Chapter 9 Overview: Parametric and Polar Coordinates

Transcription

1 Chapter 9 Overview: Parametric and Polar Coordinates As we saw briefly last year, there are axis systems other than the Cartesian System for graphing (vector coordinates, polar coordinates, rectangular coordinates--for Complex Numbers and others). The Calculus should apply to them as well. Previously, we saw the relationship between parametric motion and vectors. We will explore this further and investigate how derivatives relate the concepts of tangent line slopes, increasing/decreasing, extremes, and concavity to vectorfunction (parametric) graphs. Polar coordinates are, at their base, drastically different from Cartesian or Vector coordinates. We will review the basics we learned last year about the conversions between polar and Cartesian coordinates and look at graphing in polar mode, with an emphasis on the calculator. As with vector-functions, we will investigate how derivatives relate the concepts of tangent line slopes, increasing/decreasing, extremes, and concavity. Further, we will see how integrals relate to area in polar mode. Finally, we will revisit motion in parametric mode and see how motion could be described in polar mode. There are two contexts within which we will consider parametric and polar functions. I. Motion II. Graphing 583

2 9.: Parametrics and Motion Vocabulary:. Parameter dummy variable that determines x- and y-coordinates independent of one another. Parametric Motion movement that occurs in a plane 3. Vector a directed line segment. As such, it can be used to represent anything that has magnitude and direction (e.g., force, velocity, etc.) 4. Magnitude the length/size of a vector 5. Direction of a Vector the standard position angle when a vector is placed with its tail at the origin In Chapter 3, we considered motion in a parametric context. At the time, we only had the derivative as a tool for analysis, but now we have the integral also. Recall from earlier: Remember: Position = x( t), y t Velocity = x' ( t), y' t Acceleration = x" ( t), y" t Speed = v t ( ) = x' ( t) Therefore: ( ) Position = x' t ( ) Velocity = x" t ( ) ( ) + ( y' ( t) ) ( )dt, y' ( t)dt ( )dt, y" ( t)dt Ex Find the velocity vector for the particle whose position is described by t t, t +. At t = 3, find the speed of the particle. First, notice that this problem has a slightly different notation than we are used to. You may remember that chevrons (the angled brackets) are one of 584

3 the ways that we write vectors. Since this is a position vector, we really have the following: Taking the derivatives, we get x = t t and y = t + dx dt = t and dy dt = t Since the derivatives of x and y are actually velocities, we can simply write the velocity vector using chevrons. t, t Of course, we could still write vectors using the xi! + y "! j form, but this can become cumbersome when x and y are equations in terms of t. In general for this particle, speed = ( t ) + ( t) and at t = 3, speed = ( 3) ( ) + ( ( 3) ) = 5 Ex At t = 3, find the speed of the particle whose velocity vector is t, t. You may recall that speed is the magnitude of velocity, and we found magnitude of vectors last year using the Pythagorean Theorem. We can simply apply those two pieces of information to conclude the following: speed = dx dt + dy dt speed = ( t ) + ( t) for this particular velocity vector, and at t = 3, speed = ( 3) ( ) + ( ( 3) ) = 5 When dealing with derivatives of parametric equations, it is important to realize we have gotten a little sloppy by talking about THE derivative --as if there were only 585

4 one. It was not that important before because, in Cartesian mode, x was the independent variable and y was the dependent variable. We did not deal with derivatives in terms of different variables within the same problem. That comes back to haunt us now. There is still the issue of displacement vs. distance traveled. We recall: Key Idea #: Displacement = b v( t)dt = a b a b a x' ( t)dt y' ( t)dt Distance traveled = b a v( t) dt Note that v t ( ) = x' ( t) ( ) + ( y' ( t) ), therefore: distance traveled β = x' t α ( ( )) + y' ( t) ( ) dt OBJECTIVES Find the position of an object in motion in two dimensions from its velocity. Find the arc length of a curve expressed in parametric mode. Key Idea #: Position can be found two ways a) Indefinite integration of the velocity equation, with an initial value, will yield the position equation. Position can be determined by substituting the time value. b) Definite integration (by calculator) will yield displacement. Adding displacement to the initial value will yields the position 586

5 Ex Given that an object in motion has v( t) = t +, t 3 and the initial position, 3 (that is, t = ), find the position at t =. ( ) = t +, t 3 x' ( t ) = t + y' ( t) = t 3 v t Considering these separately, x' ( t) = t + x( t) = t + ( ) dt = 3 t 3 + t + c x( ) = = ( 3 ) 3 + ( ) + c c = and ( ) y' ( t) = t 3 y( t) = t 3 dt = 4 t 4 t + c y( ) = 3 3 = c c = 3 So, x( t) = 3 t 3 + t and y( t) = 4 t 4 t + 3 and x( ), y( ) = 7 3, 5 It should be mentioned that the fact that c and c equaled the initial values is due to the equations being polynomials and the initial t =. The constants would not equal the initial values if this had not been true. Ex 3 Given that an object in motion has v( t) = sin 3 t, cos t and the initial position, 3, find the position at t =. It this case, we cannot take the anti-derivatives of the velocity, because these functions cannot be integrated by any techniques we have. We can find the definite integrals by out calculator, though, and these would be displacements from the initial point. 587

