ds am 17m² so you 2 1+m? 14. a) You have proved in #5 that, if s= arctan m, 5 lie s is arclength on unit circle, m is sl

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Ds Am 17m So You 2 1 M 14 A You Have Proved In 5 That If S Arctan M 5 Lie S Is Arclength On Unit Circle M Is Sl 1 (55.21 KiB) Viewed 22 times

Ds Am 17m So You 2 1 M 14 A You Have Proved In 5 That If S Arctan M 5 Lie S Is Arclength On Unit Circle M Is Sl 2 (64.47 KiB) Viewed 22 times

ds am 17m² so you 2 1+m? 14. a) You have proved in #5 that, if s= arctan m, 5 lie s is arclength on unit circle, m is slope), then In #131b) you found an infinite series for 1 have an infinite series for ds. Use this to find an infinite IM Serses for s as a function of m ! b) Graph the first few terms of your formula in lal, and compare to the graph of s-asctan m, on a computer. Con the computer, you will have to write x form, y for s, s, y = tanix rather than su arctan m. Just different notation) 15. a) If s and m are arclength and slope on the unit cilue as above, then if m= 1, what is the value of s? b) Put the values from#isla) into the infinite series from # 14 (a). You should get an expression for ✓ ! and

3 1 11 1 13 S- + + + 1719 + + 10 12! . 1 3 3x2x1 1 - + and 0 = 31 ( bottom 10. I've tried to give hints in the problem, so fill just give final answers: ya sins e sosten : 31 so 15! x = cos s = 10 ks - $8 11. a) Tale derivative of formula for y above ferm by term. I'll show you just the fist three terms. 1- $(35) + (55") - s! S? + 5 S + 5x4x3221 4x3281 the same three first terms of x= COS S. 12. 2- +z = (1+ ++*++'+ t'+++...) - t{1++++++++++7.) (1+*+&+t't t'+tt) - (t+txt' t*4+...) 1 i z-tz 1 so (1-1) = = 1 So Hit *4 = 1+ ++t'+t'++++ (This works formally for the variable t, but it only works -1<t <1 .) if to 1, we get 17******* 1-X 2 13. 6) Answer: : 1- m + m^ - ^ + m +. slots' Sa din = S(1-in + m" - mb + m² - ...) dm Sendom = m- m*- *** + žm?... that is, is something whose derivative is 1 - m*+m-m' me.... 1-t numerically if 1 IL 1 + 32 - 1 1+ m2 14. (a) S = - m+ m