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, $dy/dx$ , gives the gradient of the graph of the function at any point $(x , y)$ 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
$\frac{dy}{dx}>0$	Positive	Increasing
$\frac{dy}{dx}<0$	Negative	Decreasing
$\frac{dy}{dx}=0$	Zero	Stationary

If the derivative is positive at the point $(x , y)$ , then the functional value is said to be increasing at the point $(x , y)$ . Similarly, if the derivative is negative at the point $(x , y)$ , then the function is said to be decreasing its value or simply decreasing at the point $(x , y)$ . Last and most importantly, if the gradient is zero (that is, the tangent is horizontal) at the point $(x , y)$ , 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 $x$ -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 $x$ for which the function $f(x)=x^{2}-3x+2$ 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,

$f$ ‘ $(x)<0$

$=> 2x-3<0$

$x<\frac{3}{2}$

Hence for $x<3/2$ , $f(x)$ is decreasing.

Example 2

Find and determine the nature of the stationary points of the function $f(x)=2x^{3}+3x^{2}-36x+5$ .

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,

$f(x)=2x^{3}+3x^{2}-36x+5$

$f$ ‘ $(x)=6x^{2}+6x-36$

So, to find the stationary points we let $f$ ‘ $(x)=0$ , and solve for $x$ .

$6x^{2}+6x-36=0$

$x^{2}+x-6=0$

$=> (x+3)(x-2)=0$

$x=-3 , 2$

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

$f(-3)=86 f(2)=-39$

Thus we have the stationary points being $(-3 , 86)$ and $(2 , -39)$ . So, to determine the nature of the stationary points, we set up a table as below.

$x$	$-4$	$-3$	$-2$	$1$	$2$	$3$
$f$ ‘ $(x)$	Positive	$0$	Negative	Negative	$0$	Positive
Gradient

Hence we have that the stationary point at $(-3 , 86)$ is a maximum and the stationary point at $(2 , -39)$ 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 $x=1$ , as if there were to be a change of the sign of the gradient between $x=-2$ and $x=1$ then there must exist a stationary point. Ergo, we simply do not bother testing the function’s derivative at $x=1$ since the gradient is the same as that at $x=-2$ .

Example 3

Find any stationary points on the curve $y=(x+3)^{5}+4$ , and determine the nature of any such points.

Solution 3

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

$\frac{dy}{dx}=5(x+3)^{4}=0 <=> x=-3$

Hence at $x=-3$ , we have that $y=4$ .

Now, using a table of values gives us,

$x$	$-4$	$-3$	$-2$
$dy/dx$	Positive	Zero	Positive
Gradient

Hence the point $(-3 , 4)$ is a horizontal point of inflection on the curve $y=(x+3)^{5}+4$ .

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
$\frac{d^{2}y}{dx^{2}}>0$	The Curve is Concave Up at the Point
$\frac{d^{2}y}{dx^{2}}<0$	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 $y=x^{4}$ . Both the first and second derivative are equal to zero at $x=0$ , however, the point $(0 , 0)$ is a minimum stationary point, not a horizontal point of inflection.

Example 4

Determine the nature of the stationary points of the curve $y=x^{3}+2x^{2}-4x+2$ .

Solution 4

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

$\frac{dy}{dx}=0 <=> 3x^{2}+4x-4=0$

That is,

$x=\frac{2}{3} , -2$

Now, at $=2/3$ , $y=14/27$ . At $x=-2$ , $y=10$ . Hence we have the points $(2/3 , 14/27)$ and $(-2 , 10)$ .

Now, we have that,

$\frac{d^{2}y}{dx^{2}}=6x+4$

Now at $x=2/3$ , we have that $d^{2}y/dx^{2}=8>0$ hence the point $(2/3 , 14/27)$ is a minimum stationary point.

At $x=-2$ , we have that $d^{2}y/dx^{2}=-8<0$ hence the point $(-2 , 10)$ is a maximum stationary point.

Example 5

Find the point(s) of inflection of the curve $f(x)=x^{4}-24x^{2}+70$ .

Solution 5

So, we firstly find the second derivative.

$f(x)=x^{4}-24x^{2}+70$

$f$ ‘ $(x)=4x^{3}-48x$

$f^{''}(x)=12x^{2}-48$

We now must solve the equation $f^{''}(x)=0$ for all $x$ .

$12x^{2}-48=0$

$x=\pm 2$

So there may exist points of inflection at $x=\pm 2$ .

