Geometrical Applications of Differentiation
This topic introduces students to some of the uses of the differential calculus within mathematics and shows some of the applications in real world problems. To encompass a firm understanding of this topic, students should ensure that the “Introductory Calculus” topic from the preliminary course is well understood.
The sign of the First Derivative
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Recall that the derivative of a function, , gives the gradient of the graph of the function at any point
lying on the curve. From this we have the following table describing the signs of the derivative, the gradient and the action of the function.
Sign of Derivative | Gradient | Action of Function |
Positive | Increasing | |
Negative | Decreasing | |
Zero | Stationary |
If the derivative is positive at the point , then the functional value is said to be increasing at the point
. Similarly, if the derivative is negative at the point
, then the function is said to be decreasing its value or simply decreasing at the point
. Last and most importantly, if the gradient is zero (that is, the tangent is horizontal) at the point
, then the functional value is said to be stationary. The graph below of the function, indicates the described situations.
Points | Action of Function |
A, E, G, K | Increasing |
C, I | Decreasing |
B, D, F, H, J | Stationary |
Stationary points are of particular interest in this course. There are three types of stationary points. Namely;
Maxima (Maximum Stationary Points)
As the name suggests, these points are the maximum either within their locality, or throughout the whole curve. The former type of maximum stationary points are called local maxima, the latter are called globular maxima. An example of such points is shown below.
Minima (Minimum Stationary Points)
Such points form a minimum either within their locality, or throughout the whole curve. As for the maxima, the former type are named local minima and the latter are named globular minima. An example of such points is shown below.
Horizontal Points of Inflection
These points are stationary points where the concavity of the graph changes. They essentially have one of the forms shown below.
Stationary points may be found by setting the first derivative to zero, and solving the subsequent equation for the respective -values.
The type of stationary point (or the nature of the stationary point) is easily determined by simply testing the derivative on either side of the stationary point. Example 2 illustrates this method.
Example 1Find the values of Solution 1Recall that for a curve to be decreasing, the gradient or first differential must be less than zero (negative). So we have,
Hence for |
Example 2Find and determine the nature of the stationary points of the function Solution 2Firstly, the statement “determine the nature” simply means to find out whether the stationary point is a maxima, minima or a horizontal point of inflection. So we have that,
So, to find the stationary points we let
We are required to find the stationary point. To do so requires one to find the appropriate
Thus we have the stationary points being
Hence we have that the stationary point at |
Note: One does not usually need to determine whether the stationary point is a globular or local maxima/minima. As a general rule, only find out whether the maxima/minima is globular or local when asked specifically to do so by the question.
Note: For the above question, notice that due to the continuity of the function, we could have simply skipped the testing process at , as if there were to be a change of the sign of the gradient between
and
then there must exist a stationary point. Ergo, we simply do not bother testing the function’s derivative at
since the gradient is the same as that at
.
Example 3Find any stationary points on the curve Solution 3So, firstly we differentiate, using the power rule (chain rule for powers of functions).
Hence at Now, using a table of values gives us,
Hence the point |
The sign of the Second Derivative
The second derivative turns out to be just as functional as the first derivative. As the first derivative indicates the gradient of the function, the second derivative indicates the concavity of the function. The below table summarises two key points regarding the second derivative.
On a point on a curve;
Value of the Second Derivative | Concavity at the Point |
The Curve is Concave Up at the Point | |
The Curve is Concave Down at the Point |
Consider the below examples of such curves, and their second differential values.
Now, if the value of the second derivative is equal to zero at the point, then there might exist a point of inflection at the point (this is an example of a necessary but insufficient condition). A point of inflection is a point where the concavity of the graph changes. If the point of inflection coincides with a stationary point then the point is said to be a horizontal point of inflection.
To test for the existence of a point of inflection, simply test the concavity on either side of the point. If there is a change in the sign of the second derivative (or change in concavity), then there exists a point of inflection at the point.
Note: A common mistake is to suppose that non horizontal-points of inflection look like horizontal points of inflection. This is definitely not true. A point of inflection may in general have any gradient at the point itself. Consider the below example of the point of inflection.
Now, the second derivative is also very useful for determining the nature of stationary points. If the value of the second derivative is positive at a stationary point, then it follows that the point is a minimum (Since there exists a stationary point when the curve is concave up). Similarly, if the second derivative is negative at a stationary point, then the point is a maximum (Since there exists a stationary point where the curve is concave down). If the second derivative is zero at a stationary point, then one needs to go back to the method of testing the first derivative on either side of the stationary point.
Note: If both the first derivative and the second derivative are both zero at some point, you must go back and do further testing on either side of the first derivative. You cannot assume that this point will be a horizontal point of inflection. Consider the curve . Both the first and second derivative are equal to zero at
, however, the point
is a minimum stationary point, not a horizontal point of inflection.
Example 4Determine the nature of the stationary points of the curve Solution 4We firstly differentiate and solve the resulting equation upon setting the first derivative to zero.
That is,
Now, at Now, we have that,
Now at At |
Example 5Find the point(s) of inflection of the curve Solution 5So, we firstly find the second derivative.
We now must solve the equation
So there may exist points of inflection at Now setting up a table of values to test the second derivative at the points
Hence as there is a change of sign for the second derivative at both |
The above method of determining the existence of the stationary point is vital if one is required to find the existence of any points of inflection.
