11 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS

Transcription

1 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS. Parametric Equations Preliminar Questions. Describe the shape of the curve = cos t, = sin t. For all t, + = cos t + sin t = 9cos t + sin t = 9 = 9, therefore the curve is on the circle + = 9. Also, each point on the circle + = 9 can be represented in the form cos t, sin t for some value of t. We conclude that the curve = cos t, = sin t is the circle of radius centered at the origin.. How does = + cos t, = 5 + sin t differ from the curve in the previous question? In this case we have + 5 = cos t + sin t = 9cos t + sin t = 9 = 9 Therefore, the given equations parametrize the circle of radius centered at the point, 5.. What is the maimum height of a particle whose path has parametric equations = t 9, = t? solve: The particle s height is = t. To find the maimum height we set the derivative equal to zero and d dt = d dt t = t = or t = The maimum height is = =.. Can the parametric curve t, sin t be represented as a graph = f? What about sin t,t? In the parametric curve t, sin t we have = t and = sin t, therefore, = sin. That is, the curve can be represented as a graph of a function. In the parametric curve sin t,t we have = sin t, = t, therefore = sin. This equation does not define as a function of, therefore the parametric curve sin t,t cannot be represented as a graph of a function = f. 5. Match the derivatives with a verbal description: a d b d dt dt i Slope of the tangent line to the curve ii Vertical rate of change with respect to time iii Horizontal rate of change with respect to time c d d a The derivative d is the horizontal rate of change with respect to time. dt b The derivative d is the vertical rate of change with respect to time. dt c The derivative d is the slope of the tangent line to the curve. d Hence, a iii, b ii, c i 8 April,

2 SECTION. Parametric Equations 8 Eercises. Find the coordinates at times t =,, of a particle following the path = + t, = 9 t. Substituting t =, t =, and t = into = + t, = 9 t gives the coordinates of the particle at these times respectivel. That is, t = = + =, = 9 = 9, 9 t = = + = 9, = 9 = 9, t = = + = 65, = 9 = 9 65, 9.. Find the coordinates at t =,, of a particle moving along the path ct = cos t,sin t. Setting t =, t =, and t = in ct = cos t,sin twe obtain the following coordinates of the particle: t = : cos, sin =, t = : cos, sin =, t = : cos, sin =,. Show that the path traced b the bullet in Eample is a parabola b eliminating the parameter. The path traced b the bullet is given b the following parametric equations: = t, = t 6t We eliminate the parameter. Since = t, we have t =. Substituting into the equation for we obtain: = t 6t = 6 = 5 The equation = + is the equation of a parabola. 5. Use the table of values to sketch the parametric curve t, t, indicating the direction of motion. t We mark the given points on the -plane and connect the points corresponding to successive values of t in the direction of increasing t. We get the following trajector there are other correct answers: t = 6 t = t = 5 5 t = t = t = t = 5. Graph the parametric curves. Include arrows indicating the direction of motion. a t, t, <t< b sin t,sin t, t c e t,e t, <t< d t,t, t a For the trajector ct = t, t, <t< we have =. Also the two coordinates tend to and as t and t respectivel. The graph is shown net: April,

3 8 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS b For the curve ct = sin t,sin t, t, we have =. sin t is increasing for t, decreasing for t and increasing again for t. Hence the particle moves from c =, to c =,, then moves back to c =, and then returns to c =,. We obtain the following trajector: t =, t =, t = t =, t =, <t t t< These three parts of the trajector are shown together in the net figure: t =, t = t = t =, c For the trajector ct = e t,e t, <t<, we have =. However since lim t et = and lim t et =, the trajector is the part of the line =, <. d For the trajector ct = t,t, t, we have again =. Since the function t is increasing the particle moves in one direction starting at, =, and ending at, =,. The trajector is shown net: t =, t =, 6. Give two different parametrizations of the line through, with slope. The equation of the line through, with slope is = or = 7. One parametrization is obtained b choosing the coordinate as the parameter. That is, = t. Hence = t 7 and we get = t, = t 7, <t<. Another parametrization is given b = t, = t 7, <t<. April,

4 SECTION. Parametric Equations 85 In Eercises 7, epress in the form = fb eliminating the parameter. 7. = t +, = t We eliminate the parameter. Since = t +, we have t =. Substituting into = t we obtain 8. = t, = t = t = = From = t, we have t =. Substituting in = t we obtain = t = = =, =. 9. = t, = tan t + e t Replacing t b in the equation for we obtain = tan + e.. = t, = t + From = t we get t =±. Substituting into = t + we obtain = t + = ± + =± +,. Since we must have a function of, we should probabl choose either the positive or negative root.. = e t, = 6e t we get We eliminate the parameter. Since = e t, we have t = ln or t = ln. Substituting in = 6et = 6e t = 6e ln = 6e ln = 6e ln = 6 = 6, >.. = + t, = t From = + t,wegett = ort =. We now substitute t = in = t to obtain = t = =, =.. = ln t, = t Since = ln t we have t = e. Substituting in = t we obtain = e.. = cos t, = tan t We use the trigonometric identit sin t =± cos t to write We now epress in terms of : = tan t = sin t cos cos t =± t cos t. = tan t =± =±, =. Since we must have a function of, we should probabl choose either the positive or negative root. In Eercises 5 8, graph the curve and draw an arrow specifing the direction corresponding to motion. 5. = t, = t Let ct = t, t = t,t. Then c t = t, t so the curve is smmetric with respect to the -ais. Also, the function t is increasing. Hence there is onl one direction of motion on the curve. The corresponding function is the parabola = = 8. We obtain the following trajector: t = April,

5 86 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS 6. = + t, = + t We find the function b eliminating the parameter. Since = + t we have t =, hence = + or = +. Also, since + t and + t are increasing functions, the direction of motion is the direction of increasing t. We obtain the following curve: 6 t = t = 6, 5, = t, = sin t We find the function b eliminating t. Since = t, we have t =. Substituting t = get = sin. We obtain the following curve: into = sin t we,, 8. = t, = t From = t we have t = ± /. Hence, = ± /. Since the functions t and t are increasing, there is onl one direction of motion, which is the direction of increasing t. Notice that for ct = t,t we have c t = t, t = t, t. Hence the curve is smmetric with respect to the ais. We obtain the following curve: 9. Match the parametrizations a d below with their plots in Figure, and draw an arrow indicating the direction of motion I II FIGURE III IV a ct = sin t, t b ct = t 9, 8t t c ct = t,t 9 d ct = t +, 5 t a In the curve ct = sin t, t the -coordinate is varing between and so this curve corresponds to plot IV. As t increases, the -coordinate = t is decreasing so the direction of motion is downward. April,