Now setting up a table of values to test the second derivative at the points $x=\pm 2$ gives,

$x$	$-3$	$-2$	$0$	$2$	$3$
$f^{''}(x)$	Positive	Zero	Negative	Zero	Positive

Hence as there is a change of sign for the second derivative at both $x=2$ and $x=-2$ , it follows that there exists points of inflection at both $(2 , -10)[latex/] and [latex](-2 , -10)$ .

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 $x$ .

$\frac{dy}{dx}=4x^{3}=0$

For stationary points.

$x=0$

Now, differentiating a second time gives,

$\frac{d^{2}y}{dx^{2}}=12x^{2}=0$

At $x=0$ . So we must do some further testing involving the first derivative. (That is, this test is inconclusive) Setting up a table of values gives,

$x$	$-1$	$0$	$1$
$dy/dx$	Negative	Zero	Positive
Gradient

Thus, we have that the point $(0 , 2)$ 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 $y$ -axis.
Find the point(s) of intersection with the $x$ -axis if it is relatively easy to do so.
Find the stationary points (You needn’t find the $y$ -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 $y=x^{3}+3x^{2}+x+3$ , showing all important features.

Solution 7

We firstly note that the curve $y=x^{3}+3x^{2}+x+3$ 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 $x\to \pm \infty$ ).

Now, at $x=0$ , we have that $y=3$ implying that the point of intersection with the $y$ -axis is $(0 , 3)$ .

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

$x^{3}+3x^{2}+x+3=0$

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.

$x^{2}(x+3)+(x+3)=0$

$=>(x+3)(x^{2}+1)=0$

Hence the only point of intersection with the $x$ -axis is $(-3 , 0)$ . We now differentiate the curve to obtain the coordinates of the stationary points.

$\frac{dy}{dx}=3x^{2}+6x+1=0$

For possible stationary points. Solving this equation gives,

$x=\frac{-6\pm \sqrt{36-12}}{2.3}=-1\pm \frac{\sqrt{6}}{3}$

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.

$x$	$-2$	$-1-\sqrt{6}/3$	$-1$	$-1+\sqrt{6}/3$	$0$
$dy/dx$	Positive	Zero	Negative	Zero	Positive
Gradient

Hence we have that at $x=-1+\sqrt{6}/3$ there exists a minimum and at $x=-1-\sqrt{6}/3$ there exists a maximum.

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

$\frac{d^{2}y}{dx^{2}}=6x+6=0 <=> x=-1$

Now, we have that $(-1 , 4)$ is a possible point of inflection. To determine whether it is or not, we simply test on either side of the second derivative.

$x$	$-2$	$-1$	$0$
$d^{2}y/dx^{2}$	Negative	Zero	Positive

Hence as there is a change of concavity, then it follows that there exists a point of inflection at the point $(-1 , 4)$ .

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 $y=f(x)$ where $f(x)=x^{2}+1/x$ , showing all important features.

Solution 8

Here, we first notice that $x=0$ is not included within the domain of the function, and that subsequently, $x=0$ becomes a vertical asymptote.

Now as $x?\pm \infty$ we have that $y?x^{2}$ . Hence we have that $y=x^{2}$ is an oblique asymptote. Now, to find the $x$ -intercepts, we simply set $f(x)=0$ . This gives,

$x^{2}+\frac{1}{x}=0<=>x^{3}=-1?x=-1$

Hence we have that the intersection with the $x$ -axis is at the point $(-1 , 0)$ . To find the stationary points, we simply differentiate. Doing so gives,

$f$ ‘ $(x)=2x-\frac{1}{x^{2}}=0$

To find the stationary points. Now this gives,

$x^{3}=\frac{1}{2}$

Thus implying that,

$x=\frac{1}{\sqrt[3]{2}}$

Finding the second derivative gives,

$f^{''}(x)=2+\frac{2}{x^{3}}$

Now,

$f^{''}(\frac{1}{\sqrt[3]{2}})>0$

Thus we have that $(1/\sqrt[3]{2} , 3/\sqrt[3]{4})$ (or $(0.793 , 1.89)$ ) is a minimum stationary point. Now, we simply solve $f^{''}(x)=0$ for $x$ to obtain any possible points of inflection. So, solving the resultant equation gives,

$2+\frac{2}{x^{3}}=0 <=>x^{3}=-1 ?x=-1$

Now, using a table of values gives,

$x$	$-2$	$-1$	$-1/2$
$f^{''}(x)$	Positive	Zero	Negative

As there is a change in the sign of the second derivative, it follows that $(-1 , 0)$ 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.

