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**Unformatted text preview: **560 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS PROJECT 9.6 Parabolic Trajectories
In the early part of the seventeenth century Galileo Galilei observed the motion of stones projected from the tower of Pisa and concluded that their trajectory was parabolic. Using calculus, together with some simplifying assumptions, we obtain results that agree with Galileo’s observations. Consider a projectile ﬁred at an angle θ , 0 < θ < π/2, from a point (x0 , y0 ) with initial velocity v0 (Figure A). The horizontal component of v0 is v0 cos θ , and the vertical component is v0 sin θ (Figure B.) Let x = x(t ), y = y(t ) be parametric equations for the path of the projectile.
y v2 The only vertical acceleration is due to gravity; therefore y (t ) = − g Problem 1. Show that the path of the projectile (the trajectory) is given by: x(t ) = (v0 cos θ )t − x0 , y(t ) = − 1 gt 2 + (v0 sin θ )t + y0 . 2 Problem 2. Show that the rectangular equation of the trajectory can be written g x = − 2 sec2 θ (x − x0 )2 + tan θ (x − x0 ) + y0 . 2v 0 Problem 3. Measure distance in feet, time in seconds and set g = 32 (ft/ sec2 ). Take (x0 , y0 ) as the origin and the x-axis as ground level. Consider a projectile ﬁred at an angle θ with initial velocity v0 . a. b. c. d. e. f. Give parametric and rectangular equations for the trajectory. What is the maximum height attained by the projectile? Find the range (horizontal distance) of the projectile. How many seconds after ﬁring does the impact take place? How should θ be chosen to maximize the range? How should θ be chosen to have the projectile impact at the point x = b? θ
(x0, y0) Ground level x Figure A
v0 θ
v0 cos θ v0 sin θ c Problem 4.
a. Use a graphing utility to draw the path of a projectile ﬁred at an angle of 30◦ with an initial velocity of v0 = 1500 ft/ sec. Determine the height and range of the projectile and compare with the theoretical results of Problem 3. b. Keeping v0 = 1500 ft/ sec, experiment with several values of θ . Conﬁrm that θ = 45◦ maximizes the range. What angle maximizes the height? Figure B We neglect air resistance and the curvature of the earth. Under these circumstances there is no horizontal acceleration; therefore x (t ) = 0. 9.7 TANGENTS TO CURVES GIVEN PARAMETRICALLY
Let C be a curve parametrized by the functions x = x(t ), y = y (t ) where x and y are deﬁned on some interval I . Since a curve can intersect itself, at any given point C can have (i) one tangent, (ii) two or more tangents, or (iii) no tangent at all. We illustrate these possibilities in Figure 9.7.1. As before, we are assuming that x (t ) and y (t ) exist, at least on the interior of I . To make sure that at least one tangent line exists at each point of C , we will make the additional assumption that
(9.7.1) [x (t )]2 + [ y (t )]2 = 0. 9.7 TANGENTS TO CURVES GIVEN PARAMETRICALLY
l l2 l1 561 one tangent two tangents no tangent Figure 9.7.1 This assumption is equivalent to requiring that x (t ) and y (t ) are not simultaneously equal to 0. Without this assumption almost anything can happen. See Exercises 31–35. Now choose a point (x0 , y0 ) on the curve C and a time t0 at which x(t0 ) = x0 and y(t0 ) = y0 . We want the slope of the curve as it passes through the point (x0 , y0 ) at time t0 . † To ﬁnd this slope, we assume that x (t0 ) = 0. With x (t0 ) = 0, we can be sure that, for h sufﬁciently small, h = 0, x(t0 + h) − x(t0 ) = 0. For such h we can form the quotient y(t0 + h) − y(t0 ) . x ( t 0 + h) − x ( t 0 )
C (explain) secant line (x (t0 + h), y (t0 +h)) tangent line (x (t0), y (t0)) Figure 9.7.2 This quotient is the slope of the secant line pictured in Figure 9.7.2. The limit of this quotient as h tends to zero is the slope of the tangent line and thus the slope of the curve. Since (1/h)[ y(t0 + h) − y(t0 )] y ( t0 ) y(t0 + h) − y(t0 ) = → x(t0 + h) − x(t0 ) (1/h)[x(t0 + h) − x(t0 )] x ( t0 ) as h → 0, † It could pass through the point (x0 , y0 ) at other times also. 562 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS you can see that
(9.7.2) m= y ( t0 ) . x (t0 ) As an equation of the tangent line, we can write y − y(t0 ) = Multiplication by x (t0 ) gives y (t0 )[x − x(t0 )] − x (t0 )[ y − y(t0 )] = 0, and thus
(9.7.3) y (t0 ) [x − x(t0 )]. x ( t0 ) (point-slope form) y (t0 )[x − x0 ] − x (t0 )[ y − y0 ] = 0. We derived this equation under the assumption that x (t0 ) = 0. If x (t0 ) = 0, Equation (9.7.3) still makes sense. It is simply y (t0 )[x − x0 ] = 0, which, since y (t0 ) = 0, † can be simpliﬁed to read
(9.7.4) x = x0 . In this instance the line is vertical, and we say that the curve has a vertical tangent.
