Grade 11/12 Math Circles Conics & Applications The Mathematics of Orbits Dr. Shahla Aliakbari November 18, 2015

Transcription

1 Faculty of Mathematics Waterloo, Ontario N2L 3G1 Centre for Education in Mathematics and Computing Grade 11/12 Math Circles Conics & Applications The Mathematics of Orbits Dr. Shahla Aliakbari November 18, 2015 Parabolas Recall that a conic section is a curve formed by the intersection of a plane and a double cone. By changing the angle and location of this intersection, we can produce different conics. A parabola can be formed by the intersection of a cone and a tilted plane - the plane is parallel to the edge of the cone. A parabola is the set of all points (x, y) that are equidistant from a point (called the focus) and a line (called the directrix) 1

2 Equation of a parabola If P (x, y) is any point on the parabola, the distance from P (x, y) to the directrix is then: F P = (x 0) 2 + (y p) 2 (x 0)2 + (y p) 2 = (y + p) x 2 + (y p) 2 = (y + p) 2 x 2 +y 2 2yp+p 2 = y 2 +2yp+p 2 x 2 = 4yp If we write a = 1 4p then the equation becomes y = ax 2 Below are some examples of parabolas and their equations 2

3 Example Find the vertex, focus, and directrix of the parabola and sketch it. x 2 10x 8y + 33 = 0 (x 2 10x + 25) 8y + 33 = 25 (x 5) 2 8y + 8 = 0 (x 5) 2 8(y 1) = 0 (x 5) 2 = 8(y 1) Therefore we have: Vertex: (5, 1) 4p = 8 p = 2 Focus: (5, 3) Directrix: y = 1 The equation of the parabola is then: y = 1 8 (x 5)

4 Projectile Motion Projectile motion is the motion of an object that is thrown into the air at an angle. There is only acceleration in the vertical direction: a x = 0, a y = g where a x and a y are the acceleration values in the horizontal and vertical direction respectively and g represents gravity. The components of the velocity are: v x = v 0 cos θ, At any time t, Eliminating t gives: x = v 0 t cos θ y = v o t sin θ 1 2 gt2 y = x tan θ v y = v 0 sin θ gt g 2v 2 0 cos 2 θ x2 4

5 Harmonic Oscillator The system experiences a restoring force when displaced from its equilibrium position. F = kx The motion can be described by the following differential equation: where x + ω 2 x = 0 ω 2 = k m Note that x(t) = A cos (ωt) satisfies this equation, as well as the initial conditions x(0) = A and v(0) = 0 The period of oscillation is: T = 2π ω = 2π m k Suppose k = 4(N/m) and m = 1kg. Then, ω = 2 or T = π 5

6 Energy of the harmonic oscillator The potential energy for a simple harmonic oscillator at position x is: U(x) = 1 2 kx2 The kinetic energy is given by K = 1 2 mv2 = 1 m( Aω sin ωt)2 2 The total energy is constant, as expected. Harmonic approximation In applications, it is often very useful to approximate the potential of an object with a harmonic potential. In other words: U(x) = U(x 0 ) k(x x 0) 2 Simple pendulum Suppose a mass m is hanging from a string of length L and is fixed at the point O. If we displace it to an initial angle and release, the pendulum will swing back and forth with a periodic motion. This means: m(lθ ) = mg sin theta θ + g L sin θ = 0 0.1rad = π 5.7 sin sin θ θ for small θ θ + ω 2 θ = 0 where ω 2 = g L 6

7 This is similar to the differential equation for the spring-mass system: L T = 2π g 1 Suppose L = 1. then, T = 2π 9.8 2(s) Hyperbola The hyperbola is also formed by the intersection of a plane and a double cone - the plane intersects both halves. A hyperbola is the set of points that have a constant difference of distances between two foci and a point on the hyperbola. Equation of a hyperbola where x 2 a 2 y2 b 2 = 1 a 2 + b 2 = c 2 7

8 y 2 a 2 x2 b 2 = 1 where a 2 + b 2 = c 2 Example Sketch the conic 9x 2 4y 2 72x + 8y = 0 We complete the square and re-write the equation as follows: 9(x 2 8x) + 4(y 2 2y) = 176 9(x 2 8x = 16) + 4(y 2 2y + 1) = (x 4) 2 + 4(y 1) 2 = 36 (y 1) 2 9 (x 4)2 4 = 1 The hyperbola is shifted 4 units to the right and 1 unit up. The asymptotes are: y 1 = ± 3 (x 4) 2 8

9 Reflection property of the parabola Suppose P (x 1, y 1 ) is a point on a parabola as in the figure below. If α is the angle between the parabola and the line segment F P, and β is the angle between the horizontal line y = y 1, then α = β. Light from a source placed at F will be reflected parallel to the x-axis. The law of reflection for a curved mirror states that the angles made by the incident and reflected rays with the tangent to the mirror must be equal. Reflection property of the hyperbola Let P (x 1, y 1 ) be a point on a hyperbola as in the figure below. If α and β are the angles between the lines P F 1, P F 2 and the hyperbola, the α = β. This shows that light aimed at a focus of a hyperbolic mirror is reflected toward the other focus. 9