T12-Der Coor Pol-Asqui

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ESCUELA SUPERIOR POLTÉCNICA DE 
CHIMBORAZO 
 
 
FACULTAD: MECÁNICA 
 
CARRERA: INGENIERÍA MECÁNICA 
 
 
 
PRIMER SEMESTRE 
 
PARALELO ¨B 
 
ANÁLISIS MATEMÁTICO 
 
DEBER DE DERIVADAS CON COORDENADAS POLARES 
 
ASQUI VACA BORIS JOSUE 
 
• 𝑟 = 2 + cos 𝜃 
 
𝜃 𝑟 = 2 + cos 𝜃 𝑃𝑢𝑛𝑡𝑜 𝑑𝑒 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛 
𝑟 𝜃 
 
0 𝑟 = 2 + cos 0 3 0 
 
𝜋/2 𝑟 = 2 + cos 90 2 𝜋/2 
 
𝜋 𝑟 = 2 + cos 180 1 𝜋 
 
3𝜋/2 𝑟 = 2 + cos 270 2 3𝜋/2 
 
 
 
 
 
 
tan 𝜑 =
−sin2 𝜃 + 2 cos 𝜃 + cos2 𝜃
−2sin 𝜃 cos 𝜃 − 2sin 𝜃
 
 
𝐻𝑜𝑟𝑖𝑧𝑜𝑡𝑎𝑙 𝑉𝑒𝑟𝑖𝑐𝑎𝑙 
−sin2 𝜃 + 2 cos 𝜃 + cos2 𝜃 = 0 
−1 + cos2 𝜃 + 2 cos 𝜃 + cos2 𝜃 = 0 
2cos2 𝜃 + 2 cos 𝜃 − 1 = 0 
cos 𝜃 = 0.36 
𝜃
= (1)𝑘 . 70 . +2𝜋𝑘 
𝑘 = 0; 𝜃 = 70 
𝑘 = ±1; 𝜃 = 790 
cos 𝜃 = −1.36 
≠ 
 
−2sin 𝜃 cos 𝜃 − 2sin 𝜃 = 0 
−2sin 𝜃 (cos 𝜃 + 1) = 0 
 
−2sin 𝜃 = 0 
𝜃 = 0 + 𝜋𝑘 
𝑘 = 0; 𝜃 = 0 
𝑘 = 1; 𝜃 = 𝜋 
 
cos 𝜃 = −1 
𝜃 = 𝜋 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 𝜋 
𝑘 = 1; 𝜃 = 3𝜋 
 
 
 
 
 
 
 
 
r 𝜃 
3 15o 
2.90 30o 
2.70 45o 
2.50 60o 
2.20 75o 
2 90o 
𝜃 = 𝜋/4 → 45𝑜 
tan 𝜇 =
2 + cos 𝜃
− sin 𝜃
= −3.82 
𝜇 = 285𝑜 
𝜑 = 330𝑜 
𝜃 = 𝜋/3 → 60𝑜 
tan 𝜇 =
2 + cos 𝜃
− sin 𝜃
= −2.88 
𝜇 = 290𝑜 
𝜑 = 350𝑜 
 
• 𝑟 = 2 + sin 𝜃 
 
𝜃 𝑟 = 2 + sen 𝜃 𝑃𝑢𝑛𝑡𝑜 𝑑𝑒 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛 
𝑟 𝜃 
 
0 𝑟 = 2 + sin 0 2 0 
 
𝜋/2 𝑟 = 2 + sin 90 3 𝜋/2 
 
𝜋 𝑟 = 2 + sin 180 2 𝜋 
 
3𝜋/2 𝑟 = 2 + sin 270 1 3𝜋/2 
 
 
 
 
 
 
tan 𝜑 =
2sin 𝜃 cos 𝜃 + 2cos 𝜃
−sin2 𝜃 − 2 sin 𝜃 + cos2 𝜃
 
 
𝐻𝑜𝑟𝑖𝑧𝑜𝑡𝑎𝑙 𝑉𝑒𝑟𝑖𝑐𝑎𝑙 
2sin 𝜃 cos 𝜃 + 2cos 𝜃 = 0 
2cos 𝜃 (sin 𝜃 + 1) = 0 
2cos 𝜃 = 0 
𝜃 =
𝜋
2
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
𝜋
2
 
