is tangent to c at C(t), and, is called the unit tangent to c. = 0. [Hint Differentiate TUE) -T(t) = 1.] (a) Show that T'(t} - T(t) for T’ (t). (b) Write down a formula in terms of c ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern