# 10.1 Curves defined by parametric equations

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1 Outline Section 1: Parametric Equations and Polar Coordinates 1.1 Curves defined by parametric equations 1.2 Calculus with Parametric Curves 1.3 Polar Coordinates 1.4 Areas and Lengths in Polar Coordinates 1.5 Conic Sections 1.6 Conic Sections in Polar Coordinates

2 1.1 Curves defined by parametric equations Suppose that both a and y are functions of a third parameter t: x f (t), y g(t) As t varies the point (x, y) (f (t), g(t)) traces out a curve, called parametric curve. Example What curve is represented by the following parametric equations? x cos t y sin t t 2π Obviously it is a circle x 2 + y 2 1. What if the range of t would be t 4π?

3 1.2 Calculus with Parametric Curves If y and x in a parametric curve equation are functions of t then dy Assuming dx/ we get dy dx dx dy dx dy dx if dx To compute d 2 y/dx 2 replace y with dy/dx: ( ) d dy d 2 y dx 2 d ( ) dy dx dx dx dx

4 Example Assume a curve C is defined by x t 2, y t 3 3t. Since y t(t 2 3) the curve crosses itself at t and t ± 3. dy dy/ dx/ 3t2 3 3 ( t 1 ) 2t 2 t For t ± 3 the slopes of the tangent lines dy/dx are ±6/(2 3) ± 3, so the tangent lines are y 3(x 3) and y 3(x 3) To determine concavity we calculate d 2 y/dx 2 : ( ) d dy 3 (1 + 1t ) dx 2 2 d 2 y dx 2 dx 2t 3(t2 + 1) 4t 3 Thus, C is concave upward for t > and downward for t <.

5 Areas Since the area under a curve y F(x) on [a, b] is b a F(x) dx, for a curve defined by parametric equations x f (t), y g(t), α t β we have A b a y dx β α g(t)f (t) Example For the cycloid x r(θ sin θ), y r(1 cos θ), θ 2π: A 2πr y dx 2π r(1 cos θ)r(1 cos θ) dθ 2π r 2 (1 2 cos θ + cos 2 θ) dθ r 2 2π 3πr 2 (1 2 cos θ + 1 (1 + cos 2θ)) dθ 2

6 Arc Length For the length of a curve C y F(x) on [a, b] we have L b a 1 + (dy/dx) 2 dx If C is defined parametrically with x f (t), y g(t) on [α, β] with f (x) > we get L b a β α β 1 + (dy/dx) 2 dx (dx ) 2 + ( ) dy 2 α 1 + ( ) dy/ 2 dx dx/ This formula is also valid if C cannot be expressed in the form y F(x). To show this we subdivide the interval [α, β] with points t 1, t 2,..., t n on equal-size subintervals of length t.

7 This way we get points P 1, P 2,..., P n on C so its length L is L lim n n P i 1 P i By the Mean Value Theorem applied to f on [t i 1, t i ] we have i1 f (t i ) f (t i 1 ) x i x i 1 f (t i )(t i t i 1 ) f (t i ) t Similar equation is also valid for g and some t i [t i 1, t i ], so x i f (t i ) t y i g (ti ) t Hence, P i 1 P i ( x i ) 2 + ( y i ) 2 [f (ti )] 2 + [g (ti )] 2 t So, the length becomes L lim n n i1 β [f (ti )] 2 + [g (ti )] 2 t [f (t)] 2 + [g (t)] 2 α

8 Example Find the curve length given by equations x cos t, y sin t, t 2π. We have dx/ sin t, dy/ cos t, so L 2π 2π 2π (dx ) 2 + sin 2 x + cos 2 x dx 2π ( ) dy 2

9 Example Find the length of one arch of the cycloid x r(θ sin θ), y r(1 cos θ), θ 2π. We have L r r 2r 2π 2π 2π 2π (dx ) 2 ( ) dy 2 + r 2 (1 cos θ) 2 + r 2 sin 2 θ dθ 2π 2r[2 + 2] 8r 1 2 cos θ + cos 2 θ + sin 2 θ dθ 2(1 cos θ) dθ r 2π sin(θ/2) dθ 2r[ 2 cos(θ/2)] 2π 4 sin 2 (θ/2) dθ

10 Surface Area For a curve x f (t), y g(t), α t β, with f, g continuous and g(t), rotated about the x-axis, the area of the resulting surface is given by S β α (dx ) 2 2πy + ( ) dy 2 Example To compute the surface of sphere we rotate a semicircle x r cos t, y r sin t, t π: S π 2πr 2πr sin t ( r sin t) 2 + (r cos t) 2 π sin t r 2 (sin 2 t + cos 2 t) π 2πr 2 sin t 4πr 2

11 1.3 Polar Coordinates Polar coordinates of a point P (x, y) is the distance between P and (, ) and the angle θ between the ray OP and the x-axis. That is, P (r, θ). The angle θ is measured in radians. We extend the coordinates to the case r < as follows: ( r, θ) (r, θ + π) The same point can be represented in multiple ways: (r, θ + 2nπ) ( r, θ + (2n + 1)π) Conversion formulas: Polar to Cartesian : x r cos θ y r sin θ Cartesian to Polar : r 2 x 2 + y 2 tan θ y x