$A=$ Area of Land (square metres)

$x=$ Length of one side (metres)

$y=$ Length of other side (metres)

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

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

$A=xy$

Now, since $A$ , the dependent variable is in terms of two variables, we must find a relationship between $x$ and $y$ . The perimeter is $50$ metres. Using this and the general expression for the perimeter of a rectangle we have that,

$2x+2y=50$

$=>y=25-x$

Hence we have that,

$A=x(25-x)=25x-x^{2}$

So, to obtain the maximum value, we simply differentiate $A$ with respect to $x$ .

$\frac{dA}{dx}=25-2x$

We shall set $dA/dx$ to zero to find any possible maxima.

$25-2x=0 <=> x=\frac{25}{2}$

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

$\frac{d^{2}A}{dx^{2}}=-2<0$

Hence we have that there exists a maxima at $x=25/2$ .

(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 $x=25/2$ makes physical sense, and is definitely the globular maxima on the domain of $A(x)$ , it follows that the maximum area is given by,

$\max (A)=(\frac{25}{2})(25-\frac{25}{2})=\frac{625}{4}=156.25 m^{2}$

Hence the maximum area enclosed by the rope is $156.25$ 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 $x$ 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 $V$ , expressed as a function of $x$ is given by,

$V(x)=4x(x-8)^{2}$

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

Solution 10

We can see from the diagram that the length in this case is $16-2x$ as is the breadth, the $2$ accounting for the two squares on each side. The height in this case is $x$ , upon bending along the broken lines. Hence we have that the volume (length $\times$ breadth $\times$ height) is given by,

$V(x)=(16-2x)^{2}x$

Now, upon removing the factor of $2$ from the squared expression we obtain,

$V(x)=x(2(8-x))^{2}=x(4)(8-x)^{2}=4x(8-x)^{2}$

Recall that $a^{2}=(-a)^{2}$ and hence we have that,

$V(x)=4x(x-8)^{2}$

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

$\frac{dV}{dx}=4(x-8)^{2}+8x(x-8)$

$\frac{dV}{dx}=4(x-8)(x-8+2x)=4(x-8)(3x-8)$

Now, setting $dV/dx$ to zero and solving the resultant equality gives,

$x=8 , 8/3$

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

$\frac{d^{2}V}{dx^{2}}=4(3x-8)+12(x-8)=24x-128$

Now,

$(\frac{d^{2}V}{dx^{2}}\vert _{x=8}>0 , (\frac{d^{2}V}{dx^{2}}\vert _{x=\frac{8}{3}}<0$

Hence we have a maximum stationary point at $(8/3 , 8192/27)$ .

Now, to find out whether $x=8/3$ is a globular maxima or not, we simply sketch $V(x)$ , between the possible values of $x$ . Now, it is obvious that $x\ge 0$ since it is a physical scalar (non-signed) value, and also, $x\le 8$ 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 $x=0$ and $x=8$ (This is the domain of $V(x)$ ). Now, note that at $x=8$ there exists a minima and a zero (since $V(x)=4x(x-8)^{2}$ ) and that at $x=0$ there exists a zero as well. Sketching this gives,

It is now obvious that the globular maximum value of $V$ occurs at the point $(8/3 , 8192/27)$ . Also, it is obvious that the answer makes physical sense, as it lies within the domain of $V$ .

Hence we have that the maximum volume of the box is $303\frac{11}{27}$ cm³ and this occurs at $x=8/3$ cm.

Primitive Functions

In this section of the notes, we answer the question, can we obtain the function $f(x)$ given $f$ ‘ $(x)$ ? Now, suppose that we have $f$ ‘ $(x)=2x$ . Can we find $f(x)$ that satisfies this? The answer is yes and the process is called finding the “primitive” where $f$ is the primitive of $f'$ . Such an example would be $f(x)=x^{2}$ . However, this is not the only example. Another would be $f(x)=x^{2}+1$ , another would be $f(x)=x^{2}+4$ and so on. In fact the value of $f(x)$ may be any one of a family of curves given by $f(x)=x^{2}+C$ , where $C$ 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,

$f$ ‘ $(x)=x^{n} => f(x)=\frac{x^{n+1}}{n+1}+C$

Where $n$ is a real number and $n\ne -1$ , and $C$ 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.

