# Mathematics (2 unit) – Geometric Applications of Calculus

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

Recall that the derivative of a function, $Latex formula$, gives the gradient of the graph of the function at any point $Latex formula$ 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 $Latex formula$ Positive Increasing $Latex formula$ Negative Decreasing $Latex formula$ Zero Stationary

If the derivative is positive at the point $Latex formula$, then the functional value is said to be increasing at the point $Latex formula$. Similarly, if the derivative is negative at the point $Latex formula$, then the function is said to be decreasing its value or simply decreasing at the point $Latex formula$. Last and most importantly, if the gradient is zero (that is, the tangent is horizontal) at the point $Latex formula$, 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 $Latex formula$ -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 1

Find the values of $Latex formula$ for which the function $Latex formula$ is decreasing.

### Solution 1

Recall that for a curve to be decreasing, the gradient or first differential must be less than zero (negative). So we have,

$Latex formula$$Latex formula$

$Latex formula$

$Latex formula$

Hence for $Latex formula$, $Latex formula$ is decreasing.

### Example 2

Find and determine the nature of the stationary points of the function $Latex formula$.

### Solution 2

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,

$Latex formula$

$Latex formula$$Latex formula$

So, to find the stationary points we let $Latex formula$$Latex formula$, and solve for $Latex formula$.

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

We are required to find the stationary point. To do so requires one to find the appropriate $Latex formula$ -values as well.

$Latex formula$

Thus we have the stationary points being $Latex formula$ and $Latex formula$. So, to determine the nature of the stationary points, we set up a table as below.

 $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$‘$Latex formula$ Positive $Latex formula$ Negative Negative $Latex formula$ Positive Gradient

Hence we have that the stationary point at $Latex formula$ is a maximum and the stationary point at $Latex formula$ 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 $Latex formula$, as if there were to be a change of the sign of the gradient between $Latex formula$ and $Latex formula$ then there must exist a stationary point. Ergo, we simply do not bother testing the function’s derivative at $Latex formula$ since the gradient is the same as that at $Latex formula$.

### Example 3

Find any stationary points on the curve $Latex formula$, and determine the nature of any such points.

### Solution 3

So, firstly we differentiate, using the power rule (chain rule for powers of functions).

$Latex formula$

Hence at $Latex formula$, we have that $Latex formula$.

Now, using a table of values gives us,

 $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ Positive Zero Positive Gradient

Hence the point $Latex formula$ is a horizontal point of inflection on the curve $Latex formula$.

## 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 $Latex formula$ The Curve is Concave Up at the Point $Latex formula$ 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 $Latex formula$. Both the first and second derivative are equal to zero at $Latex formula$, however, the point $Latex formula$ is a minimum stationary point, not a horizontal point of inflection.

### Example 4

Determine the nature of the stationary points of the curve $Latex formula$.

### Solution 4

We firstly differentiate and solve the resulting equation upon setting the first derivative to zero.

$Latex formula$

That is,

$Latex formula$

Now, at $Latex formula$, $Latex formula$. At $Latex formula$, $Latex formula$. Hence we have the points $Latex formula$ and $Latex formula$.

Now, we have that,

$Latex formula$

Now at $Latex formula$, we have that $Latex formula$ hence the point $Latex formula$ is a minimum stationary point.

At $Latex formula$, we have that $Latex formula$ hence the point $Latex formula$ is a maximum stationary point.

### Example 5

Find the point(s) of inflection of the curve $Latex formula$.

### Solution 5

So, we firstly find the second derivative.

$Latex formula$

$Latex formula$$Latex formula$

$Latex formula$

We now must solve the equation $Latex formula$ for all $Latex formula$.

$Latex formula$

$Latex formula$

So there may exist points of inflection at $Latex formula$.

Now setting up a table of values to test the second derivative at the points $Latex formula$ gives,

 $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ Positive Zero Negative Zero Positive

Hence as there is a change of sign for the second derivative at both $Latex formula$ and $Latex formula$, it follows that there exists points of inflection at both $Latex formula$.

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 6

Find the stationary point of the curve and determine its nature.

### Solution 6

So we firstly differentiate with respect to $Latex formula$.

$Latex formula$

For stationary points.

$Latex formula$

Now, differentiating a second time gives,

$Latex formula$

At $Latex formula$. So we must do some further testing involving the first derivative. (That is, this test is inconclusive) Setting up a table of values gives,

 $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ Negative Zero Positive Gradient

Thus, we have that the point $Latex formula$ 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;  Find any vertical asymptotes.  Find any horizontal or oblique asymptotes.  Find the point of intersection with the $Latex formula$ -axis.  Find the point(s) of intersection with the $Latex formula$ -axis if it is relatively easy to do so.  Find the stationary points (You needn’t find the $Latex formula$ -values if they prove to be cumbersome).  Find any points of inflection if this is relatively easy, or if specifically requested by the question.  Sketch your axes and asymptotes, using a ruler, ensuring that the size is at least a third of a page. (Use broken lines for asymptotes)  Sketch your curve neatly using a pencil, ensuring that the curve is a smooth crisp line, labelling all information mentioned above.  If you are having trouble sketching the curve, substitute some points into the function, to see the general shape of the curve.  It is also useful to check for any symmetry (is it an odd or even function?).