6 ( ) = + sin 3 t x t y t dt =.59 ( ) = 3+ cos t dt = 3.8 Ex 4 BC 5 # Again, velocity is a vector, so displacement is a vector also, as we saw in the last two examples. But v( t) is the speed and v( t) = x' ( t) ( ) + ( y' ( t) ). The distance traveled which, in most cases, is the Arc Length of the curve--is: Arc Length/Distance Formulae: Function: L = + dy dx b a dt Parametric: L = x' t β α ( ( )) + y' ( t) ( ) dt Note that the distance traveled is not always equal to the Arc Length. The particle can trace over the part or all of the path traveled. Ex 5 Find the distance traveled by the object in Ex. Distance = =.65 ( sin 3 t) + ( cos t) dt Ex 6 Find the arc length of x = cost and y = sint fort,4 the distance traveled. and compare it to 588

7 This problem highlights the difference between Arc Length and Distance. D = 4 4 = dt = t 4 = 4 ( ( )) dt ( sint ) + cos t But we saw that that this curve is the unit circle, with the motion traced over the circle twice. The distance traveled is 4, but the arc length is actually half this distance, or. This should make sense, since the circumference of the Unit Circle is. 589

8 9.: Homework Find (a) the velocity vector, (b) the speed, and (c) acceleration vector for each of the following parametric equations.. x( t) = t, y( t) = cos t. x( t) = 5sint, y( t) = t 3. x( t) = t 5t, y( t) = 4t + 4. x = sec θ, y = tan θ Find the position at t = 5 of an object moving according to the given equation with the given initial position. 5. x' ( t) = t t, y' ( t) = 5 3 t 3, x( ) =, y( ) = 3 6. x' ( t) = 3 t, y' ( t) = t t, x( ) =, y( ) = 3 7. x' ( t) = e t, y' ( t) = 3e 3t, x( ) =, y( ) = 8. x' ( t) = t cost, y' ( t) = 3sint, x( ) =, y( ) = 3 9. x' ( t) = t lnt, y' ( t) = tan 3 t, x( ) = 3, y( ) =. x' ( t) = cose t, y' ( t) = cot ( t +), x( 3) =, y( 3) = Find the distance traveled by each object below. Determine if the distance is equal to the arc length.. x( t) = t t, y( t) = 4 3 t 3, t,. x( t) = t sint, y( t) = t cost, t 3, 3 3. x( t) = 3t t 3, y( t) = 3t, t, 4. x( t) = e t sint, y( t) = e t cost, t, 59

9 5. x( t) = cos t, y( t) = cost, t, 4 6. x( t) = t, y( t) = cos t, t, 6 AP Handout 7. BC 5B # 8. BC #3 9. BC 8B #. BC 9 #3 59

10 9.: Parametrics and Graphing Parametric mode allows one to graph non-functions. The parameter can often be eliminated in order to graph a function, but some functions cannot be expressed in Cartesian coordinates. The most well-known example is the cycloid, which is created by rolling a circle along the x-axis and tracing the path traveled by a point on the circle. Ex Graph ( ) = ( t sint ) ( ) = ( cost) x t y t on your calculator. 6 y x Remember. Radian mode in Calculus (degrees in Physics).. We have to be concerned about the t-values (and the window) when graphing parametrics. Unless otherwise given, we need to include negative values of t. In the second context for parametrics, we will be asked to find the slopes of tangent lines and toe determine concavity for a function described in parametric mode. 59

11 Key Ideas.. dy dx is still the slope of the tangent line and d y still determines concavity. dx dy is different from dx is different from dy dt dt dx. As we saw, given parametric equations x( t),y( t), it is easy to find x' ( t),y' ( t). But x' ( t),y' ( t) is dx dt, dy dt, which is the velocity vector. How do we get dy dx instead of dy, so we can talk about slopes and increasing or decreasing intervals? dt This is where the advantages of Liebnitz notation become apparent. Since Liebnitz notations work the same as fractions, dy dx = dy dt dx dt and d y dx = d dy dt dx dx dt OBJECTIVES Graph relations in parametric mode. Find the slope of a tangent line to a curve in parametric mode. Find the concavity of a curve in parametric mode. Find the arc length of a curve in parametric mode. Graph relations in parametric mode. Eliminate the parameter to identify the function form of a parametric. 593