𝑘 = ±1; 𝜃 =
5𝜋
2
 
sin 𝜃 + 1 = 0 
𝜃 = −
𝜋
2
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 = −
𝜋
2
 
𝑘 = ±1; 𝜃 =
3𝜋
2
 
 
−sin2 𝜃 − 2 sin 𝜃 + cos2 𝜃 = 0 
−sin2 𝜃 − 2 sin 𝜃 + 1 − 𝑠𝑖𝑛2 𝜃 = 0 
− 2s𝑖𝑛2 𝜃 − 2 sin 𝜃 + 1 = 0 
sin 𝜃 = 0.36 
𝜃 = 21.46 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 21.46 
𝑘 = 1; 𝜃 = 381.𝑜 
 
sin 𝜃 = −1.36 
≠ 
 
 
 
 
 
 
r 𝜃 
2.25 15o 
2.50 30o 
2.70 45o 
2.86 60o 
2.96 75o 
3 90o 
𝜃 = 𝜋/4 → 45𝑜 
tan 𝜇 =
2 + sin 𝜃
cos 𝜃
= −3.82 
𝜇 = 75𝑜 
𝜑 = 120𝑜 
𝜃 = 𝜋/3 → 60𝑜 
tan 𝜇 =
2 + sin 𝜃
cos 𝜃
= 5.73 
𝜇 = 80𝑜 
𝜑 = 140𝑜 
 
• 𝑟 = 1 − 2cos 𝜃 
1 − 2cos 𝜃 = 0 
𝜃 =
𝜋
3
 
𝜃 1 − 2cos 𝜃 𝑃𝑢𝑛𝑡𝑜 𝑑𝑒 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛 
𝑟 𝜃 
 
0 𝑟 = 1 − 2cos 0 -1 0 
 
𝜋/2 𝑟 = 1 − 2cos 90 1 𝜋/2 
 
𝜋 𝑟 = 1 − 2cos 180 3 𝜋 
 
3𝜋/2 𝑟 = 1 − 2cos 270 1 3𝜋/2 
 
 
 
 
 
 
tan 𝜑 =
2sin2 𝜃 + cos 𝜃 − 2cos2 𝜃
4sin 𝜃 cos 𝜃 − sin 𝜃
 
 
𝐻𝑜𝑟𝑖𝑧𝑜𝑡𝑎𝑙 𝑉𝑒𝑟𝑖𝑐𝑎𝑙 
2sin2 𝜃 + cos 𝜃 − 2cos2 𝜃 = 0 
2 − 2cos2 𝜃 + cos 𝜃 − 2cos2 𝜃 = 0 
−4cos2 𝜃 + cos 𝜃 + 2 = 0 
cos 𝜃 = 0.84 
𝜃 =
9𝜋
50
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
9𝜋
50
 
𝑘 = ±1; 𝜃 =
109𝜋
50
 
cos 𝜃 = −0.59 
0 =
7𝜋
10
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
7𝜋
10
 
𝑘 = ±1; 𝜃 =
27𝜋
10
 
 
4sin 𝜃 cos 𝜃 − sin 𝜃 = 0 
sin 𝜃 (4cos 𝜃 − 1) = 0 
 
sin 𝜃 = 0 
𝜃 = 0 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 0 
𝑘 = 1; 𝜃 = 2𝜋 
 
cos 𝜃 =
1
4
 
𝜃 = 5𝜋/12 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 5𝜋/12 
𝑘 = 1; 𝜃 = 29𝜋/12 
 
 
 
