The Mathematics of Star Trek

1. The Mathematics of Star Trek Lecture 3: Equations of Motion and Escape Velocity

2. Topics • Antiderivatives • Integration • Differential Equations • Equations of Motion • Escape Velocity

3. Antiderivatives • We’ve already seen the idea of the derivative of a function. • A related idea is the following: • Given a function f(x), find a function F(x) such that F’(x) = f(x). • If such a function F(x) exists, we call F an antiderivative of f.

4. Antiderivatives (cont.) • For example, given f(x) = 3x2, an antiderivative of f is F(x) = x3, since F’(x) = 3x2. • Question: Can you think of any other antiderivatives of f(x) = 3x2? • Possible answers: G(x) = x3 + 1, H(x) = x3 + 4, K(x) = x3 - 5, etc. • Notice that all these antiderivatives of f(x) = 3x2 differ by a constant!

5. Antiderivatives (cont.) • Graphically, this should make sense, since for a fixed x-value, all of the antiderivatives given above have the same slope! • This is true in general! • If F(x) and G(x) are antiderivatives of f(x) on an interval, then F(x) = G(x) + C, for some constant C.

6. Integration • The process of finding an antiderivative of a given function f(x) is called antidifferentiation or integration. • If F’(x) = f(x), then we denote this by writing ∫ f(x) dx = F(x) + C. • We call the symbol ∫ an integral sign, f(x) the integrand, and C the constant of integration. • Thus, we can write ∫ 3x2 dx = x3 + C. • This notation is due to Leibnitz!

7. Integration (cont.) • Using the fact that integration is differentiation “backwards”, we can apply derivative shortcuts to get integration shortcuts! • ∫k f(x) dx = k ∫ f(x) dx for any constant k. • ∫[f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx. • ∫xn dx = xn+1/(n+1) + C for any rational number n  -1. • ∫ek x dx = (1/k)ek x + C.

8. Integration (cont.) • For practice, evaluate each integral! • ∫x4 dx • ∫2u7 du • ∫(t2 + 3 t + 1) dt • ∫3 e3 x dx

9. Differential Equations • One branch of mathematics is differential equations. • Many applications that involve rates of change can be modeled with differential equations. • A differential equation is an equation involving one or more derivatives of an unknown function. • When solving a differential equation, the goal is to find the unknown function(s) that satisfy the given equation.

10. Differential Equations (cont.) • Here are some differential equations:

11. Differential Equations (cont.) • We can use integration to help solve certain differential equations! • A differential equation is said to be separable if it can be put into the form: • In this case, rewrite the equation in the form s 1/g(y) dy = s f(x) dx and integrate each side with respect to the appropriate variable.

12. Differential Equations (cont.) • Here is an example! • An object moving along a straight line with under the influence of a constant acceleration a is described by the differential equation: • dv/dt = a, • where v is the object’s velocity at time t. • We can use separation of variables to solve for velocity v!

13. Differential Equations (cont.) • ∫dv = ∫ a dt • v = a t + C (general solution) • If the object has initial velocity v0 at time t = 0, then we can find C. • v0 = a (0) + C • v0 = C • Thus, v = a t + v0 (particular solution).

14. Differential Equations (cont.) • Recall that velocity is the derivative of the object’s position function s(t). • It follows that the object’s position function satisfies the differential equation: • ds/dt = a t + v0. • If the object has initial position s0 at time t = 0, then separation of variables can be used to show that: • s = ½ a t2 + v0 t + s0. (HW: Show this!)

15. Equations of Motion • In this last example, we have derived the Equations of Motion for an object moving along a straight line, under the influence of a constant acceleration a, with initial position s0 and initial velocity v0: • s = ½ a t2 + v0 t + s0 • v = a t + v0.

