Converting Equations Polar & Cartesian - MathHands
Trigonometry Sec. 03 notes
MathHands.com M´ arquez
Converting Equations Polar & Cartesian The IDEA In the last couple sections we’ve established, among other things, a ’dictionary’ to convert coordinate points from cartesian to polar and vise-varsa. In this section, we would like to exploit that same dictionary to convert equations. As a reminder, the dictionary has two components, ’from’ polar, and ’to’ polar. The goal in converting, in most cases, is to eliminate all x’s and y’s and replace them with appropriate expressions involving r’s and θ’s if we are translating to polar. When converting from polar, the idea is to eliminate the r’s and θ’s. We summarize the ’dictionary’ below. Translating
Polar 90◦ 120◦
Cartesian
y = r sin θ x = r cos θ
60◦ 30◦
150◦ b
b
(r, θ)
(x, y)
0◦
180◦
210◦
330◦
θ = tan−1 300◦
240◦
r2 = x2 + y
270◦
y x 2
Example: Converting Equation of a line to Polar Convert the following equation to polar coordinates: y=x Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
y y x tan θ θ
=x
(given)
=1
(div by x, setting up to use translating dictionary, tan θ = y/x)
=1 = 45◦
(used tan θ = y/x) (notice: this there are infinite possibilities for θ (ie −135 , 225 ...), this is just one of them.) ◦
◦
Thus, one way to translate the equation y = x to polar is to convert to θ = 45◦ Example: Converting Equation of a circle to Polar Convert the following equation to polar coordinates: x2 + y 2 = 25 Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
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pg. 1
Trigonometry Sec. 03 notes
MathHands.com M´ arquez x2 + y 2 = 25
(given)
2
2
(used the dictionary r = x2 + y 2 )
r = 25
at this point we have eliminated all x’s and y’s from the equation, thus the converting is complete. It is sometimes desirable to solve for r explicitly, thus we could also say r = 5 or r = −5, in fact all three of these equations, r2 = 25, r = 5, AND r = −5 all yield the same graph as x2 + y 2 = 25. Example: Converting the equation of a circle to Polar Convert the following equation to polar coordinates: x2 + 3x + 5 + y 2 − 4y + 2 = 25 Solution:
x2 + 3x + 5 + y 2 − 4y + 2 = 25 2
(given)
2
(x + y ) + 3(x) − 4(y) = 18
(just cleaning up, getting ready to use the dictionary)
(r ) + 3(r cos θ) − 4(r sin θ) = 18
(used the dictionary)
2
2
r + 3r cos θ − 4r sin θ = 18
(algebra)
Example: Converting the equation of a circle from Polar Convert the following equation of a circle to polar coordinates: r = 12 cos θ Solution: r = 12 cos θ r · r = r(12 cos θ)
(given) (clever little idea, mult both sides by r)
r2 = 12(r cos θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 12x 2
2
y + x + − 12x = 0 2
2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
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(used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
pg. 2
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Converting Equations Polar & Cartesian 1. Convert the following equation to polar coordinates: y = 2x
Solution: Note: the ’translating’ of this equation is not unique.
y = 2x y =2 x tan θ = 2 θ ≈ 63.435
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) ◦
(notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 63.435◦
2. Convert the following equation to polar coordinates: 2 y= − x 3
Solution: Note: the ’translating’ of this equation is not unique. 2 y= − x 3 2 y = − x 3 2 tan θ = − 3 θ ≈ −33.69◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ −33.69◦
3. Convert the following equation to polar coordinates: y=
2 x 5
Solution: Note: the ’translating’ of this equation is not unique.