 
 
 
r 𝜃 
-0.93 15o 
-0.73 30o 
-0.41 45o 
0 60o 
0.48 75o 
1 90o 
𝜃 = 5𝜋/6 → 150𝑜 
tan 𝜇 =
1 − 2cos 𝜃
2 sin 𝜃
= 2.73 
𝜇 = 70𝑜 
𝜑 = 220𝑜 
𝜃 = 7𝜋/6 → 210𝑜 
tan 𝜇 =
1 − 2cos 𝜃
2 sin 𝜃
= −2.73 
𝜇 = 290𝑜 
𝜑 = 140𝑜 
 
• 𝑟 = 1 + 2cos 𝜃 
1 + 2cos 𝜃 = 0 
𝜃 =
2𝜋
3
 
𝜃 1 + 2cos 𝜃 𝑃𝑢𝑛𝑡𝑜 𝑑𝑒 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛 
𝑟 𝜃 
 
0 𝑟 = 1 + 2cos 0 3 0 
 
𝜋/2 𝑟 = 1 + 2cos 90 1 𝜋/2 
 
𝜋 𝑟 = 1 + 2cos 180 -1 𝜋 
 
3𝜋/2 𝑟 = 1 + 2cos 270 1 3𝜋/2 
 
 
 
 
 
 
tan 𝜑 =
−2sin2 𝜃 + cos 𝜃 + 2cos2 𝜃
−4sin 𝜃 cos 𝜃 − sin 𝜃
 
 
𝐻𝑜𝑟𝑖𝑧𝑜𝑡𝑎𝑙 𝑉𝑒𝑟𝑖𝑐𝑎𝑙 
−2sin2 𝜃 + cos 𝜃 + 2cos2 𝜃 = 0 
−2 + 2cos2 𝜃 + cos 𝜃 + 2cos2 𝜃 = 0 
4cos2 𝜃 + cos 𝜃 − 2 = 0 
cos 𝜃 = −0.84 
𝜃 =
49𝜋
60
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
9𝜋
50
 
𝑘 = ±1; 𝜃 =
169𝜋
60
 
cos 𝜃 = 0.59 
0 =
11𝜋
36
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
7𝜋
10
 
𝑘 = ±1; 𝜃 =
83𝜋
36
 
 
− 4sin 𝜃 cos 𝜃 − sin 𝜃 = 0 
sin 𝜃 (−4cos 𝜃 − 1) = 0 
 
sin 𝜃 = 0 
𝜃 = 0 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 0 
𝑘 = 1; 𝜃 = 2𝜋 
 
cos 𝜃 = −
1
4
 
𝜃 = 7𝜋/12 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 7𝜋/12 
𝑘 = 1; 𝜃 = 31𝜋/12 
 
 
 
 
 
r 𝜃 
2.93 15o 
2.73 30o 
2.41 45o 
2 60o 
1.51 75o 
1 90o 
𝜃 = 𝜋/3 → 60𝑜 
tan 𝜇 =
1 + 2cos 𝜃
−2 sin 𝜃
= −1.15 
𝜇 = 310𝑜 
𝜑 = 10𝑜 
𝜃 = 14𝜋/9 → 280𝑜 
tan 𝜇 =
1 + 2cos 𝜃
−2 sin 𝜃
= 0.68 
𝜇 = 34𝑜 
𝜑 = 312𝑜 
 
 
• 𝑟 = 1 + 2sin 𝜃 
1 + 2sin 𝜃 = 0 
𝜃 =
11𝜋
6
 
𝜃 1 + 2sin 𝜃 𝑃𝑢𝑛𝑡𝑜 𝑑𝑒 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛 
𝑟 𝜃 
 
0 𝑟 = 1 + 2sin 0 1 0 
 
𝜋/2 𝑟 = 1 + 2sin 90 3 𝜋/2 
 
𝜋 𝑟 = 1 + 2sin 180 -1 𝜋 
 
3𝜋/2 𝑟 = 1 + 2sin 270 1 3𝜋/2 
 
 
 