12 Ex Find the equation of the line tangent to x( t) = t, y( t) = t 3 t at t =. As we remember, finding the equation of any line requires having a point and a slope. t = is the parameter, not the point (x, y). The point comes from substituting t = into x t ( ) = t, y( t) = t 3 t : ( ) = 4 ( ) = 3 = 6 ( x,y) = ( 4,6) x y The slope of the tangent line, of course, would come from substituting t = into dy dx : dy dy dx = dt = 3t dx t dt dy dy = dt = 3( ) dx t = dx dt Therefore, the equation of the tangent line is ( ) y 6 = ( 4 x 4 ) = 4 Ex 3 Find the equation of the line tangent to x( t) = t, y( t) = t 3 3t at ( 3, ). This time, we have the point, but not the parameter. We can find t by setting the point equal to the equations: ( ) = t = 3 t = ± 3 ( ) = t 3 3t = t =, ± 3 x t y t t = gives the correct y-value but not the right x-value, so t = ± 3 are the correct parameters. Yes, there are two parameters, meaning there are two tangent lines involved. A quick look at the calculator shows why: 594

13 5 y 4 3 x dy dx = 3t 3 t The two tangent lines are ( ) 3 t = 3 dy dx = =.73 t = 3 dy dx = 3( 3) 3 3 =.73 y =.73( x 3) and y =.73( x 3) 595

14 Ex 4 Find the points on the curve defined by x t tangent line is either vertical or horizontal. ( ) = t and y( t) = t 3 3t where the From the last example, we know dy dx = 3t 3. The tangent line would be t horizontal where dy dx = ; that is, where 3t 3 =, or t = ±. This is the parameter, though, not the point. t = ( x,y) = (, ) t = ( x,y) = (, ) The tangent line would be vertical where dy = dne ; that is, t =. dx t = ( x,y) = (, ) As we saw earlier, parametrics allow one to graph non-functions. The parameter can often be eliminated in order to graph a function, but some functions cannot be expressed in Cartesian coordinates. The most well-known example is the cycloid, which is created by rolling a circle along the x-axis and tracing the path traveled by a point on the circle. Ex 5: BC 6 #3 596

15 Ex 6 Show, algebraically, that the cycloid down on t (, ). ( ) = ( t sint ) ( ) = ( cost) x t y t in Ex is concave 6 y x Notice that, in the x-equation, the t is both inside and outside the trig function. We would not be able to isolate t so as to eliminate the parameter, despite the fact that the graph clearly shows a function (that is, it passes the vertical line test.) Concavity is determined by d y dy. To find this, we need dx dx : ( ) = ( t sint ) ( ) = ( cost) x t y t x' t y' t ( ) = ( cost) ( ) = sint dy dx = sint cost 597

16 d dy d y dt dx = dx dx dt d sint dt cost = sint ( cost)cost sint sint ( cost) = cost ( ) = cost cos t sin t ( cost) cost cost = ( cost) 3 = cost ( ) ( ) ( ) Clearly, d y is always negative (since the numerator is a negative constant dx and the denominator is always non-negative). Therefore, the cycloid is always concave down. 598

17 Eliminating the parameter refers to the process of getting rid of the variable that is defining our x and y. Because we are more familiar with function mode, sometimes it is easier to convert parametric into function mode. We do this by isolating t in either equation and substituting into the other equation. Ex 7 Eliminate the parameter for the parametric curve, by the equations x = t t and y = t +. y = t + t = y x = t t ( ) ( y ) x = y x = y 4y + 3 So the parametric equation is a parabola opening to the right, as we suspected from our sketch. Ex 8 Sketch the curve represented by the parametric equations x = sint and y = cost fort,. y x 599

18 Notice that the parameter, t, in this case represents an angle in radians. Finding values for x and y, we can see that we trace out a circle starting at the point (,) performing complete revolution clockwise. Ex 9 Eliminate the parameter in example 3. The way to eliminate the parameter when trig functions are involved is to use the Pythagorean identity: sin t + cos t =. Since x = sint and y = cost in example 5, sin t + cos t = x + y = It should make sense that we got the equation of the unit circle give the graph we have. 6

19 9. Homework Find the equation of the line tangent to the given curve at the given point.. x( t) = e t and y( t) = ( t ) at (,). x( t) = t + and y( t) = 3 t 3 3 at t = 3 3. x( t) = e t and y( t) = t Ln t at t= 4. x( t) = t sint and y( t) = t cost at (, ) Find dy dx and d y dx. 5. x( t) = t 4 and y( t) = t t 3 6. x( t) = + tant and y( t) = cost 7. x( t) = e t and y( t) = te t 8. x( t) = + t and y( t) = t Ln t Find the points (a) where the tangent line is vertical and (b) where the tangent line is horizontal. 9. x( t) = t ( t 3) and y( t) = 3( t 3). x( t) = 3t t 3 + and y ( t ) = 3t t 3 + Sketch each of the following parametric curves (with a calculator).. x = t and y = cos t for t,6. x = cos t and y = t for t,6 3. x = t 5t and y = 4t + for t, 6