6 SECTION. Parametric Equations 87 IV ct = sin t, t b The curve ct = t 9, t 8 intersects the -ais where = t 8 =, or t =. The -intercept is 5,. The -intercepts are obtained where = t 9 =, or t =±. The -intercepts are, 5 and, 9.As t increases from to, and decrease, and as t increases from to, increases and decreases. We obtain the following trajector: 9 t =, 9, 8 5 II c The curve ct = t,t 9 intersects the -ais where = t =, or t =. The -intercept is, 8. The -intercepts are obtained where t 9 = ort =±. These are the points, and,. Setting t = we get = t 9 = 9 = 8. As t increases the coordinate decreases and we obtain the following trajector: 5 III d The curve ct = t +, 5 t is a straight line, since eliminating t in = t + and substituting in = 5 t gives = 5 = + which is the equation of a line. As t increases, the coordinate = t + increases and the -coordinate = 5 t decreases. We obtain the following trajector: 5 5 I April,

7 88 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS. A particle follows the trajector t = t + t, t = t t with t in seconds and distance in centimeters. a What is the particle s maimum height? b When does the particle hit the ground and how far from the origin does it land? a To find the maimum height t, we set the derivative of t equal to zero and solve: d = d dt dt t t = t = t =. The maimum height is = = cm. b The object hits the ground when its height is zero. That is, when t =. Solving for t we get t t = t t = t =,t =. t = is the initial time, so the is t =.At that time, the object s coordinate is = + =. Thus, when it hits the ground, the object is cm awa from the origin.. Find an interval of t-values such that ct = cos t,sin t traces the lower half of the unit circle. For t =, we have c =,.Ast increases from to, the -coordinate of ct increases from to, and the -coordinate decreases from to at t = / and then returns to. Thus, for t in [, ], the equation traces the lower part of the circle.. Find an interval of t-values such that ct = t +, t 5 parametrizes the segment from, 7 to 7, 7. Note that t + = att = /, and t + = 7att =. Also, t 5 takes on the values of 7and7at t = / and t =. Thus, the interval is [ /, ]. In Eercises 8, find parametric equations for the given curve.. = 9 This is a line through P =, 9 with slope m =. Using the parametric representation of a line, as given in Eample, we obtain ct = t, 9 t.. = 8 Letting t = ields the parametric representation ct = t, 8t t. 5. = 5 We define the parameter t =. Then, = 5 + = 5 + t, giving us the parametrization ct = 5 + t,t = 9 The curve + = 9 is a circle of radius 7 centered at the origin. We use the parametric representation of a circle to obtain the representation ct = 7 cos t,7 sin t = 9 This is a circle of radius 7 centered at 9,. Using the parametric representation of a circle we get ct = cos t, + 7 sin t = 5 This is an ellipse centered at the origin with a = 5 and b =. Using the parametric representation of an ellipse we get ct = 5 cos t, sin t for t. 9. Line of slope 8 through, 9 Using the parametric representation of a line given in Eample, we get the parametrization ct = + t,9 + 8t.. Line through, 5 perpendicular to = The line perpendicular to = has slope m =. We use the parametric representation of a line given in Eample to obtain the parametrization ct = + t,5 t. April,

8 SECTION. Parametric Equations 89. Line through, and 5, We use the two-point parametrization of a line with P = a, b =, and Q = c, d = 5,. Then ct = 8t, + t for <t<.. Line through, 6 and 7 6, 5 We use the two-point parametrization of a line with P = a, b =, 6 and Q = c, d = 7 6, 5. Then ct = t, 6 + t for <t<.. Segment joining, and, We use the two-point parametrization of a line with P = a, b =, and Q = c, d =,. Then ct = + t, + t; since we want onl the segment joining the two points, we want t.. Segment joining, and, We use the two-point parametrization of a line with P = a, b =, and Q = c, d =,. Then ct = + t,t; since we want onl the segment joining the two points, we want t. 5. Circle of radius with center, 9 Substituting a, b =, 9 and R = in the parametric equation of the circle we get ct = + cos t,9 + sin t. 6. Ellipse of Eercise 8, with its center translated to 7, Since the center is translated b 7,, so is ever point. Thus the original parametrization becomes ct = cos t, + sin t for t. 7. =, translated so that the minimum occurs at, 8 We ma parametrize = b t, t for <t<. The minimum of = occurs at,, so the desired curve is translated b, 8 from =. Thus a parametrization of the desired curve is ct = + t, 8 + t. 8. = cos translated so that a maimum occurs at, 5 A maimum value of = cos occurs at =. Hence, the curve = cos, or = + cos has a maimum at the point, 5. We let t =, then = t + and = + cos t. We obtain the representation ct = t +, + cos t. In Eercises 9, find a parametrization ct of the curve satisfing the given condition. 9. =, c =, Let t = t + a and t = = t + a. We want =, thus we must use a =. Our line is ct = t, t = t +, t + = t +, t +.. =, c =, Let t = t + a; since = we have = + a so that a =. Then = = t = t 7, so that the line is ct = t, t 7 for <t<.. =, c =, 9 Let t = t + a and t = = t + a. We want =, thus we must use a =. Our curve is ct = t, t = t +,t + = t +,t + 6t =, c =, This is a circle of radius centered at the origin, so we are looking for a parametrization of that circle that starts at a different point. Thus instead of the standard parametrization cos θ,sin θ, θ = must correspond to some other angle ω. We choose the parametrization cosθ + ω, sinθ + ω and must determine the value of ω. Now, Since =, so = cos + ω = cos ω and ω = cos = or 5 =, we have = sin + ω = sin ω and ω = sin = or Comparing these results we see that we must have ω = so that the parametrization is ct = cos θ +, sin θ + April,