Here we look at an example to illustrate the algorithm above.

### Example 7

Sketch the curve $Latex formula$, showing all important features.

### Solution 7

We firstly note that the curve $Latex formula$ 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 $Latex formula$).

Now, at $Latex formula$, we have that $Latex formula$ implying that the point of intersection with the $Latex formula$ -axis is $Latex formula$.

Now, to find the points of intersection with the $Latex formula$ -axis, one must set $Latex formula$ to zero and solve the resulting equation in $Latex formula$. Doing so gives,

$Latex formula$

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.

$Latex formula$

$Latex formula$

Hence the only point of intersection with the $Latex formula$ -axis is $Latex formula$. We now differentiate the curve to obtain the coordinates of the stationary points.

$Latex formula$

For possible stationary points. Solving this equation gives,

$Latex formula$

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.

 $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ Positive Zero Negative Zero Positive Gradient

Hence we have that at $Latex formula$ there exists a minimum and at $Latex formula$ there exists a maximum.

Now, it is relatively simple to find the points of inflection in this case, so we shall do so.

$Latex formula$

Now, we have that $Latex formula$ is a possible point of inflection. To determine whether it is or not, we simply test on either side of the second derivative.

 $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ Negative Zero Positive

Hence as there is a change of concavity, then it follows that there exists a point of inflection at the point $Latex formula$.

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 8

Sketch the curve $Latex formula$ where $Latex formula$, showing all important features.

### Solution 8

Here, we first notice that $Latex formula$ is not included within the domain of the function, and that subsequently, $Latex formula$ becomes a vertical asymptote.

Now as $Latex formula$ we have that $Latex formula$. Hence we have that $Latex formula$ is an oblique asymptote. Now, to find the $Latex formula$ -intercepts, we simply set $Latex formula$. This gives,

$Latex formula$

Hence we have that the intersection with the $Latex formula$ -axis is at the point $Latex formula$. To find the stationary points, we simply differentiate. Doing so gives,

$Latex formula$$Latex formula$

To find the stationary points. Now this gives,

$Latex formula$

Thus implying that,

$Latex formula$

Finding the second derivative gives,

$Latex formula$

Now,

$Latex formula$

Thus we have that $Latex formula$ (or $Latex formula$ ) is a minimum stationary point. Now, we simply solve $Latex formula$ for $Latex formula$ to obtain any possible points of inflection. So, solving the resultant equation gives,

$Latex formula$

Now, using a table of values gives,

 $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ Positive Zero Negative

As there is a change in the sign of the second derivative, it follows that $Latex formula$ 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.

 Draw a realistic and accurate diagram. Find the variable that is to be maximised/minimised (dependent variable), and any other related variables (independent variables). Allocate appropriate pronumerals to these variables. Express the maximised/minimised variable in terms of the other variables. If the expression for your maximised/minimised variable is in terms of several variables, look through the question to find a further relationship between the independent variables involved within the expression for the maximised/minimised variable. Apply calculus by maximising/minimising the dependent variable. Check that your answer is in fact a maxima or minima (by using the second derivative test or a table of values). Finally check that the value is indeed the global maxima/minima for the domain of the dependent variable (check that your answer makes physical sense, and that there are no values which are larger/smaller than that value – often a good method to check is to draw a quick sketch (if possible) of the relation between the independent and dependent variables).

We shall illustrate this algorithm in the example below.

### Example 9

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)?

### Solution 9

Firstly, we draw a diagram, depicting the situation.

Firstly we note the differing variables within this problem and allocate variables to them.

$Latex formula$ Area of Land (square metres)

$Latex formula$ Length of one side (metres)

$Latex formula$ Length of other side (metres)

We also have that the perimeter ($Latex formula$) is equal to 50 metres.

Now, since the area of a rectangle is length by breadth, we have that

$Latex formula$

Now, since $Latex formula$, the dependent variable is in terms of two variables, we must find a relationship between $Latex formula$ and $Latex formula$. The perimeter is $Latex formula$ metres. Using this and the general expression for the perimeter of a rectangle we have that,

$Latex formula$

$Latex formula$

Hence we have that,

$Latex formula$

So, to obtain the maximum value, we simply differentiate $Latex formula$ with respect to $Latex formula$.

$Latex formula$

We shall set $Latex formula$ to zero to find any possible maxima.

$Latex formula$

Now, we find the second derivative to show that the above is a maxima.

$Latex formula$

Hence we have that there exists a maxima at $Latex formula$.

(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 $Latex formula$ makes physical sense, and is definitely the globular maxima on the domain of $Latex formula$, it follows that the maximum area is given by,

$Latex formula$

Hence the maximum area enclosed by the rope is $Latex formula$ m2.

Having looked at the process involved with a simple question, we shall now look at a slightly more difficult question.