12 Example Convert (2, π/3) from polar to Cartesian coordinates. One has: x r cos θ 2 cos π y r sin θ 2 sin π Example Convert (1, 1) from Cartesian to polar coordinates. One has r x 2 + y ( 1) 2 2 tan θ y x 1 We get the following possible representations: ( 2, π 4 ) ( 2, 7π 4 )

13 Polar Curves The graph of polar equation r f (θ) of F (r, θ) consists of all points whose at least one polar expression (r, θ) satisfies the equation. Example Find a Cartesian equation of the curve r 2 cos θ. Since x r cos θ, cos θ x/r. Using polar equation we get cos θ r/2, so r/2 x/r, or 2x r 2 x 2 + y 2. So, the curve equation is x 2 + y 2 2x, or (x 1) 2 + y 2 1

14 Symmetry If a polar equation is unchanged when θ is replaced with θ, the curve is symmetric about the x-axis. If a polar equation is unchanged when r is replaced with r or θ is replaced with θ + π, the curve is symmetric about the pole. If a polar equation is unchanged when θ is replaced with π θ, the curve is symmetric about the y-axis.

15 Tangents to Polar Curves Rewrite the parametric equation r f (θ) as x r cos θ f (θ) cos θ y r sin θ f (θ) sin θ One has dy dx dy dθ dx dθ dr sin θ + r cos θ dθ dr cos θ r sin θ dθ For tangent line at the pole when r we get dy dx tan θ provided dr dθ

16 Example Find the points on the cardioid r 1 + sin θ where the tangent line is horizontal or vertical. and So, x r cos θ (1 + sin θ) cos θ cos θ + 1 sin 2θ 2 y r sin θ (1 + sin θ) sin θ sin θ + sin 2 θ The case θ 3π 2 dy dx dy/dθ dx/dθ cos θ + 2 sin θ cos θ sin θ + cos 2θ dy dθ, θ π 2, 3π 2, 7π 6, 11π 6 dx dθ, θ 3π 2, π 6, 5π 6 needs a special treatment.

17 1.4 Areas and Lengths in Polar Coordinates Recall the area or a circle sector: A 1 2 r 2 θ. Let R be a polar region bounded by the polar curve r f (θ) and rays θ a and θ b with b a 2π. We divide it into subintervals of equal wih θ with endpoints θ 1,..., θ n. For the sector bounded with θ i 1 and θ i one has A i 1 2 [f (θ i )]2 θ θ i [θ i 1, θ i ] So the area of R can be obtained as A lim n n i1 1 b 2 [f 1 (θ i )]2 θ a 2 [f (θ i )]2 dθ 1 b r 2 dθ 2 a

18 Example Find the are of one loop of the 4-leaved rose r cos 2θ. A 1 2 π/4 π/4 π/4 r 2 dθ 1 2 cos 2 2θ dθ [θ + 14 sin 4θ ] π/4 π/4 π/4 π/4 cos 2 2θ dθ 1 (1 + cos 4θ) dθ π 8 Area between two polar curves If the curves are r f (θ) and r g(θ), a θ b, the following formula is easy to derive: A 1 2 b a ( [f (θ)] 2 [g(θ)] 2) dθ

19 Example Find the area which is inside the circle r 3 sin θ and outside of cardioid r 1 + sin θ. The intersecting points are determined by 3 sin θ 1 + sin θ θ π 6, 5π 6 Note that the area is symmetric about the vertical axis. A π π/2 π/6 π/2 π/6 π/2 π/6 π/2 π/6 (3 sin θ) 2 dθ 9 sin 2 θ dθ π/2 π/6 π/2 π/6 (8 sin 2 θ 1 2 sin θ) dθ (1 + sin 2 θ) 2 dθ (1 + 2 sin θ + sin 2 θ) dθ (3 4 cos 2θ 2 sin θ) dθ 3θ 2 sin 2θ + 2 cos θ] π/2 π/4

20 Arc Length To find the arc length of the polar curve r f (θ), a θ b, we treat θ as a parameter in parametric curve equations: x r cos θ f (θ) cos θ y r sin θ f (θ) sin θ Differentiating them we get Therefore, dx dθ dr dy cos θ r sin θ dθ dθ dr sin θ r cos θ dθ ( dx dθ ) 2 + ( ) dy 2 dθ + ( ) dr 2 cos 2 θ 2r dr dθ dθ cos θ sin θ + r 2 sin 2 θ ( ) dr 2 sin 2 θ + 2r dr dθ dθ sin θ cos θ + t2 cos 2 θ ( ) dr 2 + r 2 dθ

21 Hence, the formula for the length becomes b (dx ) 2 ( ) dy 2 b L + dθ r a dθ dθ 2 + a Example Find the length of the cardioid r 1 + sin θ. 2π ( ) dr 2 2π L r 2 + dθ dθ 2π 2π sin θ dθ 2π 2 cos θ dθ 2 2 sin θ d(1 z) 1 z π/2 ( ) dr 2 dθ dθ (1 + sin θ) 2 + cos 2 θ dθ sin θ 2 2 sin θ 2 2 sin θ π/2 d(1 sin θ) 1 sin θ ] z 8 1 dθ

