Name of School:
ROUTINE INFORMATION Hillview Secondary School
Student Surname and Name:
Maphumulo Mthokozisi
Student Number:
207503878
Grade:
10, 11, 12 and Universities.
Subject:
Mathematics
Topic:
Differentiation : the in-depth explanation of a relationship of a tangent to curve approach of a
Content /Concept Area: CAPS page number:
Gradients of the secants and tangents to the curve a circle. ( and rarely limits). 190
Duration of Lesson:
45 minutes
Date
2018 UKZN – Wartburg: - Edgewood (SA).
Specific Aims: To develop understanding of calculus differentiation, slopes and gradient of a curve. To develop understanding of concept of gradient and meaning of change in a slope To integrate understanding of the curve gradient to calculus To set up appropriate formula for dealing with limits. To synthesise relationship of a tangent to curve approach – calculus approach Prepare children for being capable of dealing with cubic function and their expression in other applied maths expressions. Prepare students for future development of graphic function and cubic function and differentiation. Lesson Objectives:
Knowledge
Learners should gain
Skills Learners should be able to do
Value /Attitude Learners should acquire
knowledge and
following :
values and attitudes conducive to :
understanding of the following : Define, determine and analyse
Draw the gradient of a curve
Determine the value of the
gradient and curve gradient.
graph and illustrate the curve
gradient or slope of a curve
gradient definition and their
graph (e.g.: hyperbola or cubic
function broadly.
function).
Determine simple gradient of
Analyse the function of cubic
Acknowledge
quadratic function or cubic
function
the
integrate the curve gradient
function to demonstrate the use
differentiation of a gradient or
knowledge to derivatives of a
of differentiation for gradient
slope now using limits.
tangent and differentiation in
function or slope.
To analyse calculus function
limits.
given by limits and curve
appreciate skills being pro-
graph using differentiation.
active in using derivatives in
within
ability
Therefore
their cubic function
to
student
to solve
problems expressed from the graph,
algebraic
expression
involving quadratic formula, cubic function, gradient of a curve and calculus at large.
Describe
an
equation
of
Differentiate between gradient
Develop ability Draw, label
function involving slope and
function equation and
and
derivatives (cumulus)
general
I
analyse
function-involving
differentiation.
Role plays. Case studies Class discussions. Questioning. Whole class-discussion Group Work Resources: OHP/ Transparency Calculators Chalkboard Newspaper Worksheet Lesson Phases: Introduction:
situation
involving gradient or slope and calculus
Approaches / Teaching Strategies:
a
To kick start this lesson I will bring a football in the class and an interactive chart of a circle. Students must watch as one student kicks the ball up and down. The aim of doing that is to make a sense of out of the mark that touches the ground as soon as student fail to handle the ball on his feet. Then I will use the interactive chart for questioning Students will be questioned to explain what is tangent of a circle. The chalkboard ruler will be used to demonstrate how a tangent of a circle will stand against the ball. The very same ideas will be used to draw a Cartesian Plan in the chalkboard so that the more or less circular shaped objects are now used to demonstrate gradient or slope standing against the curve graph. Then we shall let students to interact with the drawing briefly through observing the various functions which are likely to have line passing close to the graph to act as tangents. Following that a teacher takes a head to introduce a topic of differentiation and calculus. Basically students are likely to know calculus through a graph of a straight line in which a Cartesian plane is used to express gradient. These includes cubic function of a graph with two turning points, hyperbola, parabola, rectangular hyperbola, tan and cot function graphs. There will be no need to stick very much on the function of those graphs but we just highlight them to enable students to view a point of a graph on which there is a potential for a development of tangent.
y = x3-2x
You should have been worried why limits are often uses to find derivatives. This time around students are going to learn how rules used to differentiate develop. We would use the concepts of gradient or average gradient to demonstrate understanding of development and application of differentiation and limits.
In their general knowledge students understand gradient from Cartesian plane where it is defined as a pint of change between height and a distance. In pure applied mathematics we express gradient using G and M. M=
y 2− y 1 x 2−x 1
According to a Cartesian plen drawing on which a traight line has a function of y = mx + c , a gradient (m) is defined by a change in y divided a change in x. Then gradient of is a slope measures the rate at which y changes as x-changes. In a normal straight line you will have to points where line segments meets. In each line segment there is A( x: y ) and B(x: y) from say point A to B. Each dot represents segment , then line A to B is joined to produce a straight line. This indicates the steppness of the line drawn against the y-axis and x-axis in relation to horizontal axis . horizontal line is a line running on the distance direction of surface . The vertical line is elevated ti face the sky and it is perpendicular to horizontal line , so they meet at 900. Students are going to observe how how the gradient is developed in the the curve graph other than straight line. What is secant to the curve : We may define the secant to the curve In relation to how we define the secant into the circle. You should rememeber the trigonometric exercise taught in the past semester whereby grade 10, 11 and 12 were taught to express pie interms of angles . That was a very useful
exercise because it developed an idea that circles behave similar to trigonometric graphs (e.g.: sin , cos , tan and logarithmic graphs. For example circle has a cenral point in which all angles rotate to fom a full cycle of 3600. This gave mathematitian a pride to question and express radian of a circle to give
the
value
of
circle
points
and
π . Circumference of a cirlc is C= 2 π r
angles
mainly
soecial
angles
in
relation
to
. This refers to a perimeter of a circle . The tangent of
a circle :a line that to pas one TANGENT OFgoes CURVE AND ORsingle CIRLEpoint in a circle .