20 4. x = 5sin t and y = t for t, 5. x( t) = t 3 t and y( t) = t t for t [, ] 6. x( t) = 4cott and y( t) = 4sin t for t [, ] 7. x( t) = sin( t + sint ) and y( t) = cos( t + cost) for t [, ] 8. x( t) = 3cost + 3t sint and y( t) = 3sint + 3t cost for t [, ] Eliminate the parameter for the following parametric equations 9. x( t) = t and y( t) = cos t. x( t) = e t and y( t) = e t. x( t) = ln t and y( t) = t. x( t) = sect and y( t) = tant AP Handout 3. BC #4 4. BC 6 #3 5. BC #3 6. BC # 7. BC # 6

21 9.3: Polar and Cartesian Graphs Lat year, we briefly looked at vectors in Polar Form. Here is a quick review: Vocabulary Polar Coordinates Defn: System where points ( r, θ ) are defined by distance from the origin (pole) and at what angle that distance is measured. [Note that the points are ( r, θ ). That is, the dependent variable comes first in the pair.] Because of the definition, a Polar axis system looks like a radar screen: Points on the graph carry coordinates that describe how far from the origin (Pole) and in which direction the point is. One of the confusing things is that the radar screen is not usually drawn because of the time involved so polar graphs are drawn on what looks like but is not a Cartesian system. OBJECTIVES Graph curves in polar form. Recognize certain polar equations as having particular graphs. Determine the θ which would give a specific x- or y-value. We are so used to the Cartesian System and its perspective of points being measurement horizontally and vertically from the Origin, that sometimes the perspective shift to Polar Coordinates and the idea of distance in a specific direction is difficult. This is especially true when the r-value can now be 63

22 negative. While point-wise plotting can be tedious, it can really help with the change in perspective. EX. Plot the following functions: a) r = 4cosθ b) r = + cosθ 64

23 c) r = + 3cosθ Summary of Standard Pole Graphs. Circles: r = c, r = asinθ or r = acosθ. The first is a circle with radius c and center at the origin. The second and third are circles through the origin with diameter a. r = asinθ has the diameter on the 9 radial line and r = acosθ has the diameter on the radial line. 65

24 Note that the calculator lists points as it did in parametric mode, with θ determining x and y.. Limacons: r = c + asinθ or r = c + acosθ. As with circles, theses have their center-lies on the 9 and radial line, respectively. The relative size of a and c determine if there is a loop or not. The loop results from r being a negative value, because the trig value is negative and a is greater than c. This is a huge difference from polar vectors. Last year, we said r is always positive, and it was in that context. Now, r can be negative. 3. Lines: r = acscθ, r = asecθ, r = asec( θ α ), and θ = α. The first two are lines a units from the pole and perpendicular to the 9 and radial line, respectively. The third is for any line a units from the origin where a is measured along the α radial line. The last is a line through the pole and lying on the α radial line. These do not come up as often on the AP test as they used to. They are now generally distracter answers in multiple-choice questions. r = secθ r = cscθ 66

25 r = sec θ 3 Other than memorizing the equations above, this section is mostly about familiarizing yourself with polar graphs. The best way to do that is to play with your calculator. Recall from PreCalculus that we learned formulas to convert between the two systems: Conversions: x = r cos θ and y = r sin θ r = x + y and θ = ±cos x r EX Convert r = 4sinθ to Cartesian and identify the curve. r = 4sinθ r = 4 y r r = 4y x + y = 4y This is a circle. 67

26 Ex 3 Convert 3x + 4y = to Polar form. 3x + 4y = 3( rcosθ )+ 4 rsinθ r 3cosθ + 4sinθ r = ( ) = ( 3cosθ + 4sinθ ) ( ) = Just as there are specific Cartesian equations for which we are expected to know the shape ( x + y = is a circle, y = mx + b is a line, etc.), there are certain polar equations that AP expects you to know on sight: Ex 4 Find the values of θ on r = + 3sinθ where x =. 4 x = rcosθ = ( + 3sinθ )cosθ = 68

27 x =,.33 69

28 9.3 Homework Graph the following polar equations.. r = sin θ. r = 4sin3θ 3. r = sin( 4θ ) 4. r = ( sinθ ) 5. r = 4cos4θ 6. r = 9cosθ 7. r = cos θ sinθ 8. r = 3+ sinθ Convert the following polar equations to Cartesian and identify the curve where possible. 9. rsinθ =. r =. r = sinθ Convert the following Cartesian equations to polar.. x y = 3. x = 4y 4. x y = + cosθ For the following, find the value(s) of θ where the curve has the given Cartesian value. 5. r = 4sinθ and x = 6. r = 4sinθ and y = 3 7. r = 4sin3θ and x = 3 8. r = 4sin3θ and y = 9. r = cos θ sinθ and x =.4. r = cos θ sinθ and y =.4. r = 6cosθ and x =. r = 6cosθ and y =.4 6