Example 1 Find an equation for each tangent to the curve x (t ) = t 3 , y (t ) = 1 − t , t ∈ (−∞, ∞) at the point (8, −1).
SOLUTION Since the curve passes through the point (8, −1) only when t = 2, there can be only one tangent line at that point. With x (t ) = t 3 , y (t ) = 1 − t , y (t ) = −1, y (2) = −1. we have and therefore The tangent line has equation x (t ) = 3t 2 , x (2) = 12, (−1)[x − 8] − (12) [ y − (−1)] = 0. This reduces to x + 12y + 4 = 0.
Example 2 [by (9.7.3)] Find the points on the curve x(t ) = 3 − 4 sin t , y(t ) = 4 + 3 cos t , t ∈ (−∞, ∞) at which there is (i) a horizontal tangent, (ii) a vertical tangent.
† We are assuming that [x (t )]2 + [ y (t )]2 is never 0. Since x (t0 ) = 0, y (t0 ) = 0. 9.7 TANGENTS TO CURVES GIVEN PARAMETRICALLY
SOLUTION 563 Observe ﬁrst of all that the derivatives x (t ) = −4 cos t and y (t ) = −3 sin t are never 0 simultaneously. To ﬁnd the points at which there is a horizontal tangent, we set y (t ) = 0. This gives t = nπ , n = 0, ±1, ±2, . . . . Horizontal tangents occur at all points of the form (x(nπ ), y(nπ )). Since x(nπ ) = 3 − 4 sin nπ = 3 and y(nπ ) = 4 + 3 cos nπ = 7, n even 1, n odd, there are horizontal tangents at (3, 7) and (3, 1). To ﬁnd the vertical tangents, we set x (t ) = 0. This gives t = 1 π + nπ , 2 n = 0, ± 1, ± 2, . . . . Vertical tangents occur at all points of the form (x( 1 π + nπ ), y( 1 π + nπ )). Since 2 2 x( 1 π + nπ ) = 3 − 4 sin ( 1 π + nπ ) = 2 2 −1, n even 7, n odd
(–1, 4) y (3, 7) and y( 1 π + nπ ) = 4 + 3 cos ( 1 π + nπ ) = 4, 2 2 (3, 4) (7, 4) there are vertical tangents at (−1, 4) and (7, 4).
(3, 1) Remark We leave it to you to verify that the functions of Example 2 parametrize the ellipse (Figure 9.7.3)
Figure 9.7.3 x ( y − 4)2 (x − 3)2 + =1 16 9
Example 3 The curve parametrized by the functions
y x (t ) = 1−t , 1 + t2
2 y (t ) = t (1 − t ) , 1 + t2
2 t ∈ (−∞, ∞) x = –1 y=x is called a strophoid. The curve is shown in Figure 9.7.4. Find equations for the lines tangent to the curve at the origin. Then ﬁnd the points at which there is a horizontal tangent. The curve passes through the origin when t = −1 and when t = 1 (verify this). Differentiating x(t ) and y(t ), we have
SOLUTION (1, 0) x y = –x x (t ) = (1 + t 2 )( − 2t ) − (1 − t 2 )(2t ) −4t = 2 )2 (1 + t (1 + t 2 )2 Figure 9.7.4 and y (t ) = (1 + t 2 )(1 − 3t 2 ) − t (1 − t 2 )(2t ) 1 − 4t 2 − t 4 = . (1 + t 2 )2 (1 + t 2 )2 At time t = −1, the curve passes through the origin with slope −1 y (−1) = = −1. x (−1) 1 564 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS Therefore the tangent line has equation y = −x. At time t = 1, the curve passes through the origin with slope −1 y (1) = = 1. x (1) −1 Therefore the tangent line has equation y = x. To ﬁnd the points at which there is a horizontal tangent, we set y (t ) = 0. This implies that 1 − 4t 2 − t 4 = 0. This equation is a quadratic in t 2 . By the quadratic formula, √ √ −4 ± 16 + 4 2 = −2 ± 5. t= 2 √ √ Since t 2 ≥ 0, it follows that t 2 = 5 − 2 and t = ± 5 − 2. Therefore, √ √ √ 1 − ( 5 − 2) 5−1 ∼ 5 − 2) = x( ± = √ = 0. 62 2 1 + ( 5 − 2) and y( ± √ 5 − 2) = ( ± √ 5 2)x = ( ± √ 5 − 2) √ 5−1 2 ∼ ± 0. 