9 9 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS. Describe ct = sec t,tan t for t< in the form = f. Specif the domain of. The function = sec t has period and = tan t has period. The graphs of these functions in the interval t, are shown below: p p p p p p p p = sec t = tan t = sec t = sec t = tan t = tan t = sin t cos t = cos t cos = sec t = t Hence the graph of the curve is the hperbola =. The function = sec t is an even function while = tan t is odd. Also has period and has period. It follows that the intervals t<, <t< and <t< trace the curve eactl once. The corresponding curve is shown net: p t = t = p t = p t = t = p +,, p t = + ct = sec t,tan t. Find a parametrization of the right branch > of the hperbola = a b using the functions cosh t and sinh t. How can ou parametrize the branch <? We show first that = cosh t, = sinh t parametrizes the hperbola when a = b = : then = cosh t sinh t =. using the identit cosh sinh =. Generalize this parametrization to get a parametrization for the general hperbola a b = : = a cosh t, = b sinh t. We must of course check that this parametrization indeed parametrizes the curve, i.e. that = a cosh t and = b sin t satisf the equation a b = : a cosh t b sinh t = = cosh t sinh t =. a b a b The left branch of the hperbola is the reflection of the right branch around the line =, so it clearl has the parametrization = a cosh t, = b sinh t. 5. The graphs of t and t as functions of t are shown in Figure 5A. Which of I III is the plot of ct = t, t? Eplain. t t t A I FIGURE 5 II III April,

10 SECTION. Parametric Equations 9 As seen in Figure 5A, the -coordinate is an increasing function of t, while t is first increasing and then decreasing. In Figure I, and are both increasing or both decreasing depending on the direction on the curve. In Figure II, does not maintain one tendenc, rather, it is decreasing and increasing for certain values of t. The plot ct = t, t is plot III. 6. Which graph, I or II, is the graph of t and which is the graph of t for the parametric curve in Figure 6A? t t A I FIGURE 6 II As indicated b Figure 6A, the -coordinate is decreasing and then increasing, so plot I is the graph of. Figure 6A also shows that the -coordinate is increasing, decreasing and then increasing, so plot II is the graph for. 7. Sketch ct = t t,t following the steps in Eample 7. We note that t = t t is odd and t = t is even, hence c t = t, t = t, t. It follows that c tis the reflection of ct across -ais. That is, c tand ct are smmetric with respect to the -ais; thus, it suffices to graph the curve for t. For t =, we have c =, and the -coordinate t = t tends to as t. To analze the -coordinate, we graph t = t t for t : 8 6 = t t We see that t < and decreasing for <t</, t < and increasing for / <t< and t > and increasing for t>. Also t tends to as t. Therefore, starting at the origin, the curve first directs to the left of the -ais, then at t = / it turns to the right, alwas keeping an upward direction. The part of the path for t is obtained b reflecting across the -ais. We also use the points c =,, c =,, c =, to obtain the following graph for ct: t =, t = t = t =, t = t = t = t = Graph of ct for t. Graph of ct for all t. 8. Sketch ct = t t,9 t for t. The graphs of t = t t and t = 9 t for t are shown in the following figures: April,

11 9 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS t = t t t = 9 t The curve starts at c =, 7. For <t<, t is decreasing and t is increasing, so the graph turns to the left and upwards to c =, 9. Then for <t<, t is decreasing and so is t, hence the graph turns to the left and downwards towards c =, 5. For <t<, t is increasing and t is decreasing, hence the graph turns to the right and downwards, ending at c = 6, 9. The intercept are the points where t t = tt = or9 t =, that is t =,, ±. These are the points c =, 9, c =, 7, c =,, c =,. These properties lead to the following path: t =, 9 t =,, 5, t =,, t = t =, 7 t =, 7 In Eercises 9 5, use Eq. 7 to find d/d at the given point. 9. t,t, t = B Eq. 7 we have d d = t t = t t = t t = t Substituting t = weget d d = t = t= = t + 9, 7t 9, t = We find d d : d 7t 9 = d t + 9 = 7 d d = 7 t=. 5. s s, s, s = Using Eq. 7 we get d d = s s = s s s = s s = s s Substituting s = we obtain d d = s s = s= =. April,

12 SECTION. Parametric Equations 9 5. sin θ,cos θ, θ = 6 Using Eq. 7 we get d d = θ sin θ = θ cos θ Substituting θ = 6 we get d d = sin θ cos θ = sin θ=/6 cos = = In Eercises 5 56, find an equation = ffor the parametric curve and compute d/d in two was: using Eq. 7 and b differentiating f. 5. ct = t +, 9t Since = t +, we have t =. Substituting in = 9t we have = 9 = 9 + Differentiating = ct = t, t t d gives d = 9 d. We now find using Eq. 7: d d d = t 9t = t t + = 9 Since = t we have t =. Substituting in = t t ields = =. We differentiate = : Now, we find d d using Eq. 7. Thus, d d = t t = Since t =, then this t is the same as. 55. = s, = s 6 + s d d = t t t = t = t. We find as a function of : = s 6 + s = s + s = +. We now differentiate = +. This gives d d =. Alternativel, we can use Eq. 7 to obtain the following derivative: d d = s s 6 + s s = s = 6s5 s s = s s 6. Hence, since = s, d d =. April,

13 9 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS 56. = cos θ, = cos θ + sin θ To find as a function of, we first use the trigonometric identit sin θ = cos θ to write = cos θ + cos θ. We substitute = cos θ to obtain = +. Differentiating this function ields d d =. Alternativel, we can compute d d using Eq. 7. That is, Hence, since = cos θ, d d = θ θ = cos θ + sin θ cos θ = d d =. sin θ + sin θ cos θ sin θ = cos θ. 57. Find the points on the curve ct = t t,t 6t where the tangent line has slope. We solve d d = t 6 6t = or t 6 = 8t 6, or t 6t =, so the slope is at t =, 6 and the points are, and 96, Find the equation of the tangent line to the ccloid generated b a circle of radius at t =. The ccloid generated b a circle of radius can be parametrized b ct = t sin t, cos t Then we compute d d = sin t t=/ cos t = t=/ = so that the slope of the tangent line is and the equation of the tangent line is cos = sin or = + 8 In Eercises 59 6, let ct = t 9,t 8t see Figure 7. 6 FIGURE 7 Plot of ct = t 9,t 8t. 59. Draw an arrow indicating the direction of motion, and determine the interval of t-values corresponding to the portion of the curve in each of the four quadrants. We plot the functions t = t 9 and t = t 8t: 6 t t = t 9 = t 8t April,