29 9.4 Polar Coordinates, Arc Length, and Simple Area OBJECTIVES Find the arc length of a shape describe in polar coordinates. Find the area of a shape described in polar coordinates. Arc Length There are two main geometric uses for integrals in polar coordinates Area and Arc Length. The Arc Length is straightforward: β L = r + dr dθ α dθ The Arc Length formula is straight-forward and easy to use. Ex 4 Find the arc length of r = 5 + cosθ on θ [, ]. β L = r + dr dθ α = 5 + cosθ = 3.73 dθ ( ) dθ ( ) + sinθ Ex 5 Find the perimeter of the region bounded by r =θ and r = ( θ + )on θ [, ]. 6

30 L = θ + dθ = 6. L = θ + ( ) + L 3 = r ( ) r ( ) = =.573 P = L + L + L 3 =5.56 dθ = Simple Area Area, on the other hand, appears straightforward but can be tricky. The formula is simple: A = β α r dθ With Cartesian coordinates, we considered area as the sum of Riemann Rectangles. With polar coordinates, we would have triangles. 6

31 f( θ) = r 3 rdθ The coefficient comes area of said triangles, namely, bh, where h = r and b = r dθ. Ex 6 Find the area enclosed by the limacon r = + cosθ. We can see by grapher that the limacon completes its graph on θ [, ]

32 A = = ( + cosθ ) dθ ( ( + cosθ )) dθ ( ) dθ = 4 + cosθ = ( + cosθ + cos θ )dθ = θ + sinθ + θ + 4 sinθ = 6 The real difficulty is in finding the boundaries for the integral if they are not obvious or not given. Sometimes, they can be found by calculator. Other times, they are best found either by the r-values or as points of intersection. Ex 7 Which of the following equations gives the area of the region enclosed by the graph of the polar curve r = + cosθ? (note: This is a non-calculator question.) (A) ( + cosθ )dθ (B) ( + cosθ ) dθ (C) ( + cosθ )dθ (D) ( + cosθ ) dθ (E) + cos θ ( )dθ β A = r dθ = + cosθ α ( ) dθ Note that we can eliminate (A), (C) and (E) because the Polar area formula calls for r to be squared. But the formula also calls for, which neither (B) nor (D) have. The issue is that, because of the symmetry of the curve, we can integrate half the region and double the result to get the whole area. In other words, 64

33 + cosθ ( ) dθ = + cosθ ( ) dθ ( ) dθ = + cosθ Ex 8 Find the area enclosed by one leaf of the four-leaf rose r = cosθ. It is fairly clear from the picture what one leaf means, but it is not clear what ANGLES define that area, even after tracing. Algebraically, each leaf begins and ends at the pole, so we can find the boundaries by setting r = and solving for θ : = cosθ θ = ± ± n θ = ± 4 ±n Our trace does help us now see that the leaf on the right is bounded by θ 4, 4. So 65

34 A = β α = 4 = 4 = 4 = 4 r dθ 4 4 cosθ ( ) dθ cosθ ( ) dθ 4 = 8 cosu ( ) du ( u) 4 sinu A third twist, which often appears on the multiple choice part of the AP Exam is, using the symmetry of the graph, combine the coefficient with the boundaries. For instance, a question might ask for the set-up of this problem, but 4 ( cosθ ) dθ will not be an option. 4 ( cosθ ) dθ might be an option and 4 they want to see if you recognize this is the same integral. Because of the symmetry of the curve, you can find the area of half the leaf θ, 4 and double the integral, which cancels the in the formula. 66

35 9.4 Homework Find the area enclosed by each of these curves.. r = sin θ. r = 4sin3θ 3. r = sin( 4θ ) 4. r = ( sinθ ) 5. r = 4cos4θ 6. r = 9cosθ 7. r = cos θ sinθ 8. r = Find the area. 9. Inside r = θ on θ [, ].. Inside r = + sinθ on θ,.. Enclosed by one leaf of r = sin4θ. Inside r = 4cosθ Find the arc length. 3+ sinθ 3. r = 5cosθ on θ, r = e θ on θ [, ]. 5. r = θ on θ [, ]. 6. r = 5cosθ on θ,. 7. r = + sinθ on θ [, ]. 67

36 9.5: Area Involving Two Polar Equations There are two different kinds of figures involving two or more polar curves that will be explored here. The difference is based on which part of the shape is shaded. OBJECTIVES Find the area of a shape described by two polar coordinate equations. Area Between Two Polar Curves As with Cartesian coordinate problems dealing with area between two curves, there can be a region bounded by two polar curves. The shaded region will be inside one curve and outside the other. Instead of top minus bottom in Cartesian, it would be outer minus inner. Better yet, both systems an be combined as further minus nearer. Ex Find the area inside r = + sinθ and outside r =. The boundaries would, again, be the problem, but the boundaries are found by setting the equations equal to each other. 68