 
 
 
tan 𝜑 =
4sin 𝜃 cos 𝜃 + cos 𝜃
2c𝑜𝑠2 𝜃 − sin 𝜃 − 2𝑠𝑖𝑛2 𝜃
 
 
𝐻𝑜𝑟𝑖𝑧𝑜𝑡𝑎𝑙 𝑉𝑒𝑟𝑖𝑐𝑎𝑙 
4sin 𝜃 cos 𝜃 + cos 𝜃 = 0 
cos 𝜃 (4sin 𝜃 + 1) = 0 
 
cos 𝜃 = 0 
𝜃 =
𝜋
2
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
𝜋
2
 
𝑘 = ±1; 𝜃 =
5𝜋
2
 
sin 𝜃 = −
1
4
 
0 =
23𝜋
12
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
23𝜋
12
 
𝑘 = ±1; 𝜃 =
47𝜋
10
 
 
2c𝑜𝑠2 𝜃 − sin 𝜃 − 2𝑠𝑖𝑛2 𝜃 = 0 
2 − 2𝑠𝑖𝑛2 𝜃 − sin 𝜃 − 2𝑠𝑖𝑛2 𝜃 = 0 
−4sin2 𝜃 − sin 𝜃 + 2 = 0 
sin 𝜃 = 0.59 
𝜃 =
𝜋
5
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 = 𝜋/5 
𝑘 = 1; 𝜃 = 11𝜋/5 
 
sin 𝜃 = −0.84 
𝜃 =
5𝜋
3
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 = 5𝜋/3 
𝑘 = 1; 𝜃 = 11𝜋/3 
 
 
 
 
 
r 𝜃 
1.51 15o 
2 30o 
2.41 45o 
2.73 60o 
2.93 75o 
3 90o 
𝜃 = 𝜋/3 → 60𝑜 
tan 𝜇 =
1 + 2sin 𝜃
2 cos 𝜃
= 2.73 
𝜇 = 70𝑜 
𝜑 = 130𝑜 
𝜃 = 𝜋/6 → 30𝑜 
tan 𝜇 =
1 + 2sin 𝜃
2 cos 𝜃
= 1.154 
𝜇 = 50𝑜 
𝜑 = 80𝑜 
 
• 𝑟 = 1 − 2sin 𝜃 
1 − 2sin 𝜃 = 0 
𝜃 =
𝜋
6
 
𝜃 1 − 2sin 𝜃 𝑃𝑢𝑛𝑡𝑜 𝑑𝑒 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛 
𝑟 𝜃 
 
0 𝑟 = 1 − 2sin 0 1 0 
 
𝜋/2 𝑟 = 1 − 2sin 90 -1 𝜋/2 
 
𝜋 𝑟 = 1 − 2sin 180 1 𝜋 
 
3𝜋/2 𝑟 = 1 − 2sin 270 3 3𝜋/2 
 
 
 
 
 
 
tan 𝜑 =
− 4sin 𝜃 cos 𝜃 + cos 𝜃
−2c𝑜𝑠2 𝜃 − sin 𝜃 + 2𝑠𝑖𝑛2 𝜃
 
 
 
𝐻𝑜𝑟𝑖𝑧𝑜𝑡𝑎𝑙 𝑉𝑒𝑟𝑖𝑐𝑎𝑙 
− 4sin 𝜃 cos 𝜃 + cos 𝜃 = 0 
cos 𝜃 (− 4sin 𝜃 + 1) = 0 
 
cos 𝜃 = 0 
𝜃 =
𝜋
2
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
𝜋
2
 
𝑘 = ±1; 𝜃 =
5𝜋
2
 
sin 𝜃 =
1
4
 
0 =
1𝜋
12
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
𝜋
12
 
𝑘 = ±1; 𝜃 =
25𝜋
12
 
 
−2c𝑜𝑠2 𝜃 − sin 𝜃 + 2𝑠𝑖𝑛2 𝜃 = 0 
−2 + 2𝑠𝑖𝑛2 𝜃 − sin 𝜃 + 2𝑠𝑖𝑛2 𝜃 = 0 
4sin2 𝜃 − sin 𝜃 − 2 = 0 
sin 𝜃 = −0.59 
𝜃 =
16𝜋
9
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 = 16𝜋/9 
𝑘 = 1; 𝜃 = 34𝜋/9 
 
sin 𝜃 = 0.84 
𝜃 =
𝜋
3
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 = 𝜋/3 
𝑘 = 1; 𝜃 = 7𝜋/3 
 