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pg. 3
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez 2 x 5 2 = 5 2 = 5 ≈ 21.801◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 21.801◦ 4. Convert the following equation to polar coordinates: y=
4 x 3
Solution: Note: the ’translating’ of this equation is not unique. 4 x 3 4 = 3 4 = 3 ≈ 53.13◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 4x to polar is to convert to θ ≈ 53.13◦ 5. Convert the following equation to polar coordinates: 4 y= − x 3 Solution: Note: the ’translating’ of this equation is not unique. 4 y= − x 3 4 y = − x 3 4 tan θ = − 3 θ ≈ −53.13◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = − 4x to polar is to convert to θ ≈ −53.13◦ 6. Convert the following equation of a circle to polar coordinates: 2x2 + 3x + 2y 2 + − 5y = 7
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pg. 4
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution:
2x2 + 3x + 2y 2 + − 5y = 7 2 x2 + y 2 + 3(x) + − 5(y) = 7 2 r2 + 3(r cos θ) + − 5(r sin θ) = 7
(given) (ready to use the dictionary....) (used the dictionary)
7. Convert the following equation of a circle to polar coordinates: − 5x2 + 2x + − 5y 2 + 7y = 25
Solution:
− 5x2 + 2x + − 5y 2 + 7y = 25 − 5 x2 + y 2 + 2(x) + 7(y) = 25 − 5 r2 + 2(r cos θ) + 7(r sin θ) = 25
(given) (ready to use the dictionary....) (used the dictionary)
8. Convert the following equation of a circle to polar coordinates: 3 4x2 + x + 4y 2 + 1y = 9 2
Solution: 3 4x2 + x + 4y 2 + 1y = 9 2 3 4 x2 + y 2 + (x) + 1(y) = 9 2 3 2 4 r + (r cos θ) + 1(r sin θ) = 9 2
(given) (ready to use the dictionary....) (used the dictionary)
9. Convert the following equation of a circle to polar coordinates: 3x2 + 3x + 3y 2 + 5y = 4
Solution:
3x2 + 3x + 3y 2 + 5y = 4 3 x2 + y 2 + 3(x) + 5(y) = 4 3 r2 + 3(r cos θ) + 5(r sin θ) = 4
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(given) (ready to use the dictionary....) (used the dictionary)
pg. 5
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
10. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 9 25
Solution: y2 x2 + =1 9 25 (r cos θ)2 (r sin θ)2 + =1 9 25 ! cos2 θ sin2 θ 2 + =1 r 9 25
(given) (ready to use the dictionary....) (alebra)
11. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 16
Solution: y2 x2 + =1 4 16 (r cos θ)2 (r sin θ)2 + =1 4 16 ! cos2 θ sin2 θ + =1 r2 4 16
(given) (ready to use the dictionary....) (alebra)
12. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 49
Solution: y2 x2 + =1 4 49 (r cos θ)2 (r sin θ)2 + =1 4 49 ! cos2 θ sin2 θ =1 + r2 4 49
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(given) (ready to use the dictionary....) (alebra)
pg. 6
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
13. Convert the following equation of a circle to polar coordinates: r = 2 sin θ
Solution: r = 2 sin θ
(given)
r · r = r(2 sin θ)
(clever little idea, mult both sides by r)
r2 = 2(r sin θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 2y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 2y = 0 2
(optional step: complete the square)
2
x + y + − 2y + 1 = 1 2 x2 + y + − 1 = 1
(optional to step: complete the square) (Shows its a circle, shows center/radius)
14. Convert the following equation of a circle to polar coordinates: r = 5 sin θ
Solution: r = 5 sin θ
(given)
r · r = r(5 sin θ)
(clever little idea, mult both sides by r)
2
r = 5(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 5y = 0 25 25 x2 + y 2 + − 5y + = 4 4 2 25 5 = x2 + y + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
15. Convert the following equation of a circle to polar coordinates: r = 6 sin θ
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pg. 7
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 6 sin θ r · r = r(6 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 6(r sin θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 6y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 6y = 0 2
(optional step: complete the square)
2
x + y + − 6y + 9 = 9 2 x2 + y + − 3 = 9
(optional to step: complete the square) (Shows its a circle, shows center/radius)
16. Convert the following equation of a circle to polar coordinates: r = 11 sin θ
Solution: r = 11 sin θ r · r = r(11 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 11(r sin θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 11y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 11y = 0 121 121 = x2 + y 2 + − 11y + 4 4 2 11 121 x2 + y + − = 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
17. Convert the following equation of a circle to polar coordinates: r = 8 cos θ
Solution: r = 8 cos θ
(given)
r · r = r(8 cos θ)
(clever little idea, mult both sides by r)
2
r = 8(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 8x 2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 8x = 0 2