37 + sinθ = sinθ = θ = ±n θ = and Just as we can find area between two Cartesian curves by integrating the difference, we can find the area between to polar curves by integrating the difference. Where the Cartesian set-up required we use top bottom, polar requires further closer. A = = = = ( ) dθ ( ) + sinθ + sinθ ( ) dθ ( ) dθ ( ( + sinθ + sin θ )) dθ ( sin θ sinθ ) dθ = θ 4 sinθ cosθ = 4 sin cos cos = 4 + Ex Let R be the region bounded by r =θ and r = ( θ + ) on θ,. 69

38 A = = = = 3 4 = 3 8 ( θ + ) ( ) θ 4 θ + θ + dθ ( ) θ 3 4 θ + θ + 4 θ θ + 4 θ dθ dθ Ex 3 Find the area outside the inner loop and inside the outer loop of r = + 3sinθ. 6

39 6 4 The first thing to consider is the boundaries. The graph shows that the inner loop starts and ends at r =. Therefore, = + 3sinθ sinθ = 3 θ =.73 ± n 3.87± n θ = 3.87, A = β r dθ α 3.87 ( ) dθ = sinθ = ( + 3sinθ ) dθ 6

40 Area Inside Two Polar Curves EX 4 Let R be the region inside the graph of the polar curve r = 4 and inside the polar curve r = 4 + sinθ on θ, as shown above. Find the value of the area of the region R. The key to this kind of probem is to split the picture into two pictures each of which is a region bounded by one curve. Each region can be found and then they can be added. 4 4 R R 4 + sinθ = 4 sinθ = θ = ±n θ = ± n θ =,, 6

41 So the area formula in each case yields: A = ( 4 + sinθ ) dθ and A = ( 4) dθ NB The two integrals do not have the same boundaries, therefore they must be solved separately. A = ( 4 + sinθ ) dθ = 6 +6sinθ + 4sin θ ( ) dθ ( ) = 8 + 8sinθ + sin θ dθ ( ) ( ) ( ) = 8 dθ + 4sinθ dθ + sin θ dθ = 8θ 4cosθ +θ 4 sin4θ A = = 8dθ = 8θ ] = 4 ( 4) dθ = ( ) = 9 8 Therefore, the area of region R is = 8. 63

42 Ex 5 Let R be the region inside the circles r = 3 and r = sinθ. 3 To find the boundaries, set the equations equal to each other: sinθ = 3 sinθ = 3 θ = 3 ± n 3 ± n θ = 3, 3 64

43 3 3 3 The areas of the two sections of region R are: A = 3 3 ( ) dθ = 3 3 3dθ 3 3 = 3 θ 3 = = A = 3 sinθ 3 ( ) = 4 sin θdθ = 4 θ 4 sinθ = = dθ 3 Total area =

44 Ex 6 Find the area inside of both r = 3sinθ and r = + sinθ The boundaries are often the biggest problem. We can find them by setting the equations equal to each other because the points of intersection will be where the r values are equal. 3sinθ =+ sinθ sinθ = θ = 6 ± n 5 6 ± n θ = 6, 5 6 The areas of the two sections of region R are: 66

45 3 5 6 A B A A = 6 3sinθ 6 ( ) = 9sin θdθ = 9 θ 4 sinθ = =.764 dθ 6 B = + sinθ 3 ( ) dθ ( ) = + sinθ + sin θ dθ 3 = θ cosθ + θ 4 sinθ 3 = =. Total area =

46 68

47 9.5 Homework Find the area described.. Inside both r = 4sinθ and r =. Inside both r =+ cosθ and r = 3cosθ 3. Inside both r = secθ cosθ and r = 4. Inside both r = 4cos4θ and r = 5. Inside both r = 6cosθ and r = 3 6. Which of these integrals represents the total area shared by the graph of both r = cosθ and r = sinθ? 4 ( ) a) sin θ dθ b) 4 sin θ dθ c) sin θ dθ 4 ( ) 4 ( ) d) 4 cos θ dθ e) cos θ sin θ dθ 4 7. Enclosed by the inner loop of r = + 3cosθ 8. Inside r = 6cosθ and outside r = 3 9. Inside r =+ cosθ and outside r = 3cosθ. Inside r = cosθ and outside r = secθ cosθ. Inside r = 4sinθ and outside r =. Inside r = 3cosθ and outside r = cosθ 3. Inside r = 3cosθ and outside r = + cosθ ( ) ( ) 4. The area of the region inside the polar curve r = 4cosθ and outside the polar curve r = is given by 69

48 a) ( 4cosθ ) dθ b) 3 3 ( 4cosθ ) dθ c) ( 6cos θ 4)dθ d) 3 6cos θ 4 ( )dθ e) 3 ( 6cos θ 4 )dθ 63

