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Mathematics Lecture Notes: Trigonometric Functions - Area Calculation of Regions - Prof. J, Study notes of Analytical Geometry and Calculus

University of Southern Mississippi (USM)Analytical Geometry and Calculus
Prof. James Lambers

Solutions to selected exercises from sections 9.2, 9.3, and 9.4 of a university mathematics textbook. The examples involve calculating tangents and areas using trigonometric functions. Students studying advanced mathematics, particularly those focusing on calculus and trigonometry, may find this document useful.

Typology: Study notes

2009/2010

Uploaded on 02/25/2010

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Jim Lambers
MAT 169
Fall Semester 2009-10
Lecture 38 Examples
These examples correspond to Sections 9.2, 9.3 and 9.4 in the text.
Example (Section 9.3, Exercise 67) Let ๐‘ƒbe any point (except the origin) on the curve ๐‘Ÿ=๐‘“(๐œƒ).
If ๐œ“is the angle between the tangent line at ๐‘ƒand the radial line ๐‘‚๐‘ƒ , show that
tan ๐œ“=๐‘Ÿ
๐‘‘๐‘Ÿ/๐‘‘๐œƒ
[Hint: Observe that ๐œ“=๐œ™โˆ’๐œƒin the ๏ฌgure on page 504 in the text.]
Solution We have
tan ๐œ“= tan(๐œ™โˆ’๐œƒ) = tan ๐œ™โˆ’tan ๐œƒ
1 + tan ๐œ™tan ๐œƒ.
Because ๐œ™is the angle that the tangent line makes with the positive ๐‘ฅ-axis,
tan ๐œ™=๐‘‘๐‘ฆ
๐‘‘๐‘ฅ =
๐‘‘๐‘Ÿ
๐‘‘๐œƒ sin ๐œƒ+๐‘Ÿcos ๐œƒ
๐‘‘๐‘Ÿ
๐‘‘๐œƒ cos ๐œƒโˆ’๐‘Ÿsin ๐œƒ.
It follows that if we write ๐‘Ÿโ€ฒ=๐‘‘๐‘Ÿ/๐‘‘๐œƒ, then
tan ๐œ“=
๐‘Ÿโ€ฒsin ๐œƒ+๐‘Ÿcos ๐œƒ
๐‘Ÿโ€ฒcos ๐œƒโˆ’๐‘Ÿsin ๐œƒโˆ’sin ๐œƒ
cos ๐œƒ
1 + ๐‘Ÿโ€ฒsin ๐œƒ+๐‘Ÿcos ๐œƒ
๐‘Ÿโ€ฒcos ๐œƒโˆ’๐‘Ÿsin ๐œƒ
sin ๐œƒ
cos ๐œƒ
,
we can simplify by putting all fractions over a common denominator. Because the common denom-
inators are both equal to cos ๐œƒ(๐‘Ÿโ€ฒcos ๐œƒโˆ’๐‘Ÿsin ๐œƒ), they cancel, and we obtain
tan ๐œ“=(๐‘Ÿโ€ฒsin ๐œƒ+๐‘Ÿcos ๐œƒ) cos ๐œƒโˆ’(๐‘Ÿโ€ฒcos ๐œƒโˆ’๐‘Ÿsin ๐œƒ) sin ๐œƒ
(๐‘Ÿโ€ฒcos ๐œƒโˆ’๐‘Ÿsin ๐œƒ) cos ๐œƒ+ (๐‘Ÿโ€ฒsin ๐œƒ+๐‘Ÿcos ๐œƒ) sin ๐œƒ.
Expanding, and using sin2๐œƒ+ cos2๐œƒ= 1, we obtain tan๐œ“=๐‘Ÿ/๐‘Ÿโ€ฒ.โ–ก
Example (Section 9.4, Exercise 15) Find the area of the region enclosed by one loop of the curve
๐‘Ÿ= sin 2๐œƒ.
Solution We have ๐‘Ÿ= 0 when ๐œƒ= 0 and ๐œƒ=๐œ‹/2, so the interval 0 โ‰ค๐œƒโ‰ค๐œ‹/2 corresponds to one
loop of the curve. Therefore, the area of the region enclosed by this loop is
1
2โˆซ๐œ‹/2
0
sin22๐œƒ ๐‘‘๐œƒ =1
2โˆซ๐œ‹/2
0
1โˆ’cos 4๐œƒ
2๐‘‘๐œƒ =1
4[๐œƒโˆ’sin 4๐œƒ
2]๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
๐œ‹/2
0
=๐œ‹
8,
1
pf3

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Download Mathematics Lecture Notes: Trigonometric Functions - Area Calculation of Regions - Prof. J and more Study notes Analytical Geometry and Calculus in PDF only on Docsity!

Jim Lambers MAT 169 Fall Semester 2009- Lecture 38 Examples

These examples correspond to Sections 9.2, 9.3 and 9.4 in the text.

Example (Section 9.3, Exercise 67) Let ํ‘ƒ be any point (except the origin) on the curve ํ‘Ÿ = ํ‘“ (ํœƒ). If ํœ“ is the angle between the tangent line at ํ‘ƒ and the radial line ํ‘‚ํ‘ƒ , show that

tan ํœ“ =

[Hint: Observe that ํœ“ = ํœ™ โˆ’ ํœƒ in the figure on page 504 in the text.]

