Example of a first order differential equation for those familiar with taking the derivative.
When asked how far your car can travel on a tank of gas, most people will multiply the fuel capacity by the fuel usage measured in miles per gallon. For example, if a car has a 15 gallon fuel tank and gets 25 miles per gallon, the car can travel 15 * 25 or 375 miles on a full tank. Generally speaking this is close enough, however in reality, the gas mileage is not constant and is dependent on many things that may change during the journey such as the total vehicle weight. As the fuel is consumed, the overall weight of the car is reduced and hence the gas mileage (measured in miles per gallon) increases. Generally speaking, the weight of fuel is insignificant when compared to the weight of the car, passengers and cargo. However, one can imagine a situation where the vehicle uses fuel at a high enough rate and has a sufficiently large fuel capacity that the weight of the fuel is significant.
Consider a vehicle that weighs 1000 lbs and consumes 50 lbs of fuel for each mile it travels. The fuel it uses is proportional to the product of the total vehicle weight and the distance traveled so if the weight is doubled (by adding cargo or fuel) to 2000 lbs, the fuel used per mile also doubles to 100 lbs per mile traveled. Based on this, the fuel used per mile per pound of total vehicle weight is:
1000 / 50 = 0.05 lbs
As previously stated a total weight of 2000 lbs requires 100 lbs of fuel per mile which will be consumed after:
1000 / 100 = 10 miles
suggesting a vehicle with 1000 lbs of fuel can travel 10 miles before exhausting its fuel supply.
However this is incorrect since as fuel is used, the total weight (vehicle plus fuel) decreases and hence the rate of fuel consumption per mile traveled also decreases.
A simple model for this could be to imagine that the fuel required for each mile is consumed all at once at the end of the mile. In the beginning, the total vehicle weight is 2000 lbs. Since each pound of total weight requires 0.05 lbs of fuel per mile, then at the end of the first mile:
2000 * 0.05 = 100 lbs
of fuel are used resulting in a new total weight for the second mile of 2000 – 100 = 1900 lbs. For the second mile, the fuel used is:
1900 * 0.05 = 95 lbs
Resulting in a new total weight of 1900 – 95 = 1805 lbs. Since the total weight is decreasing with each mile traveled, the amount of fuel used for each mile is also decreasing as shown in the table below:
Miles Traveled |
Starting Weight |
Fuel Used |
Ending Weight |
Total Fuel Used |
1 |
2000.0 |
100.0 |
1900.0 |
100.0 |
2 |
1900.0 |
95.0 |
1805.0 |
195.0 |
3 |
1805.0 |
90.3 |
1714.8 |
285.3 |
4 |
1714.8 |
85.7 |
1629.0 |
371.0 |
5 |
1629.0 |
81.5 |
1547.6 |
452.4 |
6 |
1547.6 |
77.4 |
1470.2 |
529.8 |
7 |
1470.2 |
73.5 |
1396.7 |
603.3 |
8 |
1396.7 |
69.8 |
1326.8 |
673.2 |
9 |
1326.8 |
66.3 |
1260.5 |
739.5 |
10 |
1260.5 |
63.0 |
1197.5 |
802.5 |
11 |
1197.5 |
59.9 |
1137.6 |
862.4 |
12 |
1137.6 |
56.9 |
1080.7 |
919.3 |
13 |
1080.7 |
54.0 |
1026.7 |
973.3 |
14 |
1026.7 |
51.3 |
975.3 |
1024.7 |
Using this model, the 1000 lbs of fuel are used somewhere between the 13^{th} and 14^{th} mile. What also can be seen is that the change in total weight is proportional to the total vehicle weight. For each mile traveled the change in total weight is 0.05 times the current weight. Said another way, the rate of change in weight is equal to -0.05 times the current weight. We know from calculus that the rate of change is the first derivative and therefore can infer the following equation:
The value is negative since the weight is decreasing.
The equation above suggests that W(x) is proportional to its first derivative. One function that when differentiated is proportional to itself is the exponential:
Thus we can try:
and
Where A and B are constants. Substituting into:
gives:
Dividing both sides by results in . The constant A can be determined using the initial total weight when x = 0:
Hence:
To determine the distance traveled when the 1000 lbs of fuel are exhausted, we need to solve for x in W(x) when the total weight is the vehicle minus the fuel:
Divide both sides by 2000:
Take the natural log of both sides:
Divide both sides by -0.05:
So if you have 1000 lbs of fuel you can travel 13.86 miles.