(optional step: complete the square)
2
y + x + − 8x + 16 = 16 2 y 2 + x + − 4 = 16
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2007-2009 MathHands.com
(optional to step: complete the square) (Shows its a circle, shows center/radius)
math hands
pg. 8
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
18. Convert the following equation of a circle to polar coordinates: r = 12 cos θ
Solution: r = 12 cos θ
(given)
r · r = r(12 cos θ)
(clever little idea, mult both sides by r)
r2 = 12(r cos θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 12x 2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 12x = 0 2
(optional step: complete the square)
2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
(optional to step: complete the square) (Shows its a circle, shows center/radius)
19. Convert the following equation of a circle to polar coordinates: r = 5 cos θ
Solution: r = 5 cos θ
(given)
r · r = r(5 cos θ)
(clever little idea, mult both sides by r)
2
r = 5(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5x 2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 5x = 0 25 25 y 2 + x2 + − 5x + = 4 4 2 25 5 = y2 + x + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
20. Convert the following equation of a circle to polar coordinates: r = 11 cos θ
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pg. 9
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 11 cos θ r · r = r(11 cos θ)
(given) (clever little idea, mult both sides by r)
r2 = 11(r cos θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 11x 2
2
y + x + − 11x = 0 121 121 = y 2 + x2 + − 11x + 4 4 2 121 11 = y2 + x + − 2 4
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(used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
pg. 10
MathHands.com M´ arquez
Converting Equations Polar & Cartesian The IDEA In the last couple sections we’ve established, among other things, a ’dictionary’ to convert coordinate points from cartesian to polar and vise-varsa. In this section, we would like to exploit that same dictionary to convert equations. As a reminder, the dictionary has two components, ’from’ polar, and ’to’ polar. The goal in converting, in most cases, is to eliminate all x’s and y’s and replace them with appropriate expressions involving r’s and θ’s if we are translating to polar. When converting from polar, the idea is to eliminate the r’s and θ’s. We summarize the ’dictionary’ below. Translating
Polar 90◦ 120◦
Cartesian
y = r sin θ x = r cos θ
60◦ 30◦
150◦ b
b
(r, θ)
(x, y)
0◦
180◦
210◦
330◦
θ = tan−1 300◦
240◦
r2 = x2 + y
270◦
y x 2
Example: Converting Equation of a line to Polar Convert the following equation to polar coordinates: y=x Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
y y x tan θ θ
=x
(given)
=1
(div by x, setting up to use translating dictionary, tan θ = y/x)
=1 = 45◦
(used tan θ = y/x) (notice: this there are infinite possibilities for θ (ie −135 , 225 ...), this is just one of them.) ◦
◦
Thus, one way to translate the equation y = x to polar is to convert to θ = 45◦ Example: Converting Equation of a circle to Polar Convert the following equation to polar coordinates: x2 + y 2 = 25 Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
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math hands
pg. 1
Trigonometry Sec. 03 notes
MathHands.com M´ arquez x2 + y 2 = 25
(given)
2
2
(used the dictionary r = x2 + y 2 )
r = 25
at this point we have eliminated all x’s and y’s from the equation, thus the converting is complete. It is sometimes desirable to solve for r explicitly, thus we could also say r = 5 or r = −5, in fact all three of these equations, r2 = 25, r = 5, AND r = −5 all yield the same graph as x2 + y 2 = 25. Example: Converting the equation of a circle to Polar Convert the following equation to polar coordinates: x2 + 3x + 5 + y 2 − 4y + 2 = 25 Solution:
x2 + 3x + 5 + y 2 − 4y + 2 = 25 2
(given)
2
(x + y ) + 3(x) − 4(y) = 18
(just cleaning up, getting ready to use the dictionary)
(r ) + 3(r cos θ) − 4(r sin θ) = 18
(used the dictionary)
2
2
r + 3r cos θ − 4r sin θ = 18
(algebra)
Example: Converting the equation of a circle from Polar Convert the following equation of a circle to polar coordinates: r = 12 cos θ Solution: r = 12 cos θ r · r = r(12 cos θ)
(given) (clever little idea, mult both sides by r)
r2 = 12(r cos θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 12x 2
2
y + x + − 12x = 0 2
2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
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2007-2009 MathHands.com
math hands
(used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
pg. 2
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Converting Equations Polar & Cartesian 1. Convert the following equation to polar coordinates: y = 2x
Solution: Note: the ’translating’ of this equation is not unique.