Tangent
Tangent
A In this diagram there is only one tangent: this is line AB in Bold.
B Tangent line has no rule , it can occur else where as long as it touches the line of a curve once and leave . 360◦ = 2π radian, so if you cross multiply this to 1 ◦ you will get 57.30 (how?). 1◦=
2π 360
=
π radiasns ≈ 57.30 180
π radians (or half a circle or semi-circle). You will get that by simple equating
Half a revolution =
that to the entire equation entire revolution, which is to say. (this is to just cross multiply by grouping the same degree factors on one side. Full circle is 3600 = 2 π 0
80 =
π
radian
1800= 1 π
radian.
radian
Wht about 900 ? Hint dive ¼ = 900 .
FACTS :- A
FACTS –B:-
60◦ = 2π radians
2π 360
10 = 65◦
=
= 65 ×
π radiasns 180 π 180
A complete revolution is defined as 360◦ or 2π radians. π stands for the number 3.14159 . . . and you can work with 0 ≈ 57.3 this if you prefer. However in many calculations you will find that you need to work directly with multiples of π.
≈ 57.30
Now things are going to change slightly:
With the introduction of curve secant and radius of a curve:180 10 radian = Degree ≈1.134 radians π When two lines from any two distinct points on any curve are joined, the secant is formed. But this line is not a tangent, 180 1.75 radian =1.75 × which could come at slippery of anyone’s tongue.
π
1.75 radian = 100.268◦ i) ii) iii) iv) v) vi)
45◦ = π /4 radians 60◦ = π/ 3 radians 90◦ = π /2 radians 135◦ = 3π /4 radians 180◦ = π /radians 30◦ = π /6 radians
Once again return to a circle and see that a secant is not a tangent. But in a cubic graph a secant is formed because a tangent does not lie inside of a circle. Likewise in any curve you cannot make a tangent with a secant because a secant mimics a point between two points of a circle lying between two points where a distance joins these points leave a space on which secant or a line must cover. In a sense even semicircles are still secants. In a curve a tangent still remains to be a line touching the point of a curve once, no matter if it joining any point or not.
Development :
Student task – group work :
Given graph of y = x2 , students are required to secant with a point of p( 1: 1) and Q ( 3: 9) i)
Culculate the average gradient between point P and Q, using the co-ordinate
ii) iii) iv)
points given above. What is the name of the line PQ and what is the graph function y = x2. Obsseve the other secant that passed P( 1:1) and F( 2: 4). Since you know how to culculate the average gradient of line PQ (or PF). Have you ever heard of a gradient of parabola , hypebola or cubic graph function ? . There is no such thing , instead we use to apply the Pythagoras thorem indeirectly through culculating the averrage gradient of a straignt line. Whereas the everage of a straight line is fixed , the average gradient of a curve graph such as parabola may vary according to where how a a straight line used to culculate
v)
the gradient stands against the y- axis . Fom the cpnversation above you can deduce that :
The gradient of a curve y = x2, depends upon the average gradient of a straight line drawn against the y-axis but touching its point (s), we mean it must touche the point(s) of a graph. So the average gradient of a straight line MP pq=
Y 2−Y 1 1−9 = = +4. X 2−X 2 1−3
Which means y = 4x+ c , where c = -3 if you subtitute (1:1) to y = mc=c or y = 4x+ c. therefore y = 4x – 3. But our interset in in gradeint and or average gradient of y = mx+ c , m=4 that is used as a gradient of a curve of y = x 2. So this just to illustrate that average gradient of a curve is given by a straight line gradient , and also that it varies depending on how a straight line stands against the y-axis – the vertical line . So the average gradient of a graph given by line PQ is not same as the one given by PF. It is a common error for all educators including professionals for a slip of the tounge to call gradient line of a curve a secant . If a supervisor observe such mistake let him stop the class and make a constructive criticism but not to make a teacher see a mistake explicitly. This is a way of reminding a demonstrator or teacher about a diffeence between seca nt and tangent. ( hint look at both the secant and tangent at the circle before you look it from any curve graph e.g x2 + x2 = r2 vs y = x2 drawn separately. Therefore the gradient for each point PQ or PF is the average gradient of the curve of y = x2. Teacher task : i) ii)
Facilitate, monitor and boost students knowledge Introduce students to limits or differentiation using the very same graph of y = x 2 but this
iii)
time ariund impose the tangent of a curve ( if interssting enough make a cirlce tangent) . Write down ideas that come up during discussion: Eg.: the value of a gradient of a strsight line secant to y = x 2- vary from different line of a graph. E.g: That secant is not a tangent , unless stated .