$f$ ‘ $(x)=(ax+b)^{n} => f(x)=\frac{(ax+b)^{n+1}}{a(n+1)}+C$

Where $n$ is a real number unequal to $-1$ , $a$ is a non zero number and $C$ 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) $x^{2}-6$

ii) $3x^{3}+5x+2$

iii) $(3x+2)^{7}$ .

Solution 11

i) So we have that $f$ ‘ $(x)=x^{2}-6$ . Using the first of the two important rules above we have $f(x)=\frac{x^{3}}{3}-6x+C$ ,

ii) We have that $f$ ‘ $(x)=3x^{3}+5x+2$ . Using the first rule above,

$f(x)=\frac{3x^{4}}{4}-\frac{5x^{2}}{2}+2x+C$

iii) We have that $f$ ‘ $(x)=(3x+2)^{7}$ . Now, using the second rule we have that,

$f(x)=\frac{(3x+2)^{8}}{3\times 8}+C$

$f(x)=\frac{(3x+2)^{8}}{24}+C$

Example 12

Given that $\frac{dy}{dx}$ and $=4x-3$ that the curve $y=f(x)$ passes through the point $(1 , 4)$ , find $y$ .

Solution 12

Performing the process of anti-differentiation gives,

$y=\frac{4x^{2}}{2}-3x+c=2x^{2}-3x+C$

Now, we have that $y$ passes through $(1 , 4)$ . Using this information gives

$4=2-3+C$

$C=5$

Hence we have that $y=2x^{2}-3x+5$ .

Example 13

A cubic curve has a minimum turning point at the point $(-1 , -15)$ and a maximum turning point at $(1 , 12)$ . 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,

$f$ ‘ $(x)=Ax^{2}+Bx+C$

Now, since we have that there are two stationary points at $x=\pm 1$ respectively, this implies that $x=\pm 1$ form the zeros of $f$ ‘ $(x)$ and hence it follows that the derivative is of the form,

$f$ ‘ $(x)=a(x+1)(x-1)$

Where $a$ is any constant. Note that the $a$ must be included, as all these curves have zeros at $x=\pm 1$ . Expressing this in terms of powers of $x$ gives,

$f$ ‘ $(x)=a(x^{2}-1)$

Now, applying the process of finding the primitive gives,

$f(x)=a(\frac{x^{3}}{3}-x)+c$

Now, to find the values of $a$ and $c$ we simply substitute these into the expression for $f$ .

$f(-1)=\frac{2a}{3}+c=-15$

Hence we have one equation,

$2a+3c=-45 (1)$

Now, doing the same for the other point gives,

$f(1)=-\frac{2a}{3}+c=12$

Hence we have another equation,

$-2a+3c=36 (2)$

Now, adding equation $(1)$ to equation $(2)$ gives,

$6c=-9$

$=>c=-\frac{3}{2}$

Now, substituting this back in gives,

$a=-\frac{81}{4}$

Hence we have that the equation of the cubic is,

$f(x)=-\frac{27x^{3}}{4}+\frac{81x}{4}-\frac{3}{2}$

Mathematics (2 unit) – Geometric Applications of Calculus

The sign of the First Derivative

Example 1

Solution 1

Example 2

Solution 2

Example 3

Solution 3

The sign of the Second Derivative

Example 4

Solution 4

Example 5

Solution 5

Example 6

Solution 6

Sketching curves using Calculus

Example 7

Solution 7

Example 8

Solution 8

Maximisation and Minimisation problems

Example 9

Solution 9

Example 10

Solution 10

Primitive Functions

Example 11

Solution 11

Example 12

Solution 12

Example 13

Solution 13

Dux College Parramatta

Dux College Bondi Junction

Mathematics (2 unit) – Geometric Applications of Calculus

The sign of the First Derivative

Example 1

Solution 1

Example 2

Solution 2

Example 3

Solution 3

The sign of the Second Derivative

Example 4

Solution 4

Example 5

Solution 5

Example 6

Solution 6

Sketching curves using Calculus

Example 7

Solution 7

Example 8

Solution 8

Maximisation and Minimisation problems

Example 9

Solution 9

Example 10

Solution 10

Primitive Functions

Example 11

Solution 11

Example 12

Solution 12

Example 13

Solution 13

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