Example 6Find the stationary point of the curve and determine its nature. Solution 6So we firstly differentiate with respect to
For stationary points.
Now, differentiating a second time gives,
At
Thus, we have that the point |
Sketching curves using Calculus
The techniques we have so far found may be used to sketch curves. Below is a table that indicates the main features that are required in a sketch of a curve.
When sketching a curve;
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Here we look at an example to illustrate the algorithm above.
Example 7Sketch the curve Solution 7We firstly note that the curve Now, at Now, to find the points of intersection with the
Now, in this case the above equation is easily solved. However the ability to solve the equation must always be checked by the student, before attempting to do so.
Hence the only point of intersection with the
For possible stationary points. Solving this equation gives,
So we have stationary points at the respective points above. We now use a table of values to find the nature of these stationary points.
Hence we have that at Now, it is relatively simple to find the points of inflection in this case, so we shall do so.
Now, we have that
Hence as there is a change of concavity, then it follows that there exists a point of inflection at the point We shall now draw the curve, labelling all the above information. |
The next example illustrates a harder curve with points at which it is undefined.
Example 8Sketch the curve Solution 8Here, we first notice that Now as
Hence we have that the intersection with the
To find the stationary points. Now this gives,
Thus implying that,
Finding the second derivative gives,
Now,
Thus we have that
Now, using a table of values gives,
As there is a change in the sign of the second derivative, it follows that |
Maximisation and Minimisation problems
We now will look at some problems involving maximisation and minimisation of a certain quantity. In this topic, we simply set up a certain dependent variable as a function of another independent variable, which allows us to obtain a maximum or minimum value of the dependent variable, for some value of the independent variable, within the domain of the independent variable. A number of these questions also require some form of extra information, given in the question, to express the independent variable in terms of a single variable, rather than several variables.
Questions in this topic typically involve real life physical problems, but may at times involve other phenomena such as economic fluctuations, and in those cases, one will be given the relation between the independent and dependent variable. The table below illustrates the steps required to complete such questions.
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We shall illustrate this algorithm in the example below.
Example 9What is the largest area of land that may be enclosed by a 50 metre length of rope, provided that the area of land must be rectangular (or square)? Solution 9Firstly, we draw a diagram, depicting the situation. Firstly we note the differing variables within this problem and allocate variables to them.
We also have that the perimeter ( Now, since the area of a rectangle is length by breadth, we have that
Now, since
Hence we have that,
So, to obtain the maximum value, we simply differentiate
We shall set
Now, we find the second derivative to show that the above is a maxima.
Hence we have that there exists a maxima at (In fact, this maxima is a globular maxima, since the curve is always concave down and continuous and hence any maxima becomes automatically a globular maxima. The same applies for the case of a minima on a concave up curve – This argument is one way to show that a maxima/minima is in fact a globular counterpart) So, since
Hence the maximum area enclosed by the rope is |
Having looked at the process involved with a simple question, we shall now look at a slightly more difficult question.
Example 10A square sheet of metal is 16cm on each side. Squares of side a) Show that the volume of the box
b) Hence find the maximum volume of the box, and the value of Solution 10a) We can see from the diagram that the length in this case is
Now, upon removing the factor of
Recall that
b) Now, to find the maximum value of
Now, setting
Now, we find the second derivative to find the maxima and minima.
Now,
Hence we have a maximum stationary point at
Now, to find out whether It is now obvious that the globular maximum value of
Hence we have that the maximum volume of the box is |
Primitive Functions
In this section of the notes, we answer the question, can we obtain the function given
‘
? Now, suppose that we have
‘
. Can we find
that satisfies this? The answer is yes and the process is called finding the “primitive” where
is the primitive of
. Such an example would be
. However, this is not the only example. Another would be
, another would be
and so on. In fact the value of
may be any one of a family of curves given by
, where
is a constant. This slightly complicates the process of finding the function, given the derivative. A general rule for finding the primitive of such terms is,
Where is a real number and
, and
is a constant. If one is given a specific value of the function at a point, then it is possible to find the value of the constant.
Another rather useful rule is the corresponding rule for the linear power rule of differentiation.
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Where is a real number unequal to
,
is a non zero number and
is a constant.
Note: The above rules form the basis for the topic Integration and should be committed to memory. It should be obvious to the student that the process above is the exact opposite (or inverse) of the process of differentiation. It is for this reason that the process of finding the primitive of a function is at times called “anti-differentiation”.
Note: Since differentiation and anti-differentiation are inverses, differentiating the primitive function should bring you back to what you started with. We shall now analyse some examples which will bring to light the process involved in finding the primitive.
Example 11Find the primitive functions for
i)
ii)
iii) Solution 11i) So we have that ii) We have that
iii) We have that
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Example 12Given that Solution 12Performing the process of anti-differentiation gives,
Now, we have that
Hence we have that |
Example 13A cubic curve has a minimum turning point at the point Solution 13Firstly, we note that the derivative of any cubic will automatically be a quadratic. Hence we have that the derivative is of the form,
Now, since we have that there are two stationary points at
Where
Now, applying the process of finding the primitive gives,
Now, to find the values of
Hence we have one equation,
Now, doing the same for the other point gives,
Hence we have another equation,
Now, adding equation
Now, substituting this back in gives,
Hence we have that the equation of the cubic is,
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