Introduction
It is better to not need units in order to understand a problem, but they work well in a pinch. Following the units helps keep track of what is happening in a problem.
Here is an example of a problem involving first and second order derivatives, with units.
Position, Velocity, and Acceleration
A man is driving on a perfectly straight highway. His position is given by
p
, a function of time
(t)
. We will graph his position in meters and seconds.
p(t) = \frac{1}{3}t^{3}
1 second after starting his drive, he is a third of a meter away from his starting point. 9 seconds after starting his drive, he is 243 meters away. Let's ignore him surpassing the speed of light after 5 hours and get to our problem.
Problem: What is his velocity at 10 seconds?
Let's slowly work through the four steps.
1. Understand what the question is asking.
We are looking for the velocity of the man. His velocity is how fast his position is changing with respect to time. In other words, we are looking for how many meters he would move in one second, using his velocity ten seconds after starting. 2. Convert the request into mathematical notation.
p(t) = \frac{1}{3}t^{3}
His velocity is equal to
p'(t)
.
p'(t) = \frac{d}{dt}p(t) = \frac{d}{dt}\frac{1}{3}t^{3}
3. Solve the differential equation(s).
p'(t) = t^2
p'(10) = 10^2 = 100
4. Interpret the solution.
10 seconds after starting his drive, the man had a velocity of 100 meters per second.
Using Units
If we got lost in the problem, and we did not know how to find velocity, we could have used units to help us.
We know that velocity is measured in meters per second. Even if we didn't, on a test the answers will usually show the units.
Using the units, we can see that the position function is in meters. Since the answers are in meters per second, the velocity must be the first derivative of the function.
Let's use that method with the next problem.
Problem: Find the man's acceleration (
m/s^2
) at 5 seconds.
Note the units. We know that velocity is
m/s
. If we derive that one more time, we get
m/s^2
. Thus, we know that we are looking for the second derivative of the position function.
1. Understand what the question is asking.
We are looking for the acceleration of the man. His velocity is how fast his position is changing with respect to time. His acceleration is how fast his velocity is changing in respect to time. In other words, we are looking for how many meters per second his speed would change in one second, using his acceleration after 5 seconds. 2. Convert the request into mathematical notation.
p(t) = \frac{1}{3}t^3
p'(t) = t^2
His acceleration is equal to
p''(t)
.
p''(t) = \frac{d^2}{dt^2}p(t) = \frac{d}{dt}p'(t) = \frac{d}{dt}t^2
3. Solve the differential equation(s).
p''(t) = 2t
p''(5) = 2(5) = 10
4. Interpret the solution.
5 seconds after starting his drive,
the man had an acceleration of 10 meters per second squared.