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
- 1 The sign of the First Derivative
- 2 The sign of the Second Derivative
- 3 Sketching curves using Calculus
- 4 Maximisation and Minimisation problems
- 5 Primitive Functions
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|
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|
|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.
Find the values of for which the function is decreasing.
Recall that for a curve to be decreasing, the gradient or first differential must be less than zero (negative). So we have,
Hence for , is decreasing.
Find and determine the nature of the stationary points of the function .
Firstly, 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 ‘, and solve for .
We are required to find the stationary point. To do so requires one to find the appropriate -values as well.
Thus we have the stationary points being and . So, to determine the nature of the stationary points, we set up a table as below.
Hence we have that the stationary point at is a maximum and the stationary point at is a minimum.
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 .
Find any stationary points on the curve , and determine the nature of any such points.
So, firstly we differentiate, using the power rule (chain rule for powers of functions).
Hence at , we have that .
Now, using a table of values gives us,
Hence the point is a horizontal point of inflection on the curve .
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.
Determine the nature of the stationary points of the curve .
We firstly differentiate and solve the resulting equation upon setting the first derivative to zero.
Now, at , . At , . Hence we have the points and .
Now, we have that,
Now at , we have that hence the point is a minimum stationary point.
At , we have that hence the point is a maximum stationary point.
Find the point(s) of inflection of the curve .
So, we firstly find the second derivative.
We now must solve the equation for all .
So there may exist points of inflection at .
Now setting up a table of values to test the second derivative at the points gives,
Hence as there is a change of sign for the second derivative at both and , it follows that there exists points of inflection 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.
Find the stationary point of the curve and determine its nature.
So we firstly differentiate with respect to .
For stationary points.
Now, differentiating a second time gives,
At . So we must do some further testing involving the first derivative. (That is, this test is inconclusive) Setting up a table of values gives,
Thus, we have that the point is a minimum stationary 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;
Here we look at an example to illustrate the algorithm above.
Sketch the curve , showing all important features.
We firstly note that the curve does not have any vertical asymptotes (Since it is a continuous polynomial) and that it doesn’t have any horizontal or oblique asymptotes (Since no term is getting sequentially smaller, or approaching a certain value as ).
Now, at , we have that implying that the point of intersection with the -axis is .
Now, to find the points of intersection with the -axis, one must set to zero and solve the resulting equation in . Doing so gives,
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 -axis is . We now differentiate the curve to obtain the coordinates of the stationary points.
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 there exists a minimum and at there exists a maximum.
Now, it is relatively simple to find the points of inflection in this case, so we shall do so.
Now, we have that is a possible point of inflection. To determine whether it is or not, we simply test on either side of the second derivative.
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.
Sketch the curve where , showing all important features.
Here, we first notice that is not included within the domain of the function, and that subsequently, becomes a vertical asymptote.
Now as we have that . Hence we have that is an oblique asymptote. Now, to find the -intercepts, we simply set . This gives,
Hence we have that the intersection with the -axis is at the point . To find the stationary points, we simply differentiate. Doing so gives,
To find the stationary points. Now this gives,
Thus implying that,
Finding the second derivative gives,
Thus we have that (or ) is a minimum stationary point. Now, we simply solve for to obtain any possible points of inflection. So, solving the resultant equation gives,
Now, using a table of values gives,
As there is a change in the sign of the second derivative, it follows that is a point of inflection. So showing all the above information diagrammatically gives,
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.
We shall illustrate this algorithm in the example below.
What 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)?
Firstly, we draw a diagram, depicting the situation.
Firstly we note the differing variables within this problem and allocate variables to them.
Area of Land (square metres)
Length of one side (metres)
Length of other side (metres)
We also have that the perimeter () is equal to 50 metres.
Now, since the area of a rectangle is length by breadth, we have that
Now, since , the dependent variable is in terms of two variables, we must find a relationship between and . The perimeter is metres. Using this and the general expression for the perimeter of a rectangle we have that,
Hence we have that,
So, to obtain the maximum value, we simply differentiate with respect to .
We shall set to zero to find any possible maxima.
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 makes physical sense, and is definitely the globular maxima on the domain of , it follows that the maximum area is given by,
Hence the maximum area enclosed by the rope is m2.
Having looked at the process involved with a simple question, we shall now look at a slightly more difficult question.
A square sheet of metal is 16cm on each side. Squares of side cm are cut from each corner. The sheet is then bent along the dotted lines to form an open box.
a) Show that the volume of the box , expressed as a function of is given by,
b) Hence find the maximum volume of the box, and the value of at which it occurs.
We can see from the diagram that the length in this case is as is the breadth, the accounting for the two squares on each side. The height in this case is , upon bending along the broken lines. Hence we have that the volume (length breadth height) is given by,
Now, upon removing the factor of from the squared expression we obtain,
Recall that and hence we have that,
Now, to find the maximum value of , we simply maximise the function . Firstly we find the first derivative by use of the product rule.
Now, setting to zero and solving the resultant equality gives,
Now, we find the second derivative to find the maxima and minima.
Hence we have a maximum stationary point at .
Now, to find out whether is a globular maxima or not, we simply sketch , between the possible values of . Now, it is obvious that since it is a physical scalar (non-signed) value, and also, since otherwise we have that the size of the cut out squares exceeds the size of the total sheet (Consider one side). Hence, we may sketch the curve between and (This is the domain of ). Now, note that at there exists a minima and a zero (since ) and that at there exists a zero as well. Sketching this gives,
It is now obvious that the globular maximum value of occurs at the point . Also, it is obvious that the answer makes physical sense, as it lies within the domain of .
Hence we have that the maximum volume of the box is cm3 and this occurs at cm.
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.
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.
Find the primitive functions for
i) So we have that ‘. Using the first of the two important rules above we have ,
ii) We have that ‘. Using the first rule above,
iii) We have that ‘. Now, using the second rule we have that,
Given that and that the curve passes through the point , find .
Performing the process of anti-differentiation gives,
Now, we have that passes through . Using this information gives
Hence we have that .
A cubic curve has a minimum turning point at the point and a maximum turning point at . What is the equation of the cubic?
Firstly, 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 respectively, this implies that form the zeros of ‘ and hence it follows that the derivative is of the form,
Where is any constant. Note that the must be included, as all these curves have zeros at . Expressing this in terms of powers of gives,
Now, applying the process of finding the primitive gives,
Now, to find the values of and we simply substitute these into the expression for .
Hence we have one equation,
Now, doing the same for the other point gives,
Hence we have another equation,
Now, adding equation to equation gives,
Now, substituting this back in gives,
Hence we have that the equation of the cubic is,