y = 2x y =2 x tan θ = 2 θ ≈ 63.435
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) ◦
(notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 63.435◦
2. Convert the following equation to polar coordinates: 2 y= − x 3
Solution: Note: the ’translating’ of this equation is not unique. 2 y= − x 3 2 y = − x 3 2 tan θ = − 3 θ ≈ −33.69◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ −33.69◦
3. Convert the following equation to polar coordinates: y=
2 x 5
Solution: Note: the ’translating’ of this equation is not unique.
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pg. 3
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez 2 x 5 2 = 5 2 = 5 ≈ 21.801◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 21.801◦ 4. Convert the following equation to polar coordinates: y=
4 x 3
Solution: Note: the ’translating’ of this equation is not unique. 4 x 3 4 = 3 4 = 3 ≈ 53.13◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 4x to polar is to convert to θ ≈ 53.13◦ 5. Convert the following equation to polar coordinates: 4 y= − x 3 Solution: Note: the ’translating’ of this equation is not unique. 4 y= − x 3 4 y = − x 3 4 tan θ = − 3 θ ≈ −53.13◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = − 4x to polar is to convert to θ ≈ −53.13◦ 6. Convert the following equation of a circle to polar coordinates: 2x2 + 3x + 2y 2 + − 5y = 7
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math hands
pg. 4
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution:
2x2 + 3x + 2y 2 + − 5y = 7 2 x2 + y 2 + 3(x) + − 5(y) = 7 2 r2 + 3(r cos θ) + − 5(r sin θ) = 7
(given) (ready to use the dictionary....) (used the dictionary)
7. Convert the following equation of a circle to polar coordinates: − 5x2 + 2x + − 5y 2 + 7y = 25
Solution:
− 5x2 + 2x + − 5y 2 + 7y = 25 − 5 x2 + y 2 + 2(x) + 7(y) = 25 − 5 r2 + 2(r cos θ) + 7(r sin θ) = 25
(given) (ready to use the dictionary....) (used the dictionary)
8. Convert the following equation of a circle to polar coordinates: 3 4x2 + x + 4y 2 + 1y = 9 2
Solution: 3 4x2 + x + 4y 2 + 1y = 9 2 3 4 x2 + y 2 + (x) + 1(y) = 9 2 3 2 4 r + (r cos θ) + 1(r sin θ) = 9 2
(given) (ready to use the dictionary....) (used the dictionary)
9. Convert the following equation of a circle to polar coordinates: 3x2 + 3x + 3y 2 + 5y = 4
Solution:
3x2 + 3x + 3y 2 + 5y = 4 3 x2 + y 2 + 3(x) + 5(y) = 4 3 r2 + 3(r cos θ) + 5(r sin θ) = 4
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math hands
(given) (ready to use the dictionary....) (used the dictionary)
pg. 5
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
10. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 9 25
Solution: y2 x2 + =1 9 25 (r cos θ)2 (r sin θ)2 + =1 9 25 ! cos2 θ sin2 θ 2 + =1 r 9 25
(given) (ready to use the dictionary....) (alebra)
11. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 16
Solution: y2 x2 + =1 4 16 (r cos θ)2 (r sin θ)2 + =1 4 16 ! cos2 θ sin2 θ + =1 r2 4 16
(given) (ready to use the dictionary....) (alebra)
12. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 49
Solution: y2 x2 + =1 4 49 (r cos θ)2 (r sin θ)2 + =1 4 49 ! cos2 θ sin2 θ =1 + r2 4 49
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(given) (ready to use the dictionary....) (alebra)
pg. 6
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
13. Convert the following equation of a circle to polar coordinates: r = 2 sin θ
Solution: r = 2 sin θ
(given)
r · r = r(2 sin θ)
(clever little idea, mult both sides by r)