14 SECTION. Parametric Equations 95 We note carefull where each of these graphs are positive or negative, increasing or decreasing. In particular, t is decreasing for t<, increasing for t>, positive for t >, and negative for t <. Likewise, t is decreasing for t<, increasing for t>, positive for t>8ort<, and negative for <t<8. We now draw arrows on the path following the decreasing/increasing behavior of the coordinates as indicated above. We obtain: 6 t = 9, t =, 5 t =, t = 7, 6 t = 8 55, 6 This plot also shows that: The graph is in the first quadrant for t< ort>8. The graph is in the second quadrant for <t<. The graph is in the third quadrant for <t<. The graph is in the fourth quadrant for <t<8. 6. Find the equation of the tangent line at t =. Using the formula for the slope m of the tangent line we have: m = d t 8t d = t= t 9 = t 8 t= = t= t t t= =. Since the slope is zero, the tangent line is horizontal. The -coordinate corresponding to t = is = 8 = 6. Hence the equation of the tangent line is = Find the points where the tangent has slope. The slope of the tangent at t is d t 8t d = t 9 = t 8 t = t The point where the tangent has slope corresponds to the value of t that satisfies d d = t = t = t = 8. We substitute t = 8int = t 9 and t = t 8t to obtain the following point: 8 = 8 9 = 55 8 = = 55, 6. Find the points where the tangent is horizontal or vertical. In Eercise 6 we found that the slope of the tangent at t is d d = t = t t The tangent is horizontal where its slope is zero. We set the slope equal to zero and solve for t. This gives t = t =. t The corresponding point is, = 9, 8 = 7, 6. The tangent is vertical where it has infinite slope; that is, at t =. The corresponding point is, = 9, 8 = 9,. April,

15 96 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS 6. Let A and B be the points where the ra of angle θ intersects the two concentric circles of radii r<rcentered at the origin Figure 8. Let P be the point of intersection of the horizontal line through A and the vertical line through B. Epress the coordinates of P as a function of θ and describe the curve traced b P for θ. B A r P R FIGURE 8 We use the parametric representation of a circle to determine the coordinates of the points A and B. That is, A = r cos θ,r sin θ, B = R cos θ,rsin θ The coordinates of P are therefore P = R cos θ,r sin θ In order to identif the curve traced b P, we notice that the and coordinates of P satisf R = cos θ and r Hence + = cos θ + sin θ =. R r = sin θ. The equation + = R r is the equation of ellipse. Hence, the coordinates of P, R cos θ,r sin θ describe an ellipse for θ. 6. A -ft ladder slides down a wall as its bottom B is pulled awa from the wall Figure 9. Using the angle θ as parameter, find the parametric equations for the path followed b a the top of the ladder A, b the bottom of the ladder B, and c the point P located ft from the top of the ladder. Show that P describes an ellipse. A P =, 6 q B FIGURE 9 a We define the -coordinate sstem as shown in the figure: A q B April,

16 SECTION. Parametric Equations 97 As the ladder slides down the wall, the -coordinate of A is alwas zero and the -coordinate is = sin θ. The parametric equations for the path followed b A are thus =, = sin θ, θ is between and. The path described b A is the segment [, ] on the -ais. b As the ladder slides down the wall, the -coordinate of B is alwas zero and the -coordinate is = cos θ. The parametric equations for the path followed b B are therefore = cos θ, =, θ is between and. The path is the segment [, ] on the -ais. c The and coordinates of P are = cos θ, = 6 sin θ. The path followed b P has the following parametrization: cθ = cos θ,6 sin θ, θ is between and. q P, 6 q As shown in Eample, the corresponding path is a part of an ellipse. Since θ is varing between part of the ellipse in the first quadrant. and, we obtain the 6 April,

17 98 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS In Eercises 65 68, refer to the Bézier curve defined b Eqs. 8 and Show that the Bézier curve with control points has parametrization P =,, P =,, P = 6, 5, P = 7, ct = + 6t + t t, + t 5t 9t Verif that the slope at t = is equal to the slope of the segment P P. For the given Bézier curve we have a =, a =, a = 6, a = 7, and b =, b =, b = 5, b =. Substituting these values in Eq. 8 9 and simplifing gives t = t + 9t t + 8t t + 7t Then = t + t t + 9t t + t + 8t 8t + 7t = t + t t + 9t 8t + 9t + 8t 8t + 7t = t + t + 6t + t = t + 6t t + 5t t + t = t + t t + 6t t + t + 5t 5t + t = t + t t + 6t 7t + 6t + 5t 5t + t = + t 5t 9t ct = + 6t + t t, + t 5t 9t, t. We find the slope at t =. Using the formula for slope of the tangent line we get d d = + t 5t 9t t 7t + 6t + t t = 6 + 6t 9t d d = t= 6 =. The slope of the segment P P is the slope of the line determined b the points P =, and P =,. That is, = 8 =. We see that the slope of the tangent line at t = is equal to the slope of the segment P P, as epected. 66. Find an equation of the tangent line to the Bézier curve in Eercise 65 at t =. We have d d = t t 7t t = 6 6 t 9t so that at t =, d d = t=/ t 7t 6 + 6t 9t = t=/ 7 and = 9 9, = Thus the tangent line is = 9 or = Find and plot the Bézier curve ct passing through the control points P =,, P =,, P = 5,, P =, Setting a =, a =, a = 5, a =, and b =, b =, b =, b = into Eq. 8 9 and simplifing gives t = t + + 5t t + t = t + t t + 5t 5t + t = 9t + t 6t April,

18 SECTION. Parametric Equations 99 t = t + 6t t + t t + t = t + t t + 6t t + t + t t + t = 6t + 6t t + 6t t + 6t + t t + t = + 6t t We obtain the following equation ct = 9t + t 6t, + 6t t, t. The graph of the Bézier curve is shown in the following figure: 68. Show that a cubic Bézier curve is tangent to the segment P P at P. The equations of the cubic Bézier curve are t = a t + a t t + a t t + a t t = b t + b t t + b t t + b t We use the formula for the slope of the tangent line to find the slope of the tangent line at P. We obtain d d = t t = b t + b t t t + b t t t + b t a t + a t t t + a t t t + a t The slope of the tangent line at P is obtained b setting t = in. That is, m = + b + b + a + a = b b a a We compute the slope of the segment P P for P = a,b and P = a,b. We get m = b b a a Since the two slopes are equal, we conclude that the tangent line to the curve at the point P is the segment P P. 69. A bullet fired from a gun follows the trajector = at, = bt 6t a, b > Show that the bullet leaves the gun at an angle θ = tan b a and lands at a distance ab/6 from the origin. The height of the bullet equals the value of the -coordinate. When the bullet leaves the gun, t = tb 6t =. The s to this equation are t = and t = 6 b, with t = corresponding to the moment the bullet leaves the gun. We find the slope m of the tangent line at t = : d d = t t = b t m = b t a a = b t= a It follows that tan θ = a b or θ = tan b a. The bullet lands at t = 6 b. We find the distance of the bullet from the origin at this time, b substituting t = 6 b in t = at. This gives b = ab 6 6 April,

