Related Rates—Position, Velocity, and Acceleration over time (Jan. 29, 2010)
NB: I wrote this for my Dad, who I hope to show that calculus is something he can learn, and actuarial science is something we can both pursue over the summer. He's a numbers guy, and he's plenty bright. This was supposed to be an introduction to derivatives.
Position, Velocity, and Acceleration (Related Rates)
Consider a baseball thrown at 10 feet per second, straight down from a building that is 1000 feet tall.
Initial velocity: -10 ft/s (negative since it's being thrown down)
Initial height: +1000 feet
The position function for falling objects is:
position over time = -16 * (time2) + (initial velocity)*(time) + (initial position)
...and we're counting time in seconds, velocity in feet per second, and position in feet.
Since mathematicians are lazy (and physicists are WORSE), they express this:
p(t) = -16t2 + v0t + p0 (v0 = velocity at time zero, p0 = position at time zero)
So at time zero, position is 1000 feet, since -16(02) + (-10)(0) + 1000 = 1000 feet. Mathematically, this is expressed:
p(0) = 1000
...which can be read: "At time zero, position is 1000 feet."
At time = 2 seconds:
p(2) = -16(22) + (-10)(2) + 1000 = -64 -20 +1000 = 916 feet above ground.
At what time (t) will it hit the ground? (i.e. What value of t makes p(t) zero?):
p(t) = -16t2 + v0t + p0 = 0
p(t) = -16t2 - 10t + 1000 = 0 (divide by -2 to make smaller numbers so you can...)
p(t) = 8t2 + 5t - 500 = 0 (...apply the Quadratic Formula, where a = 8, b = 5, c = -500)
Quadratic Formula:
What a mess, but it turns out you can discard the negative numerator (since time will always be positive), and it turns out to be (-5 + 126.6)/16 ≈ 7.6 seconds.
Well that's fine and dandy, but that's got nothing to do with derivatives. That's algebra. The position formula becomes more useful when you take derivatives of it.
Position, velocity, and acceleration are all related functions of time. It just so turns out that the derivative of the position function is the velocity function. And the derivative of the velocity function is the acceleration function.
Mathematically lazy folks describe derivatives of f(x) as f'(x), like this:
| p(t) |
= |
-16t2 + v0t + p0 (Position function) |
| v(t) |
= |
p'(t) (velocity over time is the derivative (') of the position function p(t) |
| a(t) |
= |
v'(t) (acceleration over time is the derivative (') of the velocity function (or the second derivative of the position function (e.g. p''(t)))). |
So how to take a derivative...? Easy stuff. The “Generalized Power Rule” is the secret weapon they don't show you until you've learned to figure derivatives the way they did it in the 1600's. I'm told it built character, but every time I've built character I've regretted it.
Some rules to know:
- The derivative of any constant is zero.
- Derivatives are often expressed [d/dx] (some expression)
For example, it looks syntactically like this:
d/dx [3x2 + 6x + π]
The d on top says we're taking a derivative, the dx on the bottom says we're taking it with respect to x, which is convenient because we have no other variables here. The stuff in the brackets is the function we're taking the derivative of.
The Power Rule says we take any argument with an x in it, multiply the exponent by the coefficient, and then decrement the exponent, and the result is the derivative.
So our first argument, 3x2, becomes (2)(3)x(2-1) = 6x1 or just plain 6x.
The second argument, 6x becomes (1)(6)x(1-1) = 6x0 = 6(1) = 6
The third argument is there to trick you. π is a constant, so its derivative is zero.
The derivative of that big bundle of crap is 6x + 6.
What if they throw another variable in? Well, then it's a constant and its derivative is zero.
d/dx [9x4 - 6x3 + 55a1]
First argument 9x4 = (4)(9)x(4-1) = 36x3
Second argument -6x3 = (3)(-6)x(3-1) = -18x2
Third argument has nothing to do with x and is therefore a constant (evaluates to zero.)
So back to the position/velocity/acceleration stuff, using the generalized power rule, you can find all sorts of other useful things about this ball in motion:
If position is described by:
p(t) = -16t2 + v0t + p0
...and velocity over time is described by the derivative of this function [d/dt], then:
v(t) = p'(t) = -32t1 + v00 + 0 = -32t + v0
...and acceleration over time is described by the derivative of velocity function, then:
a(t) = p''(t) = v'(t) = -32t0 + 0 = -32t
This is a simplified acceleration (-32t feet per second), but it roughly approximates the 9.8 meters per second2 we're used to on planet Earth. You can plug in any positive number for time t and find instantaneous velocity or acceleration or position.
Some examples of derivatives with respect to x (denoted d/dx):
- d/dx [-302894022 / (.555)3/4] = 0 (it's ugly, but it's a constant)
- d/dx [8x9] = 72x8
- d/dx [-x3/4] = -(3/4)x-1/4
- Second derivative of the mess directly above = (3/16)x-5/4
So you can see if you keep taking derivatives, you will eventually reduce the powers all to zero and then be left with a constant whose derivative is zero, and further derivatives are zero. This is the case unless you have a fractional power, and in that case, derivatives go on forever, but their utility does not necessary correspond with the amount of work required to figure them. For instance, the second derivative of the position function is acceleration, but what is the fifth derivative useful for? (I don't know, and I'm not convinced anyone else does.)
That's derivatives in a nutshell. Everything else is some variation. There are rules for trigonometric functions and their inverses, logarithms, e, and a bunch of other stuff, but basically the challenge of any actuarial (or biomedical, etc.) problem is creating an accurate mathematical model... some function that approximates the rates of accidents (or growth, or death (e.g. cancer-killing), etc.), and then making predictions based on the models. There's lots of software to create models with, so it's not just guesswork... the computer does the big stuff.
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