r2 = 2(r sin θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 2y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 2y = 0 2
(optional step: complete the square)
2
x + y + − 2y + 1 = 1 2 x2 + y + − 1 = 1
(optional to step: complete the square) (Shows its a circle, shows center/radius)
14. Convert the following equation of a circle to polar coordinates: r = 5 sin θ
Solution: r = 5 sin θ
(given)
r · r = r(5 sin θ)
(clever little idea, mult both sides by r)
2
r = 5(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 5y = 0 25 25 x2 + y 2 + − 5y + = 4 4 2 25 5 = x2 + y + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
15. Convert the following equation of a circle to polar coordinates: r = 6 sin θ
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pg. 7
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 6 sin θ r · r = r(6 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 6(r sin θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 6y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 6y = 0 2
(optional step: complete the square)
2
x + y + − 6y + 9 = 9 2 x2 + y + − 3 = 9
(optional to step: complete the square) (Shows its a circle, shows center/radius)
16. Convert the following equation of a circle to polar coordinates: r = 11 sin θ
Solution: r = 11 sin θ r · r = r(11 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 11(r sin θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 11y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 11y = 0 121 121 = x2 + y 2 + − 11y + 4 4 2 11 121 x2 + y + − = 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
17. Convert the following equation of a circle to polar coordinates: r = 8 cos θ
Solution: r = 8 cos θ
(given)
r · r = r(8 cos θ)
(clever little idea, mult both sides by r)
2
r = 8(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 8x 2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 8x = 0 2
(optional step: complete the square)
2
y + x + − 8x + 16 = 16 2 y 2 + x + − 4 = 16
c
2007-2009 MathHands.com
(optional to step: complete the square) (Shows its a circle, shows center/radius)
math hands
pg. 8
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
18. Convert the following equation of a circle to polar coordinates: r = 12 cos θ
Solution: r = 12 cos θ
(given)
r · r = r(12 cos θ)
(clever little idea, mult both sides by r)
r2 = 12(r cos θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 12x 2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 12x = 0 2
(optional step: complete the square)
2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
(optional to step: complete the square) (Shows its a circle, shows center/radius)
19. Convert the following equation of a circle to polar coordinates: r = 5 cos θ
Solution: r = 5 cos θ
(given)
r · r = r(5 cos θ)
(clever little idea, mult both sides by r)
2
r = 5(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5x 2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 5x = 0 25 25 y 2 + x2 + − 5x + = 4 4 2 25 5 = y2 + x + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
20. Convert the following equation of a circle to polar coordinates: r = 11 cos θ
c
2007-2009 MathHands.com
math hands
pg. 9
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 11 cos θ r · r = r(11 cos θ)
(given) (clever little idea, mult both sides by r)
r2 = 11(r cos θ)
(algebra, getting ready to use dictionary)
2
2
x + y = 11x 2
2
y + x + − 11x = 0 121 121 = y 2 + x2 + − 11x + 4 4 2 121 11 = y2 + x + − 2 4
c
2007-2009 MathHands.com
math hands
(used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
pg. 10
