10.3 Polar Coordinates

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1 .3 Polar Coordinates Plot the points whose polar coordinates are given. Then find two other pairs of polar coordinates of this point, one with r > and one with r <.. (a) (, /4) (b) (, 3/) (c) (3, /3) /4 (b) (a) (c) 3/ /3 To draw the points, we find the angle and radius. To find different pairs, for r > we can just add, since that rotates us 36 to bring us back to where we were: For (a): (, 9/4), for (b): (, 7/), for (c): (3, 5/3). To change the radius to r <, we just add a negative sign to the radius and add to the angle. Adding the negative sign to the radius sends us to the symmetric point (with respect to (, )), and adding rotates us back to our original point. So, for (a): (, 5/4), for (b): (, 5/), for (c): ( 3, /3).. (a) (, 5/6) (b) (, /3) (c) (, 5/4) 5/6 (a) (b) (c) 5/4 /3 Again, for r > another equivalent set of points is (, 7/6), (, 4/3), (, 3/4), found by adding to θ. As for r <, (, /6), (, /3), (, 9/4) found by negating r and adding to θ. Plot the points whose polar coordinates are given. Then find the Cartesian coordinates of the point. 3. (a) (, 3/) (b) (, /4) (c) (, /6) /4 (c) (b) (a) /6 3/ To find the Cartesian coordinates, we use the formulas = r cos θ and y = r sin θ to obtain (a) (, y) = (, ) (b) (, y) = (, ) (c) (, y) = (, 3 ) 4. (a) (4, 4/3) (b) (, 3/4) (c) ( 3, /3)

2 3/4 (c) (a) (b) 4/3 /3 To find the Cartesian coordinates, we use the formulas = r cos θ and y = r sin θ to obtain (a) (, y) = (, 3) (b) (, y) = (, ) (c) (, y) = ( 3, 3 3 ) The Cartesian coordinates of a point are given. Find the polar coordinates for r > and θ <. Find the polar coordinates for r < and θ <. 5. (a) ( 4, 4) (b) (3, 3 3) We use the formulas r = + y and θ = arctan ( y ) to find (a) r = 4, θ = /4. We don t like θ being negative, so we add until it becomes positive. Thus: (4, 7/4). (b) (r, θ) = (6, /3) Repeating for r <, θ <, we negate r and add (or subtract) to θ, to keep it within [, ), (a) ( 4, 3/4). Here we subtracted since adding gives us /4. (b) ( 6, 4/3). Here we added since subtracting gives /3 <. 6. (a) ( 3, ) (b) ( 6, ) (a) r =, θ = /6. We don t like θ being negative, so we add until it becomes positive. Thus: (, /6). (b) Here the point ( 6, ) lies on the left of the -ais. Thus r = 6, θ =. Repeating for r <, θ <, we negate r and add (or subtract) to θ, to keep it within [, ), (a) (, 5/6). Here we subtracted since adding gives 7/6. (b) ( 6, ). Here we subtracted since adding gives. Identify the curve by finding a Cartesian equation for the curve. Here we repeatedly use the following triangle: y r = + y θ and the equations r = + y and tan θ = y/. 7. r = 5. We have + y = 5, thus a circle of radius 5 centered at (, ). 8. r = 4 sec θ. Using the triangle, + y + y = 4 = 4. Thus a vertical line = 4.