Solution We have tan ํœ“ = tan(ํœ™ โˆ’ ํœƒ) = tan^ ํœ™^ โˆ’^ tan^ ํœƒ 1 + tan ํœ™ tan ํœƒ

Because ํœ™ is the angle that the tangent line makes with the positive ํ‘ฅ-axis,

tan ํœ™ =

ํ‘‘ํ‘Ÿ ํ‘‘ํœƒ sin^ ํœƒ^ +^ ํ‘Ÿ^ cos^ ํœƒ ํ‘‘ํ‘Ÿ ํ‘‘ํœƒ cos^ ํœƒ^ โˆ’^ ํ‘Ÿ^ sin^ ํœƒ

It follows that if we write ํ‘Ÿโ€ฒ^ = ํ‘‘ํ‘Ÿ/ํ‘‘ํœƒ, then

tan ํœ“ =

ํ‘Ÿโ€ฒ^ sin ํœƒ+ํ‘Ÿ cos ํœƒ ํ‘Ÿโ€ฒ^ cos ํœƒโˆ’ํ‘Ÿ sin ํœƒ โˆ’^

sin ํœƒ cos ํœƒ 1 + ํ‘Ÿํ‘Ÿโ€ฒโ€ฒ^ sincos^ ํœƒ ํœƒ+โˆ’ํ‘Ÿํ‘Ÿ^ cossin^ ํœƒํœƒ^ sin cos^ ํœƒํœƒ

we can simplify by putting all fractions over a common denominator. Because the common denom- inators are both equal to cos ํœƒ(ํ‘Ÿโ€ฒ^ cos ํœƒ โˆ’ ํ‘Ÿ sin ํœƒ), they cancel, and we obtain

tan ํœ“ =

(ํ‘Ÿโ€ฒ^ sin ํœƒ + ํ‘Ÿ cos ํœƒ) cos ํœƒ โˆ’ (ํ‘Ÿโ€ฒ^ cos ํœƒ โˆ’ ํ‘Ÿ sin ํœƒ) sin ํœƒ (ํ‘Ÿโ€ฒ^ cos ํœƒ โˆ’ ํ‘Ÿ sin ํœƒ) cos ํœƒ + (ํ‘Ÿโ€ฒ^ sin ํœƒ + ํ‘Ÿ cos ํœƒ) sin ํœƒ.

Expanding, and using sin^2 ํœƒ + cos^2 ํœƒ = 1, we obtain tan ํœ“ = ํ‘Ÿ/ํ‘Ÿโ€ฒ. โ–ก

Example (Section 9.4, Exercise 15) Find the area of the region enclosed by one loop of the curve ํ‘Ÿ = sin 2ํœƒ.

Solution We have ํ‘Ÿ = 0 when ํœƒ = 0 and ํœƒ = ํœ‹/2, so the interval 0 โ‰ค ํœƒ โ‰ค ํœ‹/2 corresponds to one loop of the curve. Therefore, the area of the region enclosed by this loop is

1 2

Z ํœ‹/ 2

0

sin^2 2 ํœƒ ํ‘‘ํœƒ =

Z ํœ‹/ 2

0

1 โˆ’ cos 4ํœƒ 2 ํ‘‘ํœƒ^ =

sin 4ํœƒ 2

ํœ‹/ 2

0

where a double-angle identity was used to rewrite sin^2 2 ํœƒ in such a way that it can be integrated. โ–ก

Example (Section 9.4, Exercise 21) Find the area of the region that lies inside the curve ํ‘Ÿ = 3 cos ํœƒ and outside the curve ํ‘Ÿ = 1 + cos ํœƒ.

Solution These curves intersect when 3 cos ํœƒ = 1 + cos ํœƒ, or cos ํœƒ = 1/2, which is the case when ํœƒ = ยฑํœ‹/3. Therefore, the area ํด of the region between them is given by

Z ํœ‹/ 3

โˆ’ํœ‹/ 3

[3 cos ํœƒ]^2 โˆ’ [1 + cos ํœƒ]^2 ํ‘‘ํœƒ

Z ํœ‹/ 3

โˆ’ํœ‹/ 3

9 cos^2 ํœƒ โˆ’ (1 + 2 cos ํœƒ + cos^2 ํœƒ) ํ‘‘ํœƒ

Z ํœ‹/ 3

โˆ’ํœ‹/ 3

8 cos^2 ํœƒ โˆ’ 1 โˆ’ 2 cos ํœƒ ํ‘‘ํœƒ

Z ํœ‹/ 3

โˆ’ํœ‹/ 3

8 1 + cos 2 2 ํœƒโˆ’ 1 โˆ’ 2 cos ํœƒ ํ‘‘ํœƒ

Z ํœ‹/ 3

โˆ’ํœ‹/ 3

3 + 4 cos 2ํœƒ โˆ’ 2 cos ํœƒ ํ‘‘ํœƒ

3 ํœƒ + 4 sin 2 2 ํœƒโˆ’ 2 sin ํœƒ

ํœ‹/ 3

โˆ’ํœ‹/ 3 = ํœ‹.

โ–ก

Example (Section 9.4, Exercise 23) Find the area of the region that lies inside both of the curves ํ‘Ÿ = sin ํœƒ and ํ‘Ÿ = cos ํœƒ.

Solution These curves intersect when ํœƒ = ํœ‹/4. Therefore, if we divide the region that lies inside both curves with the ray ํœƒ = ํœ‹/4, we can obtain its area ํด by computing

Z ํœ‹/ 4

0

sin^2 ํœƒ ํ‘‘ํœƒ +^12

Z ํœ‹/ 2

ํœ‹/ 4

ํ‘ํ‘œํ‘ ^2 ํœƒ ํ‘‘ํœƒ

Z ํœ‹/ 4

0

1 โˆ’ cos 2ํœƒ 2

ํ‘‘ํœƒ +^1

Z ํœ‹/ 2

ํœ‹/ 4

1 + cos 2ํœƒ 2

2 โˆ’^

sin 2ํœƒ 4

ํœ‹/ 4

0

sin 2ํœƒ 4

ํœ‹/ 2

ํœ‹/ 4 = ํœ‹ 8