ISM_T11_C06_C

# ISM_T11_C06_C - Section 6.3 Lengths of Plane Curves 17(a dy...

This preview shows pages 1–5. Sign up to view the full content.

Section 6.3 Lengths of Plane Curves 381 17. (a) 2x 4x dy dy dx dx œ Ê œ Š # # L 1 dx Ê œ ' 1 2 Ê Š dy dx # 1 4x dx œ ' 1 2 È # (c) L 6.13 ¸ (b) 18. (a) sec x sec x dy dy dx dx œ Ê œ # % # Š L 1 sec x dx Ê œ ' 3 0 È % (c) L 2.06 ¸ (b) 19. (a) cos y cos y dx dx dy dy œ Ê œ Š # # L 1 cos y dy Ê œ ' 0 È # (c) L 3.82 ¸ (b) 20. (a) dx dx dy dy 1 y y y 1 y œ Ê œ È # Š L 1 dy dy Ê œ œ ' ' 1 2 1 2 1 2 1 2 É É y 1 y 1 y a b " 1 y dy œ ' 1 2 1 2 a b # "Î# (c) L 1.05 ¸ (b) 21. (a) 2y 2 2 (y 1) œ Ê œ dx dx dy dy Š # # L 1 (y 1) dy Ê œ ' 1 3 È # (c) L 9.29 ¸ (b)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
382 Chapter 6 Applications of Definite Integrals 22. (a) cos x - cos x + x sin x x sin x dy dy dx dx œ Ê œ Š # # # L 1 x sin x dx Ê œ ' 0 È # # (c) L 4.70 ¸ (b) 23. (a) tan x tan x dy dy dx dx œ Ê œ Š # # L 1 tan x dx dx Ê œ œ ' ' 0 0 6 6 È É # sin x cos x cos x sec x dx œ œ ' ' 0 0 6 6 dx cos x (c) L 0.55 ¸ (b) 24. (a) sec y 1 sec y 1 dx dx dy dy œ Ê œ È Š # # # L 1 sec y 1 dy Ê œ ' 3 4 È a b # sec y dy sec y dy œ œ ' ' 3 3 4 4 k k (c) L 2.20 ¸ (b) 25. 2 x 1 dt, x 0 2 1 1 y f(x) x C where C is any È È Ê Ê Š Š œ   Ê œ Ê œ „ Ê œ œ „ ' 0 x dy dy dy dt dx dx # # real number. 26. (a) From the accompanying figure and definition of the differential (change along the tangent line) we see that dy f (x ) x length of kth tangent fin is œ ˜ Ê w k 1 k ( x ) (dy) ( x ) [f (x ) x ] . È È ˜ œ ˜ ˜ k k k 1 k # # # w # (b) Length of curve lim (length of kth tangent fin) lim ( x ) [f (x ) x ] œ œ ˜ ˜ n n Ä _ Ä _ ! ! È n n k 1 k 1 k k 1 k # w # lim 1 [f (x )] x 1 [f (x)] dx œ ˜ œ n Ä _ ! È È n k 1 k 1 k a b w # w # '
Section 6.3 Lengths of Plane Curves 383 27. (a) correspondes to here, so take as . Then y x C and since ( ) lies on the curve, C 0. Š È dy dy dx 4x dx x # " " # È œ "ß " œ So y x from ( ) to (4 2). œ "ß " ß È (b) Only one. We know the derivative of the function and the value of the function at one value of x. 28. (a) correspondes to here, so take as . Then x C and, since ( ) lies on the curve, C 1 Š dx dy dx y y y dy # " " " œ !ß " œ So y . œ " " x (b) Only one. We know the derivative of the function and the value of the function at one value of x. 29. (a) 2 sin 2t and 2 cos 2t ( 2 sin 2t) (2 cos 2t) 2 dx dx dt dt dt dt dy dy œ œ Ê œ œ Ê ˆ Š È # # # # Length 2 dt 2t Ê œ œ œ ' 0 2 c d 1 Î# ! 1 (b) cos t and sin t ( cos t) ( sin t) dx dx dt dt dt dt dy dy œ œ œ œ œ 1 1 1 1 1 1 1 1 1 Ê ˆ Š È # # # # Length dt t Ê œ œ œ ' 1 2 1 2 1 1 1 c d "Î# "Î# 30. x a( sin ) a(1 cos ) a 1 2 cos cos and y a(1 cos ) œ Ê œ Ê œ œ ) ) ) ) ) ) dx dx d d ) ) ˆ a b # # # a sin a sin Length d 2a (1 cos ) d Ê œ Ê œ Ê œ œ dy dy dy d d d d dx ) ) ) ) ) ) ) ) ) Š Š Ê ˆ È # # # # # # ' ' 0 0 2 2 a 2 2 d 2a sin d 2a sin d 4a cos 8a œ œ œ œ œ È È É ¸ ¸ ' ' ' 0 0 0 2 2 2 1 cos 2 # # # # ! ) ) ) ) 1 ) ) ) 31-36. Example CAS commands: : Maple with( plots ); with( Student[Calculus1] ); with( student ); f := x -> sqrt(1-x^2);a := -1; b := 1; N := [2, 4, 8 ]; for n in N do xx := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x,f(x)],x=xx)]; L := simplify(add( distance(pts[i+1],pts[i]), i=1..n )); # (b) T := sprintf("#31(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [f(x),pts], x=a..b, title=T ): # (a) end do: display( [seq(P[n],n=N)], insequence=true, scaling=constrained ); L := ArcLength( f(x), x=a..b, output=integral ): L = evalf( L ); # (c) 37-40. Example CAS commands: : Maple with( plots ); with( student ); x := t -> t^3/3; y := t -> t^2/2; a := 0;

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
384
This is the end of the preview. Sign up to access the rest of the 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