19 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS 7. Plot ct = t t,t t + 8 for t. Find the points where the tangent line is horizontal or vertical. The graph of ct = t t,t t + 8, t is shown in the following figure: 6 5 t =, 8 t =.5.,.8 t = 5, t =.5.,.8 t =, 5, t =.5.9, 5 5 t =.5.9, 5 5 We find the slope of the tangent line at t: d d = t = t t + 8 t t = t t t The tangent line is horizontal where d d =. That is, t t t = tt 6 = t =, t= 6,t= 6. We find the corresponding points b substituting these values of t in ct. We obtain: c =, 8, c 6.9,, c 6.9,. The tangent line is vertical where the slope in is infinite, that is, where t = ort =± ±.5. We find the points b setting t =± in ct. We get c.,.8, c., Plot the astroid = cos θ, = sin θ and find the equation of the tangent line at θ =. The graph of the astroid = cos θ, = sin θ is shown in the following figure: =, =, =, =, The slope of the tangent line at θ = is m = d d = sin θ θ=/ cos θ = sin θ cos θ θ=/ cos θ sin θ = sin θ θ=/ cos θ = tan θ = θ=/ / We find the point of tangenc:, = cos, sin = 8, 8 The equation of the tangent line at θ = is, thus, = = Find the equation of the tangent line at t = to the ccloid generated b the unit circle with parametric equation 5. We find the equation of the tangent line at t = to the ccloid = t sin t, = cos t. We first find the derivative d d : April,

20 SECTION. Parametric Equations d d = t cos t = t t sin t = sin t cos t The slope of the tangent line at t = is therefore: m = d d = sin t=/ cos = = We find the point of tangenc: The equation of the tangent line is, thus, =, = sin, cos = =, + 7. Find the points with horizontal tangent line on the ccloid with parametric equation 5. The parametric equations of the ccloid are = t sin t, = cos t We find the slope of the tangent line at t: d cos t = d t sin t = sin t cos t The tangent line is horizontal where it has slope zero. That is, d d = sin t cos t = sin t = cos t = t = k, k =, ±, ±,... We find the coordinates of the points with horizontal tangent line, b substituting t = k in t and t. This gives The required points are = k sink = k = cosk = = k,, k =, ±, ±, Propert of the Ccloid Prove that the tangent line at a point P on the ccloid alwas passes through the top point on the rolling circle as indicated in Figure. Assume the generating circle of the ccloid has radius. Tangent line Ccloid FIGURE The definition of the ccloid is such that at time t, the top of the circle has coordinates Q = t, since at time t = the circle has rotated eactl once, and its circumference is. Let L be the line through P and Q. To show that L is tangent to the ccloid at P it suffices to show that the slope of L equals the slope of the tangent at P. Recall that the ccloid is parametrized b ct = t sin t, cos t. Then the slope of L is cos t t t sin t = + cos t sin t and the slope of the tangent line is t cos t = t t sin t = sin t sin t + cos t sin t + cos t = cos t cos = t sin = + cos t t sin t and the two are equal. April,

21 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS 75. A curtate ccloid Figure is the curve traced b a point at a distance h from the center of a circle of radius R rolling along the -ais where h<r. Show that this curve has parametric equations = Rt h sin t, = R h cos t. h R FIGURE Curtate ccloid. Let P be a point at a distance h from the center C of the circle. Assume that at t =, the line of CP is passing through the origin. When the circle rolls a distance Rt along the -ais, the length of the arc ŜQ see figure is also Rt and the angle SCQ has radian measure t. We compute the coordinates and of P. C S P h t A C R Rt Q = Rt PA = Rt h sin t = Rt h sin t = R + AC = R + h cos t = R h cos t We obtain the following parametrization: = Rt h sin t, = R h cos t. 76. Use a computer algebra sstem to eplore what happens when h>rin the parametric equations of Eercise 75. Describe the result. Look first at the parametric equations = hsin t, = hcos t. These describe a circle of radius h. See for instance the graphs below obtained for h = and h = ct = h*sint, h*cost h =, 5 Adding R to the coordinate to obtain the parametric equations = h sin t, = R h cos t, ields a circle with its center moved up b R units: ct = h*sint, R h*cost R =, 5 h = 5 April,

22 SECTION. Parametric Equations Now, we add Rt to the coordinate to obtain the given parametric equation; the curve becomes a spring. The figure below shows the graphs obtained for R = and various values of h. We see the inner loop formed for h>r Show that the line of slope t through, intersects the unit circle in the point with coordinates = t t +, = t t + Conclude that these equations parametrize the unit circle with the point, ecluded Figure. Show further that t = / +., Slope t, FIGURE Unit circle. The equation of the line of slope t through, is = t +. The equation of the unit circle is + =. Hence, the line intersects the unit circle at the points, that satisf the equations: Substituting from equation into equation and solving for we obtain This gives = t + + = + t + = + t + t + t = + t + t + t =, = t ± t t + t + t = t ± ± t + t = + t So = and = t t. The = corresponds to the point,. We are interested in the second + point of intersection that is varing as t varies. Hence the appropriate is = t t + We find the -coordinate b substituting in equation. This gives t = t + = t t + + = t t + t + t + = t t + We conclude that the line and the unit circle intersect, besides at,, at the point with the following coordinates: = t t +, = t t + April,