49 9.6: Derivatives and Polar Graphs Polar coordinates can be considered a form of parametric mode. θ is the parameter (like t) which not only determines r, but, due to the conversions, can determine ( x, y). When considered this way, there are two derivatives of particular interest with polar coordinates: dr dy and dθ dx. OBJECTIVES Determine and interpret intervals of increasing or decreasing of a polar curve. Find slopes of lines tangent to polar curves. The derivative dr is the rate of change of r in terms of θ. We consider dθ maximums and minimums, as well as intervals of increasing and decreasing, by looking at dr. But the interpretation is not of high and low points. Rather, the dθ extremes would be points furthest and nearest the origin/pole. Ex r = θ + cosθ on θ [, ]. What are the points closest and furthest from the origin? y x 3 63

50 dr = sinθ = dθ sinθ = θ = θ = 6 ± n 5 6 ± n and 5 Don t forget the endpoints: θ = and. Now we need to find the r-values. r( ) = r =.8 r 5 =.443 ( ) = 3 r So the point nearest the pole is.443, 5 and the point furthest from the pole is ( 3, ). Ex Consider the graphs of r = θ + ( ) and r = cosθ on x, specific value of θ, what is the closest the two graphs are to one another?. For any 4 63

51 distance = r r = ( θ + ) ( cosθ ) = θ + + cosθ d dθ θ + + cosθ = sinθ = sinθ = 4 θ = sin.53 ± n = ± n The nearest they come is.45 units. 633

52 Tangent lines The derivative dy dx is still the slope of the tangent line. As we saw in the first section, dy dx = dy dt dx dt. With θ serving as t, dy dx = dy dθ dx dθ = this formula: d ( dθ rsinθ ) d ( dθ rcosθ ). This leads to dy dx = rcosθ + dr dθ sinθ rsinθ + cosθ dr dθ Often, it is easier to convert the polar equation to Cartesian in order to find dy dx. Ex 3 Find the Cartesian equation of the line tangent to the curve in r =θ + cosθ at θ = 3. First we need the point ( x, y) : x = r cosθ = ( θ + cosθ )cosθ x 3 =.74 y = rsinθ = ( θ + cosθ )sinθ y 3 =.474 Then we need the slope of the tangent line: 634

53 dy = dx θ = dy dx = = dr sinθ + r cosθ dθ cosθ dr dθ rsinθ ( sinθ )sinθ + ( θ + cosθ )cosθ cosθ ( sinθ ) ( θ + cosθ )sinθ sin 3 cos 3 sin 3 =.49 sin 3 + cos cos cos 3 sin 3 And the tangent line is: y.474 =.49( x.74) Ex might have been easier to do in either Polar coordinates, rather than mixing the two together. A line in Polar form is r = asec ( θ α ). r 3 =.547 represents the a and α. So the answer in polar form is r =.547sec θ 3 Ex 4 Let R be the region bounded by r = θ on θ,. a. Sketch and shade the region R. b. Find the area of R. c. Find the equation of the tangent line at θ =. a. 635

54 b. A = = = = θ 4 β α r dθ ( θ ) dθ θ dθ = 4 c. dy dx dr rcosθ + sinθ = dθ rsinθ + cosθ dr dθ = θ dr rcosθ + sinθ = dθ rsinθ + cosθ dr dθ Ex 4 BC4 # 636

55 9.6 Homework Find the points closest to and furthest from the pole (points that are the relative extremes of r) for the given domain.. r = + sin θ on θ,. r = +.8sin θ on θ, 3. r = 5( + cosθ ) on θ, 4. r = ( θ + ) on θ, 5. r = θ + sin3θ on θ, 6. r = sinθ + cosθ on θ, Find the slope of the line tangent to the given equation at the given value of θ 7. r = at θ = 8. r = Ln θ at θ = e θ 9. r = sin3θ at θ = 6. r = sinθ + cosθ at θ = 4 Find the points where the tangent line is (a) horizontal and (b) vertical for the domain θ,.. r = 3cosθ. r = sinθ + cosθ 3. r = e θ 4. Explain the difference between dy dx, dy dt, dy dr, and dt dθ. 5. 3BC # BC # 7. 7BC # BC # 637

56 Parametric and Polar Test x = 3t. Give the length of the curve determined by y = t 3 + 4t a).45 b).9 c).7 on t,. d).35 e).54. The area inside one petal of the polar graph r = 4sinθ is a) b) 4 c) 8 d) e) 3. The equation of the line tangent to x = e t y = t + 5t at t = is a) y = 5x b) y = 5( x ) e c) y = 5( x ) d) y = 5x 5 e) y = 5x 5 x' ( t) = t 4. A particle moves according to y' ( t) = sint when t =, then the position vector at t = 3 is. If the particle is at (, ) a) 9, b), c) 6, d) (, ) e) (, ) ( ) 638