30. = There is a horizontal tangent line at the points (0. 62, ± 0. 30) (approximately). We can apply these ideas to a curve given in polar coordinates by an equation of the form r = ρ (θ ). The coordinate transformations x = r cos θ , y = r sin θ enable us to parametrize such a curve by setting x(θ ) = ρ (θ ) cos θ , y(θ ) = ρ (θ ) sin θ . θ ∈ [0, ∞) Example 4 Take a > 0. Find the slope of the spiral r = aθ , at θ = aπ , π 22 1 π. 2 SOLUTION We write x(θ ) = r cos θ = aθ cos θ , y(θ ) = r sin θ = aθ sin θ . Now we differentiate:
polar axis x (θ ) = −aθ sin θ + a cos θ , Since x (1π) = −1πa 2 2 y (θ ) = aθ cos θ + a sin θ . and y ( 1 π ) = a, 2 the slope of the curve at θ = 1 π is 2
Figure 9.7.5 y (1π) 2 x See Figure 9.7.5. (1π) 2 =− 2∼ = −0. 64. π 9.7 TANGENTS TO CURVES GIVEN PARAMETRICALLY 565 Example 5 is vertical. Find the points of the cardioid r = 1 − cos θ at which the tangent line SOLUTION Since the cosine function has period 2π , we need only concern ourselves with θ in [0, 2π ). The parametric equations for the curve are: x(θ ) = (1 − cos θ ) cos θ , Differentiating and simplifying, we ﬁnd that x (θ ) = (2 cos θ − 1) sin θ , y(θ ) = (1 − cos θ ) sin θ . [2, π ] [, ]
1 2 π 3 polar axis y (θ ) = (1 − cos θ )(1 + 2 cos θ ). [, ] 1 5π 23 The only numbers in the interval [0, 2π ) at which x is zero and y is not zero are 1 π , π , 3 and 5 π . The tangent line is vertical at 3 [ 1 , 5 π ]. 23 √ √ These points have rectangular coordinates ( 1 , 1 3), (−2, 0), ( 1 , − 1 3). See 44 4 4 Figure 9.7.6 EXERCISES 9.7
Find an equation in x and y for the line tangent to the curve. 1. x(t ) = t , 2. x(t ) = t ,
2 [ 1 , 1 π ], 23 [2, π ], Figure 9.7.6 y(t ) = t 3 − 1; y(t ) = t + 5; y(t ) = cos πt ; y (t ) = t ;
4 t = 1. t = 2. t = 0. t = 1. t = 1. 2 t = 1. t = 1π. 4 t = 0. Find the points (x, y) at which the curve has: (a) a horizontal tangent; (b) a vertical tangent. Then sketch the curve. 19. 20. 21. 22. 23. 24. 25. 26. 27. x(t ) = 3t − t 3 , y(t ) = t + 1. x(t ) = t 2 − 2t , y(t ) = t 3 − 12t . x(t ) = 3 − 4 sin t , y(t ) = 4 + 3 cos t . x(t ) = sin 2t , y(t ) = sin t . x(t ) = t 2 − 2t , y(t ) = t 3 − 3t 2 + 2t . x(t ) = 2 − 5 cos t , y(t ) = 3 + sin t . x(t ) = cos t , y(t ) = sin 2t . x(t ) = 3 + 2 sin t , y(t ) = 2 + 5 sin t . Find the tangent(s) to the curve x(t ) = −t + 2 cos 1 πt , 4 at the point (2, 0). 28. Find the tangent(s) to the curve x (t ) = t 3 − t , y(t ) = t sin 1 πt 2 y(t ) = t 4 − 4t 2 3. x(t ) = 2t , 5. x(t ) = t 2 , 6. x(t ) = 1/t ,
3 4. x(t ) = 2t − 1, y(t ) = (2 − t )2 ; y(t ) = t 2 + 1; y(t ) = sin t
3 7. x(t ) = cos t , 8. x(t ) = et , y(t ) = 3e−t ; Find an equation in x and y for the line tangent to the polar curve. 9. r = 4 − 2 sin θ , 10. r = 4 cos 2θ , 4 , 5 − cos θ 5 12. r = , 4 − cos θ 11. r = 13. r = 14. r = θ = 0. θ = 1π. 2 θ = 1π. 2 θ = 1π. 6 θ = 0. θ = 1π. 2 sin θ − cos θ , sin θ + cos θ sin θ + cos θ , sin θ − cos θ y = y (t ) at the point (0, 1). 29. Let P = [r1 , θ ] be a point on a polar curve r = f (θ ) as in the ﬁgure. Show that, if f (θ1 ) = 0 but f (θ1 ) = 0, Parametrize the curve by a pair of differentiable functions x = x(t ), with [x (t )]2 + [ y (t )]2 = 0.
