Question

answer 1-4 using desmos or any other graphing calculator

1. (a) Investigate the family of curves defined by the polar equations r = sin nd, where n is a positive integer. How is the number of loops related to n? (b) What happens if the equation in part (a) is replaced by r = |sixí ng|? 2. A family of curves is given by the equations r=1+c sin nd, where c is a real number and n is a positive integer. How does the graph change as n increases? How does it change as c changes? Illustrate by graphing enough members of the family to support your conclusions, 3. A family of curves has polar equations 1-a cos 1 + acos Investigate how the graph changes as the number a changes. In particular, you should identify the transitional values of a for which the basic shape of the curve changes

Answer 1

4G 7:27 AM 0.1KB/s Vo LTE LTE 82 III 14 1/2 21/3 T1/3 51/6 f/6. TT 1 7/B 1111/6 41/3 51/3 + 1 w p=sin(n:0) Х 0 <O< 27 2 Х n = 2 -10 10 3 X 4 powered by desmos

4G 7:27 AM 0.OKB/s vo LTE LTE 82 III 1+ 1/2 21/3 T[/3 3 511/6 TT/ TT 1 711/6 1111/6 41/3 51/3 + 1 Х r = sin(n:0) <O< 7 0 2 Х n=3 -10 10 3 X 4 powered by desmos

4G 7:29 AM|0.OKB/s Vo LTE LTE 82 III 21/3 T/3 511/6 TT/ TT 1 AT/6 1116 41/3 51/3 + 1 Х r = sin(n·0)| 0 <o< 27 2 x n = 3 12 -10 10 3 Х 4 powered by desmos

4G 7:29 AM|1.5KB/S Vo LTE LTE 82 III 14 1/2 21/3 T1/3 51/6 f/6. TT 1 7/6 1 1/6 41/3 517/3 + 1 Х r = sin(n·0)| 0 <o< 27 2 n = 2 t2 -10 X • X 10 3 4 powered by desmos

4G 7:31 AM|1.7KB/s vo LTE LTE 82 4 2 1/2 21/3 13 51/6 TT/6 T 0 2 2 4 ZTt/6 111/6 41/3 51/3 + 1 N 7=1c sin(n :0) Х 0 <O< 26 2 n = 1 -10 10 X X . 3 c=1 -10 10 4. powered by desmos

4G 7:31 AM|0.0KB/s Vo LTE LTE 82 4 211/27 21/3 TT/3 5/6 TT/6 TT 0 2 2 4 ZTt/6 111/6 41/3 51/3 + 1 N 7=1c sin(n+ :0) Х 0 <O< 260 2 n = 3 -10 10 X X . 3 c=1 -10 10 4. powered by desmos

4G 7:33 AM | 10.2KB/S Vo LTE LTE 81 4 211/27 21/3 T1/3 5ſt/6 TT/6 2 2 4 71/6 1111/6 41/3 51/3 + 1 Х p=1+c sin(n.6 :0) 0 <o< 27 2 Х n = 3 -10 10 3 X c=2 -10 10 4. powered by desmos

4G 7:33 AM 0.0KB/s Vo LTE LTE 81 4 21/22 8/3 TT/3 51t/6 T/6 2 2 4 ZTt/6 111/6 41/3 Бт/3 + « 1 Х p=1+c sin(n.6 :0) 0 <o< 27 2 Х n = 3 -10 10 3 X c=4 -10 10 4. powered by desmos

4G 7:36 AM | 0.OKB/s Vo LTE LTE 81 III 101 T1/2 2/3 TT/3 51/6 TT/6 M 0 20 10 10 20 71/6 11T1/6 + 1 Х p= (1 - a cos(0)) os(0)) 1 + a cos 0 <O< 27 2 Х a = 1 -10 10 3 powered by desmos

4G 7:39 AM | 4.8KB/s Vo LTE LTE 81 201 T/2 21/3 T[/3 51/6 TT/6 TT OD 20 20 40 Žrt/6 111/6 + 1 Х p= (1 - a cos(0)) os(0)) 1 + a cos 0 <O< 27 2 Х a = 1.1 -10 10 3 powered by desmos

7:36 AM 1.6KB/s Vo LTE 4G JN LTE 81 101 1/2 2ít/B. TT/3) 51/6 TT/6 TT 0 20 10 10 20 71/6 1111/6 + 1 Х p= (1 – a cos(0) os(0)) 1 + a cos 0 <O< 27 2 Х a = 2 -10 10 3 powered by desmos

Oilas o= Suno n is odd. 2o. of loope equals for ton even for nis no. of loops equals to an. (b) In case of 8 - Süno) Both nis odd f even the no. of loops are an Q.2. gel + C Sivno • No. of loope inculares du There is in new loop due to Or Smallose los due to c. size of Small with increase in a constant n lineckily with with n loep inceland c for 0-3 r=1-a Caso It a coso Shape Changes as all Changer
according to the Chegg guidelines I am bound to do only first four parts of the question in case of multiple parts.

Including two sub parts of a I have already done 4 parts. If you want any specific part then please mark that one

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