 
 
 
r 𝜃 
0.48 15o 
0 30o 
-0.41 45o 
-0.73 60o 
-0.93 75o 
-1 90o 
𝜃 = 13𝜋/12 → 195𝑜 
tan 𝜇 =
1 − 2sin 𝜃
−2 cos 𝜃
= 0.78 
𝜇 = 37𝑜 
𝜑 = 232𝑜 
𝜃 = 5𝜋/4 → 225𝑜 
tan 𝜇 =
1 − 2sin 𝜃
−2 cos 𝜃
= 1.70 
𝜇 = 60𝑜 
𝜑 = 285𝑜 
 
• 𝑟 = 2 + 3cos 𝜃 
2 + 3cos 𝜃 = 0 
𝜃 =
11𝜋
15
 
𝜃 2 + 3cos 𝜃 𝑃𝑢𝑛𝑡𝑜 𝑑𝑒 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛 
𝑟 𝜃 
 
0 𝑟 = 2 + 3cos 0 5 0 
 
𝜋/2 𝑟 = 2 + 3cos 90 2 𝜋/2 
 
𝜋 𝑟 = 2 + 3cos 180 -1 𝜋 
 
3𝜋/2 𝑟 = 2 + 3cos 270 2 3𝜋/2 
 
 
 
 
 
 
tan 𝜑 =
−3sin2 𝜃 + 2cos 𝜃 + 3cos2 𝜃
−6sin 𝜃 cos 𝜃 − 2sin 𝜃
 
 
𝐻𝑜𝑟𝑖𝑧𝑜𝑡𝑎𝑙 𝑉𝑒𝑟𝑖𝑐𝑎𝑙 
−3sin2 𝜃 + 2cos 𝜃 + 3cos2 𝜃 = 0 
−3 + 3cos2 𝜃 + 2cos 𝜃 + 3cos2 𝜃 = 0 
6cos2 𝜃 + 2 cos 𝜃 − 3 = 0 
cos 𝜃 = −0.89 
𝜃 =
5𝜋
6
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
5𝜋
6
 
𝑘 = ±1; 𝜃 =
17𝜋
6
 
cos 𝜃 = 0.55 
0 =
𝜋
3
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
𝜋
3
 
𝑘 = ±1; 𝜃 =
7𝜋
3
 
 
−6sin 𝜃 cos 𝜃 − 2sin 𝜃 = 0 
2 sin 𝜃 (−3cos 𝜃 − 1) = 0 
 
2 sin 𝜃 = 0 
𝜃 = 0 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 0 
𝑘 = 1; 𝜃 = 2𝜋 
 
cos 𝜃 = −
1
3
 
𝜃 = 11𝜋/18 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 11𝜋/18 
𝑘 = 1; 𝜃 = 47𝜋/18 
 
 
 
 
 
r 𝜃 
4.89 15o 
4.59 30o 
4.12 45o 
3.5 60o 
2.77 75o 
2 90o 
𝜃 = 𝜋/6 → 30𝑜 
tan 𝜇 =
2 + 3cos 𝜃
−3 sin 𝜃
= −3.06 
𝜇 = 288𝑜 
𝜑 = 318𝑜 
𝜃 = 𝜋/4 → 45𝑜 
tan 𝜇 =
2 + 3cos 𝜃
−3 sin 𝜃
= −1.94 
𝜇 = 297𝑜 
𝜑 = 342𝑜 
 
• 𝑟 = 4 − 3cos 𝜃 
 
𝜃 4 − 3cos 𝜃 𝑃𝑢𝑛𝑡𝑜 𝑑𝑒 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛𝑟 𝜃 
 
0 𝑟 = 4 − 3cos 0 1 0 
 
𝜋/2 𝑟 = 4 − 3cos 90 4 𝜋/2 
 
𝜋 𝑟 = 4 − 3cos 180 7 𝜋 
 
3𝜋/2 𝑟 = 4 − 3cos 270 4 3𝜋/2 
 
 
 