r1 P Sketch the curve and determine the tangent line at the origin by the method of this section. 15. y = x3 . 17. y5 = x3 . 16. x = y3 . 18. y3 = x5 .
θ1
O polar axis 566 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS then the tangent line at P is perpendicular to the line segment OP . 30. If 0 < a < 1, the polar curve r = a − cos θ is a limaçon with an inner loop. Choose a so that the curve will intersect itself at the pole in a right angle. In Exercises 31–35, verify that x (0) = y (0) = 0 and that the given description holds at the point where t = 0. Sketch the graph. 31. x(t ) = t 3 , 32. x(t ) = t ,
3 5 Calculate d 2 y/dx2 at the indicated point without eliminating the parameter. 37. x(t ) = cos t , y(t ) = sin t ; t = 1π. 6 38. x(t ) = t 3 , y(t ) = t − 2; t = 1. 39. x(t ) = et , y(t ) = e−t ; t = 0. 40. x(t ) = sin2 t , y(t ) = cos t ; t = 1 π . 4 c 41. Let x = 2 + cos t , y = 2 − sin t . Use a CAS to ﬁnd d 2 y/dx2 . c 42. Use a CAS to ﬁnd an equation in x and y for the line tangent to the curve:
x = sin2 t , y = cos2 t ; t = 1π. 4 y (t ) = t 2 ; y (t ) = t ;
5 3 cusp. horizontal tangent. vertical tangent. tangent with slope 2. no tangent line. 33. x(t ) = t , 35. x(t ) = t 2 , y (t ) = t ; 34. x(t ) = t 3 − 1, y ( t ) = 2t 3 ; y(t ) = t 2 + 1; Then use a graphing utility to draw the curve and the tangent line together. 36. Suppose that x = x(t ), y = y(t ) are twice differentiable functions that parametrize a curve. Take a point on the curve at which x (t ) = 0 and d 2 y/dx2 exists. Show that d 2y x (t )y (t ) − y (t )x (t ) = 2 dx [x (t )]3 c 43. Repeat Exercise 42 with x = e−3t , y = et ; t = ln 2. (9.7.5) c 44. Use a CAS to ﬁnd an equation in x and y for the line tangent to the polar curve: 4 ; θ = 1π. r= 3 2 + sin θ Then use a graphing utility to draw the curve and the tangent line together. 9.8 ARC LENGTH AND SPEED
Arc Length
In Figure 9.8.1 we have sketched the path C traced out by a pair of continuously differentiable functions† x = x(t ),
C y = y(t ), t ∈ [a, b]. What is the length of this curve? Figure 9.8.1 We want to determine the length of C . Here our experience in Chapter 5 can be used as a model. To decide what should be meant by the area of a region , we approximated by the union of a ﬁnite number of rectangles. To decide what should be meant by the length of C , we approximate C by the union of a ﬁnite number of line segments. Each number t in [a, b] gives rise to a point P = P (x(t ), y(t )) that lies on C . By choosing a ﬁnite number of points in [a, b], a = t0 < t1 < · · · < ti−1 < ti < · · · < tn−1 < tn = b, we obtain a ﬁnite number of points on C , P0 , P1 , . . . , Pi−1 , Pi , . . . , Pn−1 , Pn . We join these points consecutively by line segments and call the resulting path, γ = P0 P1 ∪ · · · ∪ Pi−1 Pi ∪ · · · ∪ Pn−1 Pn , a polygonal path inscribed in C. (See Figure 9.8.2.) † By this we mean functions that have continuous ﬁrst derivatives. ...

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