23 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS Since these points determine all the points on the unit circle ecept for, and no other points, the equations in parametrize the unit circle with the point, ecluded. We show that t =. Using we have + + = t t + t t + + = t t + t +t + t + = t t + t + = t = t. 78. The folium of Descartes is the curve with equation + = a, where a = is a constant Figure. a Show that the line = t intersects the folium at the origin and at one other point P for all t =,. Epress the coordinates of P in terms of t to obtain a parametrization of the folium. Indicate the direction of the parametrization on the graph. b Describe the interval of t-values parametrizing the parts of the curve in quadrants I, II, and IV. Note that t = isa point of discontinuit of the parametrization. c Calculate d/d as a function of t and find the points with horizontal or vertical tangent. II I III IV FIGURE Folium + = a. a We find the points where the line = t t =, and the folium intersect, b solving the following equations: Substituting from in and solving for we get = t + = a + t = at + t at = + t at = =, = Substituting in we find the corresponding -coordinates. That is, at + t at = t =, = t + t = at + t We conclude that the line = t, t =, intersects the folium in a unique point P besides the origin. The coordinates of P are: = at at,=, t =, + t + t The coordinates of P determine a parametrization for the folium. We add the origin so t = must be included in the interval of t. We get at ct = + t, at + t, t = To indicate the direction on the curve for a>, we first consider the following limits: lim t = lim t lim t = lim t t t = lim t = lim t + lim t = t t = t t lim t = lim t t = t = t + lim t = t April,

24 SECTION. Parametric Equations 5 These limits determine the directions of the two parts of the folium in the second and fourth quadrant. The loop in the first quadrant, corresponds to the values t<, and it is directed from c = a, a to c = a, a where t = and t = are two chosen values in the interval t<. The following graph shows the directed folium: < t < < t < t < t = + t = t = t = b The limits computed in part a indicate that the parts of the curve in the second and fourth quadrants correspond to the values <t< and <t< respectivel. The loop in the first quadrant corresponds to the remaining interval t<. c We find the derivative d d, using the Formula for the Slope of the Tangent Line. We get at d d = t t = +t at = +t t = 6at+t at t +t 6at at = a+t at t a 6at = t t t +t Horizontal tangent occurs when d d =. That is, t t t = t t =, t = t =,t =. The corresponding points are: c =, =, c =, = Vertical tangent line occurs when d d is infinite. That is, t = t = The corresponding point is c =, = a +, a = a,a + a a +, + = a, a. 79. Use the results of Eercise 78 to show that the asmptote of the folium is the line + = a. Hint: Show that lim + = a. t We must show that as or the graph of the folium is getting arbitraril close to the line + = a, and the derivative d d is approaching the slope of the line. In Eercise 78 we showed that when t and when t +. We first show that the graph is approaching the line + = a as or, b showing that lim + = lim + = a. t t + For t = at at, t =, a>, calculated in Eercise 78, we obtain using L Hôpital s Rule: + t + t We now show that d d 78 to obtain lim + = lim t lim + = lim t + t t + at + at + t = lim t at + at + t = lim t + a + 6at t a + 6at t is approaching ast and as t +. We use d d d lim t d = lim 6at at t a 6at = = = 9a 9a = a 6a a 6a = a = a 6at at = computed in Eercise a 6at April,

25 6 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS d lim t + d = lim 6at at t + a 6at = 9a 9a = We conclude that the line + = a is an asmptote of the folium as and as. 8. Find a parametrization of n+ + n+ = a n n, where a and n are constants. Following the method in Eercise 78, we first find the coordinates of the point P where the curve and the line = t intersect. We solve the following equations: = t n+ + n+ = a n n Substituting = t in the second equation and solving for ields n+ + t n+ n+ = a n t n n + t n+ n+ at n n = n + t n+ at n = =, = at n + t n+ We assume that t = so + t n+ = and obtain one besides the origin. The corresponding coordinates are at n atn+ = t = t = + tn+ + t n+ at n atn+ Hence, the points =, =, t =, are eactl the points on the curve. We obtain the following + tn+ + tn+ parametrization: at n atn+ =,=, t =. + tn+ + tn+ 8. Second Derivative for a Parametrized Curve Given a parametrized curve ct = t, t, show that Use this to prove the formula d dt d = t t t t d t d d = t t t t t B the formula for the slope of the tangent line we have d d = t t Differentiating with respect to t, using the Quotient Rule, gives B the Chain Rule we have d dt d = d d dt d d = d d t = t t t t t t d = d d dt d dt d d Substituting into the above equation and using dt d = d/dt = gives t d d = t t t t t t = t t t t t April,

26 SECTION. Parametric Equations 7 8. The second derivative of = is d /d =. Verif that Eq. applied to ct = t, t ields d /d =. In fact, an parametrization ma be used. Check that ct = t,t 6 and ct = tan t,tan t also ield d /d =. For the parametrization ct = t, t, we have so that indeed For ct = t,t 6, we have so that again Finall, for ct = tan t,tan t, and t =, t =, t = t, t = t t t t t = t = t = t, t = 6t, t = 6t 5, t = t t t t t t = t t 6t 5 6t t = 5t6 7t 6 = t = sec t, t = tan t sec t, t = tan t sec t, t = 6 sec t sec t t t t t t = sec t6 sec t sec t tan t sec t tan t sec t sec 6 t = 6 sec6 t sec t sec t tan t sec 6 t = sec6 t sec 6 = t = 6 sec6 t sec t + sec t sec 6 t In Eercises 8 86, use Eq. to find d /d. 8. = t + t, = 7t, t = We find the first and second derivatives of t and t: t = t + t = + = 6 t = 6t + = 6 + = t = t = = 8 t = = Using Eq. we get d d = t t t t t= t = t= = 5 8. = s + s, = s, s = s =. Since s = s + = s, we have =. Hence, Eq. cannot be used to compute d d at 85. = 8t + 9, = t, t = We compute the first and second derivatives of t and t: t = 8 = 8 t = = t = = t = = April,

27 8 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS Using Eq. we get d d = t= = 8 8 = 86. = cos θ, = sin θ, θ = We find the first and second derivatives of θ and θ: θ = sin θ = θ = cos θ = θ = cos θ = θ = sin θ = Using Eq. we get d d = θ= = 87. Use Eq. to find the t-intervals on which ct = t,t t is concave up. = The curve is concave up where d >. Thus, d t t t t t > We compute the first and second derivatives: t = t, t = t = t, t = 6t Substituting in and solving for t gives t 6t 8 8t = 6t + 8 8t Since 6t + 8 > for all t, the quotient is positive if 8t >. We conclude that the curve is concave up for t>. 88. Use Eq. to find the t-intervals on which ct = t,t t is concave up. The curve is concave up where d >. That is, d t t t t t > We compute the first and second derivatives: Substituting in and solving for t gives t = t, t = t = t, t = t t 8t 8 8t = 6t + 8 8t = + t This is clearl positive for t>. For t<, we want + t >, which means t >, so t < b taking the reciprocal of both sides, so t<. Thus, we see that our curve is concave up for t< and for t>. April,