 
 
 
tan 𝜑 =
3sin2 𝜃 + 4cos 𝜃 − 3cos2 𝜃
6sin 𝜃 cos 𝜃 − 4sin 𝜃
 
 
𝐻𝑜𝑟𝑖𝑧𝑜𝑡𝑎𝑙 𝑉𝑒𝑟𝑖𝑐𝑎𝑙 
3sin2 𝜃 + 4cos 𝜃 − 3cos2 𝜃 = 0 
3 − 3cos2 𝜃 + 4 cos 𝜃 − 3cos2 𝜃 = 0 
−6cos2 𝜃 + 4cos 𝜃 + 3 = 0 
cos 𝜃 = −0.44 
𝜃 =
2𝜋
3
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
2𝜋
3
 
𝑘 = ±1; 𝜃 =
8𝜋
3
 
cos 𝜃 = 1.11 
≠ 
 
6sin 𝜃 cos 𝜃 − 4sin 𝜃 = 0 
sin 𝜃 (6cos 𝜃 − 4) = 0 
 
sin 𝜃 = 0 
𝜃 = 0 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 0 
𝑘 = 1; 𝜃 = 2𝜋 
 
cos 𝜃 =
2
3
 
𝜃 = 4𝜋/15 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 4𝜋/15 
𝑘 = 1; 𝜃 = 24𝜋/15 
 
 
 
 
 
 
 
r 𝜃 
1.10 15o 
1.40 30o 
1.87 45o 
2.5 60o 
3.22 75o 
4 90o 
𝜃 = 8𝜋/9 → 160𝑜 
tan 𝜇 =
4 − 3cos 𝜃
3 sin 𝜃
= 6.64 
𝜇 = 80𝑜 
𝜑 = 240𝑜 
𝜃 = 17𝜋/18 → 170𝑜 
tan 𝜇 =
4 − 3cos 𝜃
3 sin 𝜃
= 13.34 
𝜇 = 85𝑜 
𝜑 = 255𝑜 
 
• 𝑟 = −2 + 3sin 𝜃 
−2 + 3sin 𝜃 = 0 
𝜃 =
7𝜋
30
 
𝜃 −2 + 3sin 𝜃 𝑃𝑢𝑛𝑡𝑜 𝑑𝑒 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛 
𝑟 𝜃 
 
0 −2 + 3sin 0 -2 0 
 
𝜋/2 −2 + 3sin 90 1 𝜋/2 
 
𝜋 −2 + 3sin 180 -2 𝜋 
 
3𝜋/2 −2 + 3sin 270 -5 3𝜋/2 
 
 
 
 
 
 
tan 𝜑 =
6sin 𝜃 cos 𝜃 − 2 cos 𝜃
3c𝑜𝑠2 𝜃 + 2 sin 𝜃 − 3𝑠𝑖𝑛2 𝜃
 
 
𝐻𝑜𝑟𝑖𝑧𝑜𝑡𝑎𝑙 𝑉𝑒𝑟𝑖𝑐𝑎𝑙 
6sin 𝜃 cos 𝜃 − 2 cos 𝜃 = 0 
cos 𝜃 (6sin 𝜃 − 2) = 0 
 
cos 𝜃 = 0 
𝜃 =
𝜋
2
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
𝜋
2
 
𝑘 = ±1; 𝜃 =
5𝜋
2
 
sin 𝜃 =
1
3
 
0 =
𝜋
9
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
𝜋
9
 
𝑘 = ±1; 𝜃 =
19𝜋
9
 
 
3c𝑜𝑠2 𝜃 + 2 sin 𝜃 − 3𝑠𝑖𝑛2 𝜃 = 0 
3 − 3𝑠𝑖𝑛2 𝜃 − 2 sin 𝜃 − 3𝑠𝑖𝑛2 𝜃 = 0 
−6sin2 𝜃 − 2 sin 𝜃 + 3 = 0 
sin 𝜃 = 0.59 
𝜃 =
𝜋
5
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 = 𝜋/5 
𝑘 = 1; 𝜃 = 11𝜋/5 
 
sin 𝜃 = −0.84 
𝜃 =
5𝜋
3
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 = 5𝜋/3 
𝑘 = 1; 𝜃 = 11𝜋/3 
 
