**
Solved math exercises for free - How calculate a derivative graphically - Rate of growth of a seals and bears population. **

4.4 Derivatives applied to biology, medicine and economy : Population of seals and polar bears in the Arctic Ocean

The following graph shows the number of seals and polar bears in an arctic population. The growth of those two populations has been studied on a 600 day period.

Initial number of seals = 3000.

Initial number of polar bears = 400

Notice that the number of seals increases at the beginning because they reproduce very well. However, seals are being preyed on by polar bears. Therefore polar bears reproduce very well too because their food (the seals) is becoming more abundant. Then, while the number of polar bears is growing, the growth of the seal population is slowing down. Eventually the seal population declines. The following graph shows the derivative of the seal population with respect to time (this is the derivative of the blue curve hereabove).

Answer the following questions:

**a) What is the value of the derivative of the seal population with respect to time when the number of seals is the highest ? The lowest ? Explain the meaning of the result obtained.**

First, let us answer that question by looking only at the first graph (number of seals and polar bears with respect to time).
What can you observe on that graph ? You can see that the number of seals is the highest after 120 days. On that day you observe 5800 seals in the population.
Have a look at the tangent line for that 120^{th} day, it is an horizontal line. Its slope is thus 0.
We can then conclude that the value of the derivative of the curve on the 120^{th} day is zero.
In fact, let us remind ourselves that *the derivative of a function (or a curve) at a point is the slope of the tangent line at that point.*

We also need to relate the first derivative with *extrema* and *maximum* :
in other words, *when the number of seals is maximum, the first derivative with respect to time equals zero.
Indeed, the value of the derivative at a point that is an extrema (here it is a maximum) is zero.
*

The samething occurs when the number of seals is minimum,
which means that on the 450^{th} day and beyond, the tangent line to the curve is horizontal.
Thus the value of the derivative of the curve on the day 450 or beyond that day is zero.

Now, let us answer the same question but this time we will also observe the second graph : the graph of the first derivative (hereunder).

On the graph hereabove, the first curve represents the evolution of the number of seals with respect to time.
While the second curve represents the derivative (*y'*) of the number of seals with respect to time.
Because those two graphs are built with the same scale, you can place them one below the other and the solution
appears immediately.
When the number of seals is maximum (extrema on the graph above), the derivative *y'* is zero.
(Just look at the intersection with the graph below). The samething occurs for a minimum number of seals.

To conclude,
note that the graph of the derivative represents the variation of the number of seals per day within the seal population.
Thus it is *DELTA seals / DELTA time*. And if the time is counted in days, then the derivative graph
shows, for each day, the number of seals that are added or eliminated from the population (= number of seals that are born minus the number of seals dead).
Therefore, on the 120^{th} day you can observe that there are 5800 seals in the population
and on that day the population has stagnated (it has neither increased neither decreased).
This means that on the 120^{th} day, the number of seals born is equal to the number of seals dead.

**b) What is the size of the seal population when its derivative has a maximum value ? A minimum value ? Explain the meaning of the result obtained.**

You must here again have a look at both graphs *y* and *y'*. You see immediately
that when the value of the derivative is the highest, in other words when the growth of the population of seals is
the fastest, there are 3000 seals in the population. This occurs on the first day (day zero).
However when the value of the derivative is the lowest, you can observe that there are 4000 seals in the population.
This occurs on the 187^{th} day.
Note that in our case, the lowest value of the derivative is a negative value. This means that the population is decreasing:
the number of dead seals is greater than the number of seals born. To be accurate, on the 187^{th} day,
37 seals disappear from the population.

**c) Which units must be used to express the slope of the first graph ? The vertical axis of the second graph ?**

The slope of the first graph (*y*) is
*
DELTA y / DELTA x =
DELTA seals / DELTA days =
(number of seals on day i - number of seals on day j) / (day i - day j),*

thus the units for the slope are a number of seals per day:

The vertical axis (ordinate) of the second graph *y'* is the derivative of *y*. Hence its unit must be the same than the slope of the first graph : *seals/day*.

Conclusion: the second graph (*y'*) represents also, for each day, the value of the slope of the curve of the first graph (*y*) on that day.