22 1.5 Conic Sections Parabolas Parabola is a set of point in a plane which are equidistant from a fixed point F (focus) and a fixed line (directrix). If directrix is y p and focus is (, p), the distance from a point (x, y) on the parabola to F is x 2 + (y p) 2. The distance from (x, y) to the directrix is y + p, so the parabola equation becomes x 2 + (y p) 2 y + p which simplifies to x 2 4py Similarly, if directrix is x p and focus is (p, ) the parabola equation is y 2 4px

23 Ellipses An ellipse is the set of points in a plane, the sum of whose distances from two fixed points F 1, F 2 (foci) is a constant. If F 1 ( c, ), F 2 (c, ), and the constant is 2a the ellipse equation becomes (x + c) 2 + y 2 + (x c) 2 + y 2 2a which simplifies to x 2 a 2 + y 2 b 2 1, c2 a 2 b 2, a b > If the foci are of the y-axis, i.e. F 1 (, c), F 2 (, c), the ellipse equation becomes x 2 b 2 + y 2 a 2 1, c2 a 2 b 2, a b >

24 Hyperbolas A hyperbola is the set of points in a plane, the difference of whose distances from two fixed points F 1, F 2 (foci) is a constant. If F 1 ( c, ), F 2 (c, ), and the constant is 2a the hyperbola equation is x 2 a 2 y 2 b 2 1, c2 a 2 + b 2 The x-intercepts (±a, ) are called the vertices of hyperbola, and it has asymptotes y ±(b/a)x. If the foci are of the y-axis, i.e. F 1 (, c), F 2 (, c), the hyperbola equation becomes y 2 a 2 x 2 b 2 1, c2 a 2 + b 2 Its vertices are (, ±a) and asymptotes are y ±(a/b)x.

25 Shifted conics Shifted conic can be obtained by replacing x and y in its equation with x h and y k, respectively. Example Identify the conic 9x 2 4y 2 72x + 8y and find its foci. To accomplish it, we complete the squares: 4(y 2 2y) 9(x 2 8x) 176 4(y 2 2y + 1) 9(x 2 8x + 16) (y 1) 2 9(x 4) 2 36 (y 1) 2 9 (x 4)2 4 1 Hence, it is a shifted hyperbola with a 3, b 2, c 13 whose (original) foci (, ± 13) are also shifted accordingly and become (4, 1 ± 13).

26 1.6 Conic Sections in Polar Coordinates Theorem Let F be a fixed point (focus) l be a fixed line (directrix) in a plane e > be a fixed number (eccentricity). The set of all points P in the plane such that is a conic section. The conic is (i) an ellipse, if e < 1 (ii) a parabola, if e 1 (iii) a hyperbola, if e > 1 PF Pl e

27 Note that for e 1 then we get the definition of parabola. Let F (, ) and l be the line x d. For P (r, θ) one has PF r Pl d r cos θ The condition PF e Pl becomes r e(d r cos θ) or x 2 + y 2 e(d x) By squaring both parts after a little algebra we get x 2 + y 2 e 2 (d x) 2 e 2 (d 2 2dx + x 2 ) (1 e 2 )x 2 + 2de 2 x + y 2 e 2 d 2 After completing the square we obtain ( ) 2 x + e2 d 1 e 2 + y 2 1 e 2 e2 d 2 (1 e 2 ) 2

28 Proof. For e < 1 we get the ellipse equation of the form (x h) 2 a 2 + y 2 b 2 1 where h e2 d, a ed, b ed. The foci are at 1 e 2 1 e 2 1 e2 distance c from the ellipse center, where c a 2 b 2 e2 d h and e c 1 e 2 a. For e > 1 we have 1 e 2 <, so the equation represents a hyperbola. We could write its equation in the form (x h) 2 a 2 y 2 b 2 1 and derive e c a with c2 a 2 + b 2.

29 Therefore, the curve equation in polar coordinates is r ed 1 + e cos θ For directrix of the form x d, y d, or y d the equation can be obtained by rotating the graph on angles π, π/2, or π/2, respectively. For example. for y d we get the equation r ed 1+e cos(θ π/2) ed 1+e sin θ Theorem A polar equation of the form r ed 1 ± e cos θ. Thus, we the theorem: or r ed 1 ± e sin θ represents a conic setion of eccentricity e. The conic is an ellipse if e < 1, a parabola if e 1, or a hyperbola if e > 1.

30 Example 1 The conic equation r 3 2 cos θ can be rewritten as r cos θ So, e 2/3 and it is an ellipse. The directrix line is at distance d 1 3 e 5 from the origin, so its equation is x 5. Example If we replace the equation from the previous example with r cos(θ π/4) we get an ellipse rotated on angle π/4 about one of its foci. The directrix line becomes y 5 2 x.

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