16. Equations of Motion (cont.) • One application of these equations of motion is projectile motion. • For example, suppose Commander Sisko’s baseball is thrown straight up into the air with an initial velocity of 30 m/sec. • Assuming that the acceleration due to gravity is -9.8 m/sec2, find each of the following: • (a) The time at which the ball reaches its maximum height. • (b) The maximum height that the ball reaches. • (c) The time at which the ball hits the ground. • (d) The velocity with which the ball hits the ground.

17. Equations of Motion (cont.) • Solution: • (a) With initial height s0 = 0 m and initial velocity v0 = 30 m/sec, the ball’s equations of motion are: • s = -4.9 t2 + 30 t + 0 (m) • v = -9.8 t + 30 (m/sec) • At the ball’s maximum height, the velocity is zero, so using the second equation: • 0 = -9.8 t + 30 • t = 30/9.8 = 3.06 seconds is time at which maximum height is reached.

18. Equations of Motion (cont.) • (b) To find the maximum height the ball reaches, use the first equation with t = 30/9.8 sec: • s = -4.9 (30/9.8)2 + 30 (30/9.8) = 45.9 m. • Maximum height of the ball reaches is approximately 45.9 meters.

19. Equations of Motion (cont.) • (c) When the ball hits the ground, its height will be zero, so using the first equation, • 0 = -4.9 t2 +30 t = t(-4.9 t + 30), • Thus t = 0 sec or t = 30/4.9 = 6.1 sec. • The ball hits the ground approximately 6.1 seconds after it is thrown into the air.

20. Equations of Motion (cont.) • (d) Using the equation for velocity, when the ball hits the ground it’s velocity will be: • v = -9.8 (30/4.9) +30 = -30 m/sec.

21. Escape Velocity • Another question that we can use differential equations to answer is the following: • What initial velocity is required for an object to escape the Earth’s gravitational field? • To answer this question, we need Newton’s Law of Universal Gravitation and Newton’s Second Law of Motion!

22. Escape Velocity (cont.) • Newton’s Law of Universal Gravitation: The gravitational force between two masses M and m is proportional to the product of the masses and inversely proportional to the square of the distance between them, i.e. F = GMm/r2, where G is a constant. • Newton’s Second Law: The net external force on an object is equal to its mass times acceleration, i.e. F = ma.

23. Escape Velocity (cont.) • Using these laws, we find that the acceleration of an object a distance r from the Earth’s center is given by the equation: • a = dv/dt = -k/r2, • where k is a constant of proportionality. r R

24. Escape Velocity (cont.) • When r = R, then a = -g, the acceleration at the surface of the Earth, so: • -g = -k/R2, • which yields k = gR2. • Thus, a = dv/dt = -gR2/r2. • Now, since v is a function of position r and r is a function of time t, we can write: • dv/dt = dv/dr dr/rt = v dv/dt. (This is an application of the Chain Rule from Calculus.)

25. Escape Velocity (cont.) • Substituting v dv/dr for dv/dt in the equation dv/dt = -gR2/r2, we are led to the following model for an object’s velocity as a function of distance from the Earth’s center: • v dv/dr = -gR2/r2. • This differential equation can be solved via separation of variables!

26. Escape Velocity (cont.) • ∫v dv = ∫ -gR2 r-2 dr • ½ v2 = g R2 r-1 + C • v2 = (2g R2)/r + 2 C (general solution) • If the object leaving the Earth’s surface has an initial velocity of v0, then we can find constant C! • v02 = (2g R2)/R + 2 C • C = ½ (v02 - 2g R)

27. Escape Velocity (cont.) • Thus, our solution to this differential equation is: • v2 = (2g R2)/r + v02 - 2g R. • In order for the velocity v to stay positive, we need v02 - g R ≥ 0, which means that • We call the right-hand side of this last expression Earth’s escape velocity, i.e. the minimum initial velocity needed for an object to escape the Earth’s force of gravity.

28. References • Calculus: Early Transcendentals (5th ed) by James Stewart • Elementary Differential Equations (8th ed) by Rainville, Rainville, and Bedient • Hyper Physics: http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html • The Cartoon Guide to Physics by Larry Gonick and Art Huffman