### Example 10

A square sheet of metal is 16cm on each side. Squares of side $Latex formula$ 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 $Latex formula$, expressed as a function of $Latex formula$ is given by,

$Latex formula$

b) Hence find the maximum volume of the box, and the value of $Latex formula$ at which it occurs.

### Solution 10

a)

We can see from the diagram that the length in this case is $Latex formula$ as is the breadth, the $Latex formula$ accounting for the two squares on each side. The height in this case is $Latex formula$, upon bending along the broken lines. Hence we have that the volume (length $Latex formula$ breadth $Latex formula$ height) is given by,

$Latex formula$

Now, upon removing the factor of $Latex formula$ from the squared expression we obtain,

$Latex formula$

Recall that $Latex formula$ and hence we have that,

$Latex formula$

b)

Now, to find the maximum value of $Latex formula$, we simply maximise the function $Latex formula$. Firstly we find the first derivative by use of the product rule.

$Latex formula$

$Latex formula$

Now, setting $Latex formula$ to zero and solving the resultant equality gives,

$Latex formula$

Now, we find the second derivative to find the maxima and minima.

$Latex formula$

Now,

$Latex formula$

Hence we have a maximum stationary point at $Latex formula$.

Now, to find out whether $Latex formula$ is a globular maxima or not, we simply sketch $Latex formula$, between the possible values of $Latex formula$. Now, it is obvious that $Latex formula$ since it is a physical scalar (non-signed) value, and also, $Latex formula$ 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 $Latex formula$ and $Latex formula$ (This is the domain of $Latex formula$ ). Now, note that at $Latex formula$ there exists a minima and a zero (since $Latex formula$ ) and that at $Latex formula$ there exists a zero as well. Sketching this gives,

It is now obvious that the globular maximum value of $Latex formula$ occurs at the point $Latex formula$. Also, it is obvious that the answer makes physical sense, as it lies within the domain of $Latex formula$.

Hence we have that the maximum volume of the box is $Latex formula$ cm3 and this occurs at $Latex formula$ cm.

## Primitive Functions

In this section of the notes, we answer the question, can we obtain the function $Latex formula$ given $Latex formula$$Latex formula$ ? Now, suppose that we have $Latex formula$$Latex formula$. Can we find $Latex formula$ that satisfies this? The answer is yes and the process is called finding the “primitive” where $Latex formula$ is the primitive of $Latex formula$. Such an example would be $Latex formula$. However, this is not the only example. Another would be $Latex formula$, another would be $Latex formula$ and so on. In fact the value of $Latex formula$ may be any one of a family of curves given by $Latex formula$, where $Latex formula$ 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,

 $Latex formula$‘$Latex formula$

Where $Latex formula$ is a real number and $Latex formula$, and $Latex formula$ 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.

 $Latex formula$‘$Latex formula$

Where $Latex formula$ is a real number unequal to $Latex formula$$Latex formula$ is a non zero number and $Latex formula$ 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 11

Find the primitive functions for

i)$Latex formula$

ii)$Latex formula$

iii)$Latex formula$.

### Solution 11

i) So we have that $Latex formula$$Latex formula$. Using the first of the two important rules above we have $Latex formula$,

ii) We have that $Latex formula$$Latex formula$. Using the first rule above,

$Latex formula$

iii) We have that $Latex formula$$Latex formula$. Now, using the second rule we have that,

$Latex formula$

$Latex formula$

### Example 12

Given that $Latex formula$ and $Latex formula$ that the curve $Latex formula$ passes through the point $Latex formula$, find $Latex formula$.

### Solution 12

Performing the process of anti-differentiation gives,

$Latex formula$

Now, we have that $Latex formula$ passes through $Latex formula$. Using this information gives

$Latex formula$

$Latex formula$

Hence we have that $Latex formula$.

### Example 13

A cubic curve has a minimum turning point at the point $Latex formula$ and a maximum turning point at $Latex formula$. What is the equation of the cubic?

### Solution 13

Firstly, we note that the derivative of any cubic will automatically be a quadratic. Hence we have that the derivative is of the form,

$Latex formula$$Latex formula$

Now, since we have that there are two stationary points at $Latex formula$ respectively, this implies that $Latex formula$ form the zeros of $Latex formula$$Latex formula$ and hence it follows that the derivative is of the form,

$Latex formula$$Latex formula$

Where $Latex formula$ is any constant. Note that the $Latex formula$ must be included, as all these curves have zeros at $Latex formula$. Expressing this in terms of powers of $Latex formula$ gives,

$Latex formula$$Latex formula$

Now, applying the process of finding the primitive gives,

$Latex formula$

Now, to find the values of $Latex formula$ and $Latex formula$ we simply substitute these into the expression for $Latex formula$.

$Latex formula$

Hence we have one equation,

$Latex formula$

Now, doing the same for the other point gives,

$Latex formula$

Hence we have another equation,

$Latex formula$

Now, adding equation $Latex formula$ to equation $Latex formula$ gives,

$Latex formula$

$Latex formula$

Now, substituting this back in gives,

$Latex formula$

Hence we have that the equation of the cubic is,

$Latex formula$