3 9. r = 5 cos θ. Using the triangle, + y = 5 + y +y 5 =. We complete the square to obtain ( 5/) +y = 5 4. This is a circle centered at (5/, ) of radius 5/.. θ = /3. We have tan(θ) = 3 and thus y/ = 3 = y = 3.. r cos(θ) =. We don t like θ so we use the double angle identity, ( ) cos(θ) = cos () = = y + y + y Thus the equation becomes + y ( y. r sin(θ) =. We don t like θ so we use an identity, +y ) = y =. This is a hyperbola. y sin(θ) = sin θ cos θ = + y + y = y + y ( Thus the equation becomes + y ) y = y =. This is a hyperbola (you can solve for y = + y ) Find a polar equation for the curve represented by the given Cartesian equation. Here we use = r cos θ and y = r sin θ. 3. y =. Thus r sin θ = = r = sin θ. 4. y =. Here r sin θ = r cos θ = tan θ = thus the equation is θ = /4. 5. y = + 3. Here r sin θ = + 3r cos θ = r = 6. 4y =. Here 4r sin θ = r cos θ = r = cos θ 4 sin θ. sin θ 3 cos θ y = c. Here r = cr cos θ = r = c cos θ. 8. y = 4. Here r cos θ r sin 4 θ = 4 = r = cos θ sin θ..4 Areas and Lengths in Polar Coordinates 5 / /6 / Graph of cos(5θ) Graph of sin(5θ) Graph of cos(6θ) (Note for the even number we have double petals.) Find the area of the region enclosed by one loop of the curve. Graph of sin(6θ) 3

4 . r = 4 cos(3θ). /6 /6 Here we have an odd number thus 3 petals. Let us solve 4 cos(3θ) = = cos(3θ) = = 3θ = / = θ = /6 As you see, the integration is from to /6! (We still have to multiply by because that s half of the petal). r = sin(4θ). /6 A = (4 cos(3θ)) dθ = 4 3 /4 /4 Here we have an even number thus 8 petals. Let us solve We have to integrate from θ = to θ = /4. sin(4θ) = = 4θ = = θ = /4 A = /4 (sin(4θ)) dθ = 6 Find the area enclosed inside r = 5 sin(θ) but outside of r =. 3. Graph, / 3/ By solving the equation 5 sin(θ) = we obtain θ = ( ) sin.57. Call this angle a. Then by integrating from 5 a to / a we obtain the outside of one petal. 4

5 a a We multiply the result by 4 to obtain the answer. 4 [ a a ] a (5 sin(θ)) dθ a () dθ (note the answer is an approimation because we used an approimation for a.57) Find the area enclosed inside r = cos(5θ) but outside of r = /. 4. Graph, 5 ( ) By solving the equation cos(5θ) = / = θ = cos = we obtain the angle Now we integrate from to /5 and multiply by, [ /5 A = ( ) /5 cos(5θ) dθ ( ) 3 dθ] =

6 Limacon We consider the equation r = + c sin(θ) for c. There are three cases, c =, < c and c >. A circle. c = < c c > For r = + c cos(θ), it is similar, A circle. c = < c c > Moreover, if c is negative, the cases remain the same, ecept the whole shape is reflected. A circle. c = c < c < For r = + c cos(θ), it is similar, 6

7 A circle. c = c < c < Find the area of the small loop of the Limacon r = + 3 sin(θ). 5. We re looking for the area of the small loop (bold) in the following graph, First note the graph starts from θ = which corresponds to r =. Look at the drawn point to see where the graph starts from: If we continue, we ll reach r = again by solving = + 3 sin(θ) = = 3 sin(θ) = θ =. We solve + 3 sin(θ) = = sin(θ) = 3 = θ = sin ( 3 ) We want θ <, so we add to this angle, to get This angle is on the fourth quadrant. Since we reach the point (, ) twice, once on the third quadrant and once on the fourth quadrant, we have to use symmetry of the graph to figure out the other angle. It 7

8 should be Thus the area is A =.3398 = [ = ( + 3 sin θ) dθ ( ) + 6 sin θ + 9 sin θ dθ θ 6 cos θ + 9 ( θ sin(θ) ) ] Note: To deal with similar equations, like 3 cos θ for eample, write it as 3( cos θ). Here we have a magnification of 3 of 3 the shape with c =. 3 8

Arkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan. Figure 50.1

50 Polar Coordinates Arkansas Tech University MATH 94: Calculus II Dr. Marcel B. Finan Up to this point we have dealt exclusively with the Cartesian coordinate system. However, as we will see, this is