28 SECTION. Parametric Equations Area Under a Parametrized Curve Let ct = t, t, where t > and t > Figure. Show that the area A under ct for t t t is t A = t t dt t Hint: Because it is increasing, the function t has an inverse t = g and ct is the graph of = g. Appl the change-of-variables formula to A = t t gd. ct t FIGURE t Let = t and = t. We are given that t >, hence = t is an increasing function of t, so it has an inverse function t = g. The area A is given b gd. Recall that is a function of t and t = g, so the height at an point is given b = g. We find the new limits of integration. Since = t and = t, the limits for t are t and t, respectivel. Also since t = d dt, we have d = tdt. Performing this substitution gives t A = gd = g t dt. Since g = t, we have A = t t t t dt. 9. Calculate the area under = over [, ] using Eq. with the parametrizations t,t 6 and t,t. The area A under = on [, ] is given b the integral A = t t t t t dt where t = and t =. We first use the parametrization t,t 6. We have t = t, t = t 6. Hence, = t = t t = = t = t t = Also t = t. Substituting these values in Eq. we obtain A = t 6 t dt = t 8 dt = 9 t9 = 9 = Using the parametrization t = t, t = t, we have t = t. We find t and t : = t = t t = = t = t t = or t =. Equation is valid if t >, that is if t>. Hence we choose the positive value, t =. We now use Eq. to obtain A = t t dt= t 5 dt = 6 t6 = 6 = Both answers agree, as epected. 9. What does Eq. sa if ct = t, f t? In the parametrization t = t, t = ft we have t =, t = t, t = t. Hence Eq. becomes t t A = t t dt = ftdt t t We see that in this parametrization Eq. is the familiar formula for the area under the graph of a positive function. April,

29 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS 9. Sketch the graph of ct = ln t, t for t and compute the area under the graph using Eq.. We use the following graphs of t = ln t and t = t for t : t t t = ln t t = t We see that for <t<, t is positive and increasing and t is positive and decreasing. Also c = ln, =, and c = ln, = ln,. Additional information is obtained from the derivative d t = d ln t = /t = t, ielding d d t= and d d. t= We obtain the following graph:, t = ln, t = t We now use Eq. to compute the area A under the graph. We have t = ln t, t = t, t = t, t =, t =. Hence, t A = t t dt = t t t dt = t dt = lnt t = ln ln = ln Galileo tried unsuccessfull to find the area under a ccloid. Around 6, Gilles de Roberval proved that the area under one arch of the ccloid ct = Rt R sin t,r R cos t generated b a circle of radius R is equal to three times the area of the circle Figure 5. Verif Roberval s result using Eq.. R R R FIGURE 5 The area of one arch of the ccloid equals three times the area of the generating circle. This reduces to R R cos trt R sin t dt = R cos t dt = R. April,

30 SECTION. Parametric Equations Further Insights and Challenges 9. Prove the following generalization of Eercise 9: For all t>, the area of the ccloidal sector OPC is equal to three times the area of the circular segment cut b the chord PC in Figure 6. P t R P t R O C = Rt, O C = Rt, A Ccloidal sector OPC FIGURE 6 B Circular segment cut b the chord PC Drop a perpendicular from point P to the -ais and label the point of intersection T, and denote b D the center of the circle. Then the area of the ccloidal sector is equal to the area of OPT plus the area of PTC. The latter is a triangle with height t = R R cos t and base Rt Rt R sin t = R sin t, so its area is R sin t cos t. The area of OPT, using Eq., is t t t u u du = R R cos uru R sin u du = R cos u du = R t sin t + sin t cos t so that the total area of the ccloidal sector is R t sin t + sin t cos t + R sin t cos t = R t R sin t = R t sin t The area of the circular segment is the area of the circular sector DPC subtended b the angle t less the area of the triangle DPC. The triangle DPC has height R cos t and base R sin t so that its area is R cos t sin t = R sin t, and the area of the circular sector is R which is one third the area of the ccloidal sector. t = R t. Thus the area of the circular segment is R t sin t 95. Derive the formula for the slope of the tangent line to a parametric curve ct = t, t using a method different from that presented in the tet. Assume that t and t eist and that t =. Show that t lim + h t h t + h t = t t Then eplain wh this limit is equal to the slope d/d. Draw a diagram showing that the ratio in the limit is the slope of a secant line. Since t and t eist, we have the following limits: t lim + h t = t t, lim + h t = t h h h h We use Basic Limit Laws, the limits in and the given data t =, to write t t lim + h t +h t h t + h t = lim h h t +h t h = lim h t +h t h t lim +h t h h = t t Notice that the quotient t + h t t + h t is the slope of the secant line determined b the points P = t, t and Q = t + h, t + h. Hence, the limit of the quotient as h is the slope of the tangent line at P, that is the derivative d d. April,

31 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS t, h Q t P t t + h 96. Verif that the tractri curve l > ct = t l tanh t l,lsech t l has the following propert: For all t, the segment from ct to t, is tangent to the curve and has length l Figure 7. ct FIGURE 7 The tractri ct = t t l tanh t l,lsech t. l Let P = ct and Q = t,. Pt, t Q = t, The slope of the segment PQis m = t t t = l sech t l l tanh t l = sinh t l We compute the slope of the tangent line at P : m = d d = t l sech tl t = t l tanh tl = l l sech tl tanh tl l l sech t l = sech t l tanh tl sech t = sech tl tanh tl l tanh t l = sech tl tanh t = l sinh t l Since m = m, we conclude that the segment from ct to t, is tangent to the curve. We now show that PQ =l: PQ = t t + t = = = l sech l tanh t l t l l tanh t t + l sech l l + sech t = l sech t sinh t + sech t l l l l sinh t + = l sech l t cosh l t = l = l l April,

