The key to being successful in solving related rate problems is proper organization of given information at the start of the problem. By listing the given rate information, and the requested rate, labeled with the correct variable and differentials, the required equations will become clear and the process should be easier. Example:
First, identify the given and needed information. The rate that is given is a volume over time change, and the needed information is an area over time change. So:
These differentials indicate that we need the formulae for the volume and surface area of a sphere:
We can take the derivatives of these equations to get the needed differentials:
Substituting the radius and the change in radius over time into the second equations:
The key item to notice is that by setting up the rates at the top, the next step was always dictated by what variables were in use and what needed to be found next.
Extrema and the Mean Value Theorem
A useful principle related to the topics in this section is the racetrack principle:
The First Derivative Test
The first derivative tells much about the function. The temptation is for students who are raised in a graphing calculator environment to rely on the graphing or guess and check methods to answer questions that could easily be solved by testing using derivatives. Using a chart is a nice way to organize the information. Example:
First thing to do is to take the first derivative and set it equal to zero to find the critical points.
which is not factorable so applying the quadratic formula yields:
Now set up a table with the critical points with some chosen values between each point:
Setting up the table to dictate what values to choose is a key tool. I think of the critical points as being “partitions” for the real numbers. When the partitions are established then any values can be chosen inside those intervals. This is really important for some functions that may not be clear on the calculator, like functions that have critical points well outside the normal graphing window or functions that have critical points that are very close and do not appear correctly on a typical graphing window.
The Second Derivative Test
Since the minimum value is greater than zero, then all values must be greater than zero, proving the original statement.
Limits at Infinity
Now the limit that is in the exponent is ∞ over ∞ meaning that l'Hopital's rule can be applied. Taking the derivative:
Which still results in over , so l'Hopital's rule can be applied again:
Where the limit can be evaluated as going to , which means:
One thing to watch out for is the trap of using l'Hopital's rule in a circular manner. Sometimes now it may be tempting to find derivatives using the limit definition and applying l'Hopital's rule for over cases. This is circular, as a requirement for l'Hopital's rule is that the function has a derivative, and it is known. Therefore, l'Hopital's rule can't be used to find a derivative.
Analyzing the Graphs of a Function
Often times tests require an interpretation of the derivatives of a graph without the function expressed in algebraic form. This can be made easier though using the same techniques used for algebraic functions, rather than simply try to sketch directly from the graph. Example: Sketch the first and second derivatives of the following function:
First up is the first derivative. Just like when given an analytic function, first find the places where this function is going to have a critical point. There are 3 critical points on this graph, with the sign of the slope in between each critical point:
It's possible at this point to sketch a good approximation, but it could be made better by looking for the inflection points, which will show up as maxima and minima for the first derivative:
Now indicate the concavity and sketch the second derivative:
The process is exactly the same, and can provide a good way to reinforce the conceptual parts of the derivative tests, as well as practice sketching graphs based on derivative information.
A very common question is asking for optimization of a path with different rates. Example:
The next thing to do is to draw an accurate diagram with all of the quantities labeled. Any variables that can be put in the diagram will help. In this case, students should be encouraged to think of what is likely to happen. If the cost of the pipe was equal, land or water, then a straight line between the two points is the least pipe, and therefore the cheapest. It is probably also not likely that the pipe runs perpendicular to the river as this would be the most amount of pipe possible. The standard diagram for this type of problem looks something like:
The next step is to try to develop a relationship between our two variables in our cost function. Put another way, there needs to only be a single variable to take a derivative and maximize, so one variable needs to be put in terms of the other. The diagram listed gives us a huge clue, in that the hypotenuse of the right triangle is going to be the distance traveled across the water, and it can be expressed in terms of the distance traveled along the shore using the Pythagorean theorem:
Substituting for the original function and taking the derivative:
Which will only be true when the numerator is equal to zero:
Applying the quadratic formula:
Geometric relationships are the favorites of problem writers.In most circumstances for optimization problems the relationship between variables is going to come from an area, volume, or distance formula. It is useful then to have a couple of the more common ones memorized.