 
 
 
 
 
r 𝜃 
-1.22 15o 
-0.5 30o 
0.12 45o 
0.59 60o 
0.89 75o 
1 90o 
𝜃 = 𝜋/12 → 15𝑜 
tan 𝜇 =
−2 + 3sin 𝜃
3 cos 𝜃
= −0.42 
𝜇 = 337𝑜 
𝜑 = 352𝑜 
𝜃 = 𝜋/6 → 30𝑜 
tan 𝜇 =
−2 + 3sin 𝜃
3 cos 𝜃
= −0.19 
𝜇 = 350𝑜 
𝜑 = 20𝑜 
 
• 𝑟 =
1
2
− cos 𝜃 
1
2
− cos 𝜃 = 0 
𝜃 =
𝜋
3
 
𝜃 1/2 − cos 𝜃 𝑃𝑢𝑛𝑡𝑜 𝑑𝑒 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛 
𝑟 𝜃 
 
0 𝑟 = 1/2 − cos 0 -1/2 0 
 
𝜋/2 𝑟 = 1/2 − cos 90 ½ 𝜋/2 
 
𝜋 𝑟 = 1/2 − cos 180 3/2 𝜋 
 
3𝜋/2 𝑟 = 1/2 − cos 270 1/2 3𝜋/2 
 
 
 
 
 
 
tan 𝜑 =
sin2 𝜃 + cos 𝜃/2 − cos2 𝜃
2sin 𝜃 cos 𝜃 − sin 𝜃/2
 
 
𝐻𝑜𝑟𝑖𝑧𝑜𝑡𝑎𝑙 𝑉𝑒𝑟𝑖𝑐𝑎𝑙 
sin2 𝜃 + cos 𝜃/2 − cos2 𝜃 = 0 
1 − cos2 𝜃 + cos 𝜃/2 − cos2 𝜃 = 0 
−2cos2 𝜃 + cos 𝜃/2 + 1 = 0 
cos 𝜃 = 0.84 
𝜃 =
9𝜋
50
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
9𝜋
50
 
𝑘 = ±1; 𝜃 =
109𝜋
50
 
cos 𝜃 = −0.59 
0 =
7𝜋
10
+ 2𝜋𝑘 
𝑘 = 0; 𝜃 =
7𝜋
10
 
𝑘 = ±1; 𝜃 =
27𝜋
10
 
 
2sin 𝜃 cos 𝜃 − sin 𝜃/2 = 0 
sin 𝜃 (2cos 𝜃 − 1/2) = 0 
 
sin 𝜃 = 0 
𝜃 = 0 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 0 
𝑘 = 1; 𝜃 = 2𝜋 
 
cos 𝜃 =
1
4
 
𝜃 = 5𝜋/12 + 2𝜋𝑘 
𝑘 = 0; 𝜃 = 5𝜋/12 
𝑘 = 1; 𝜃 = 29𝜋/12 
 
 
 
 
r 𝜃 
-0.46 15o 
-0.36 30o 
-0.20 45o 
0 60o 
0.24 75o 
0.5 90o 
𝜃 = 13𝜋/18 → 130𝑜 
tan 𝜇 =
1/2 − cos 𝜃
sin 𝜃
= 1.49 
𝜇 = 56𝑜 
𝜑 = 186𝑜 
𝜃 = 19𝜋/36 → 145𝑜 
tan 𝜇 =
1/2 − cos 𝜃
sin 𝜃
= 2.29 
𝜇 = 66.5𝑜 
𝜑 = 210𝑜