32 SECTION. Parametric Equations 97. In Eercise 5 of Section. ET Eercise 5 of Section 9., we described the tractri b the differential equation d d = l Show that the curve ct identified as the tractri in Eercise 96 satisfies this differential equation. Note that the derivative on the left is taken with respect to, not t. Note that d/dt = sech t/l = tanh t/l and d/dt = secht/l tanht/l. Thus, Multipling top and bottom b l/l gives d d = d/dt d/dt = secht/l /l = tanht/l /l d d = l In Eercises 98 and 99, refer to Figure In the parametrization ct = a cos t,bsin t of an ellipse, t is not an angular parameter unless a = b in which case the ellipse is a circle. However, t can be interpreted in terms of area: Show that if ct =,, then t = /aba, where A is the area of the shaded region in Figure 8. Hint: Use Eq.., q FIGURE 8 The parameter θ on the ellipse + =. a b We compute the area A of the shaded region as the sum of the area S of the triangle and the area S of the region under the curve. The area of the triangle is S = = a cos tb sin t = ab sin t, S S The area S under the curve can be computed using Eq.. The lower limit of the integration is t = corresponds to a, and the upper limit is t corresponds to t, t. Also t = b sin t and t = asin t. Since t < on the interval <t< which represents the ellipse on the first quadrant, we use the positive value a sin t to obtain a positive value for the area. This gives Combining and we obtain t S = t b sin u a sin udu= ab cos u du = ab ] t = ab [ t sin t = ab = abt sin udu [ ] u sin u t ab sin t Hence, t = A ab. A = S + S = ab sin t + abt ab sin t = abt April,

33 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS 99. Show that the parametrization of the ellipse b the angle θ is ab cos θ = a sin θ + b cos θ ab sin θ = a sin θ + b cos θ We consider the ellipse a + b =. For the angle θ we have tan θ =, hence, Substituting in the equation of the ellipse and solving for we obtain a + tan θ b = b + a tan θ = a b a tan θ + b = a b = a b a tan θ + b = = tan θ a b cos θ a sin θ + b cos θ We now take the square root. Since the sign of the -coordinate is the same as the sign of cos θ, we take the positive root, obtaining ab cos θ = a sin θ + b cos θ Hence b, the -coordinate is ab cos θ tan θ = tan θ = a sin = θ + b cos θ ab sin θ a sin θ + b cos θ Equalities and give the following parametrization for the ellipse: ab cos θ ab sin θ c θ = a sin, θ + b cos θ a sin θ + b cos θ. Arc Length and Speed Preliminar Questions. What is the definition of arc length? A curve can be approimated b a polgonal path obtained b connecting points p = ct, p = ct,...,p N = ct N on the path with segments. One gets an approimation b summing the lengths of the segments. The definition of arc length is the limit of that approimation when increasing the number of points so that the lengths of the segments approach zero. In doing so, we obtain the following theorem for the arc length: b S = t + t dt, a which is the length of the curve ct = t, t for a t b.. What is the interpretation of t + t for a particle following the trajector t, t? The epression t + t denotes the speed at time t of a particle following the trajector t, t. April,

34 SECTION. Arc Length and Speed 5. A particle travels along a path from, to,. What is the displacement? Can the distance traveled be determined from the information given? The net displacement is the distance between the initial point, and the endpoint,. That is + = 5 = 5. The distance traveled can be determined onl if the trajector ct = t, t of the particle is known.. A particle traverses the parabola = with constant speed cm/s. What is the distance traveled during the first minute? Hint: No computation is necessar. Since the speed is constant, the distance traveled is the following product: L = st = 6 = 8 cm. Eercises In Eercises, use Eq. to find the length of the path over the given interval.. t +, 9 t, t Since = t + and = 9 t we have = and =. Hence, the length of the path is S = + dt = 5 dt =.. + t, + t, t We have = + t and = + t, hence = and =. Using the formula for arc length we obtain S = + dt = dt = = 6 5. t, t, t Since = t and = t, we have = t and = 6t. B the formula for the arc length we get S = t + t dt = 6t + 6t dt = 5 tdt= 5 t = 6. t,t /, t We have = t and = t /, hence = and = 6t /. Using the formula for the arc length we obtain S = t + t dt = + 6t / dt = 9 + 6t dt= + t dt Setting u = + t we get S = 5 udu= 5 u/ = 5/ t, t, t We have = t and = t. Hence = 6t and = t. B the formula for the arc length we get S = t + t dt = 6t + t dt = 6 + t tdt. Using the substitution u = + t, du = 8t dtwe obtain S = 6 65 udu= u/ = 5 65/ 5 / 56. April,

35 6 CHAPTER PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS 6. t +,t, t We have = t +, = t, hence, = t and = t. B the formula for the arc length we get S = t + t dt = 9t + t dt = t 9t + dt We compute the integral using the substitution u = + 9t. This gives S = udu= 8 8 u/ = 7 / / = 7 / sin t,cos t, t We have = sin t, = cos t, hence = cos t and = sin t. B the formula for the arc length we obtain: S = t + t dt = 9 cos t + 9 sin tdt= 9 dt = 8. sin θ θ cos θ,cos θ + θ sin θ, θ We have = sin θ θ cos θ and = cos θ + θ sin θ. Hence, = cos θ cos θ θ sin θ = θ sin θ and = sin θ + sin θ + θ cos θ = θ cos θ. Using the formula for the arc length we obtain: S = θ + θ dθ = θ sin θ + θ cos θ dθ = θ sin θ + cos θdθ = θdθ= θ = In Eercises 9 and, use the identit cos t = sin t 9. cos t cos t, sin t sin t, t We have = cos t cos t, = sin t sin t. Thus, = sin t + sin t and = cos t cos t. We get t + t = sin t + sin t + cos t cos t = sin t 8 sin t sin t + sin t + cos t 8 cos t cos t + cos t = sin t + cos t + sin t + cos t 8sin t sin t + cos t cos t = + 8 cost t = 8 8 cos t = 8 cos t We now use the formula for the arc length to obtain / / S = t + t / = 8 cos tdt = 6 sin t / dt = sin t dt = 8cos t / = 8 cos cos = 8.. 5θ sin θ,5 cos θ, θ Since = 5θ sin θand = 5 cos θ, we have = 5 cos θand = 5 sin θ. Using the formula for the arc length we obtain: S = θ + θ dθ = 5 cos θ + 5 sin θdθ = 5 cos θ + cos θ + sin θdθ= 5 cos θdθ = 5 sin θ dθ = sin θ dθ = sin udu = cos u = =. April,