57 5. A particle moves according to statements is true? x = e t + t y = t 5 t. Which of the following I. The particle is moving to the right when t= II. The particle is moving up when t= III. The particle is at rest when t= a) None of these b) I only c) I and II only d) II and III only e) I, II, and III 6. Which of the following integrals gives the length of the graph y = sin x between x = a and x = b, where < a < b? b a) x + cos b xdx b) + cos x a dx c) sin x + a 4x cos b d) + 4x cos x dx e) a b + cos x dx 4x a b a xdx 7. Which of these integrals represents the area enclosed by the graph of r = + cosθ? ( ) a) + cos θ dθ b) + cosθ ( ) c) ( + cosθ ) dθ d) + cosθ ( ) dθ dθ e) ( ) + cos θ dθ 639

58 8. A particle moves in the xy-plane so that at time t such that its acceleration vector is sint, t +. If, at t =, it velocity is,, then how fast is it moving at t=? a) 5 b) 7 c) 37 d) 4 e) If x( t) = sin t and y( t) = cost, then d y a) b) c) dx at t = is d) e) DNE. Let R be the region outside r = and inside r = 3sinθ on θ, a. Sketch and shade the region R. b. Find the area of R. c. Find the equation of the tangent line to r = ( sinθ ) at θ = 4... The velocity vector of a particle moving in the xy-plane has components given by dx dt = 4cos ( t )sin( e t ) and dy = + sin( t ) for t.5. dt At time t =, the position of the particle is (-, 3). (a) For < t <.5, find all values of t at which the line tangent to the path of the particle is vertical. (b) Write an equation for the line tangent to the path of the particle at t =. (c) Find the speed of the particle at t =. (d) Find the acceleration vector of the particle at t =. 64

59 9. Homework. (a) t, sint, (b) s t ( ) = 4t + sin t, (c) 4 t 3, cost. (a) 5cost, t, (b) s( t) = 5sin t + 4, (c) 5sint, 3. (a) t 5, 8t, (b) s( t) = 68t t + 5, (c), 8 4. (a) secθ tanθ, sec θ, (b) s( θ ) = secθ tan θ + sec θ, (c) sec 3 θ + tan θ secθ, sec θ tanθ 5. x( 5) = 3.333, y( 5) = x( 5) = 8.667, y( 5) = x( 5) = 7.38, y( 5) = x( 5) =.487, y( 5) = x( 5) =.877, y( 5) = dne. x( 5) =.5, y( 5) = L=.958, distance= See AP website 9. Homework. y = ( x ). y 6 = 3 ( 4 x 9 ) ( ) 4. y + = x 3. y = e x e dy 5. dx = 3t and d y 4t dx = 3t 4 6t dy dx = sint cos t and d y dx = cos t ( sin t 4cost cos t). dy dx = e3t ( t +) and d y dx = ( et 6t + 5). 64

60 8. dy + lnt = dx t and d y dx = lnt 4t Horizontal: (, 9) Vert: ( ±, 6) 3. Horizontal: (, ),,.587 ( ) Vert:,

61 y = cos x. y = x 3-7. See AP Central. y = e x. y = tan( sec x) 9.3 Homework.. 643

62 y =. x 3y + 4y =. x 4 + x y + y 4 = xy. r = secθ 3. r = 4 tanθ secθ 4. r = cosθ sinθ 644

63 5. θ = 7, 6. θ 3, 3, 4 3, θ =.3,.643, 3.443, θ =.53,.785,.356, θ =.4,.43,.73,.44. θ =.735,.463,.679,.446. θ =.355,.787, 3.497, θ =.69,.973, 3.3, 5.56, Homework Homework B E 9.6 Homework 645

64 . Nearest: (, ), (, ); Furthest: ( 3, ). Nearest: (, ), (, ), (, ) ; Furthest:.34, 3. Nearest: (, ); Furthest: 5, ( ), 5, ( ),.34, 3 4. Nearest:, ; Furthest: 3, 5. Nearest: ( 3.55,.69) ; Furthest: (, ) 6. Nearest:, ; Furthest:, 3 dy 7. dx = 8. dy dx =.9 9. dy dx = 3. dy dx =. Vert: ( 3, ), (.999,.5), (.,.99), ( 3, ), (., 4.9), (, 5.33), ( 3, ) Horiz: (.998,.4 ), 3,, (.966,.7), (., 3.56), ( 3, 4.7), (., 5.863). Vert: (.37,.393), (.54,.963), (.36, 3.534), (.54, 5.5) Horiz: (.37,.78), (.54,.749), (.36, 4.3), (.54, 5.89) 3. Vert: (.93, 4), ( 5.754, 5 4) Horiz: (.55, 3 4), 44.5, ( ) dy dx is the rate of change of y with respect to x - we typically refer to this as the slope of the tangent line dy dt is the rate of change of y with respect to t dx dt is the rate of change of x with respect to t 646

65 dr dθ is the rate of change of r with respect to - refers to how fast the radius of a polar graph is changing as changes See AP Central 647