iv) Introduce a tangent of a curve graph ? But How ? Fisrt a teacher removes off the previous values in the chalboard . We mean to rub off .. then totally out . If the equation of the function of a parabola is y= f(x) = x 2, where x – co-ordinate of P = 1 and y –coordinate of P (x: y) is 1 . This means that we still return P( x: y) = (1: 1) . This time around get rid of points of secant Q or secants but introduce a straight line with points P (1:1) and Q ( x: y) so that P be the secant with a point P(1:1) and Q ( x: y) .
In the diagram below : Given y = f(x) = x 2, x-co-ordinate of P is P( 1: y) while the co-ordinate of y is y to stand in the place of the uknown co-ordinte y. Since the formula of a parabola is y = f(x) = x2. So where x= 1, y = f(x) = x2 = 12 = 1, therefore y = 1. Since you can express co-ordinate of as P ( 1: 1) , so you can also express this as ( i: f (1) , because f(1) still gives you y = 1- which is a y –co-ordinate of a graph of y = x 2. You are fortunate enough that we are using a graph that was primarily tested for you , so you are evaluating this with confidence. Now prodice a a horizontal line from P to run parallel to x-axis and at the same time produce vertical line from Q to run parallel to y-axis to meet a horizontal line from P sothat a new point called M is produced. Hence this forms a new triangle PMQ is poroduced, where two dotted line - the horizontal line PM vs the vertical line QM. In a geogebra with showboard linked toa a computer or just simple plain chalkboard a teacher draw a triangle separate it out of its original form to illstrate points and lines such as Line PM , OS and OT. NB While P ( 1: y) , y = f(x) = x 2 - requoired to forst determine y where x =1 and secondly xpress y – co-ordinate of P( 1: y) interms of f(x). Solution y = f(x) = x2, where y – co-ordinate stand against x-co-ordinate of 1 as in P ( 1: x). So where x = 1 , given a formula of a function of a parabola where both points ar found , y = f(x) = x 2 mediate their values, and this value apply in any form of graph found attached points like this , so substitute x into f(x) = x2 to dtermine the actual y-co-ordinate of points P (1: y). Sos y = f(x) = x2
≡ y = f(1) = 1.
To express co-ordinates of P and Q in terms of point a or 1 , but beginning with 1 using triangle PMQ and , where we let PM = h purposefully. Hint retrieve the y value of P( 1: y) and also retrieve all values of Q( x: y) . in a well known graph of y = x 2. So for P ( x: y ) , we have ( 1: y ) then , (1 : f(1) ) , then ( 1: 1) . your fortunate is that this graph was done before it is used to exmine the application of of the functio nof a parabola f(x) to derive the vaues of the co-ordinate , distances etc. This leads us towars how differentiation , limits and tangent gradient develops.
Simlarly for point Q ( x: y) , even though they both uknown, y = f( x ) = x 2. But whiich y , things are a bit coplicated this time arouund , because you have both co-ordinates P ( x:y) uknown . Solution : To everything you do you may be eminded to realise that you are looking to determin the coordinates of Q ( x: y ) through expressing them in terms of 1 or a but this time around we are looking it through a = 1 and distance PM = unknown where we let PM = h or let unknown distance PM = h purposefully. See a triangle given above. While this has to take place , you need to be careful to students to raise them awareness that :Points of Q (x:y) are given by horizontal line from Origin O ( 0: 0) to T( x: y) and vertical line from Origin to point R. These are vertical line line OT and horizontal line OR. Line OR is parallel to line T Q Line OT // Line PM //RQ. Saying that line AT is parallel to Line PM and parallel to RQ does not guarantee that they are equal . But it means that they keep equal distance apart such that they will never meet. OT is not equal to PM . But OT = RQ because they share equal vertical apace across two vertical lines and also because they are parallel to each other with their relative angular positions likely to meet at 90 degrees . This make them either quare or parallellogram or rectangles. In this case they form rectangle. This is just to illustrate what that the word parallel lines may entais o a serious notice. So whatever we are doing to triangle P MQ we are tryin got manipulate distances in relation to a or a =1 or 1 to bring about the expression of vertical line and horizontal lines of parabola in terns of a or 1, where y = x 2. In this case we are beginning by expressing a =1 , more specifically 1 in terms of the vertical line TQ and horizontal line OT . Then OT //RQ and PM and by these we do not mean they are equal , if so you state a sound reason At the same time TQ //OR but it does not guarantee that they are equal , unless a sound reason is stated in fulll. E,g OTQR is a rectangle . Solutions : to express poits Q ( x: y) in terms of a or a =1 , or 1‌.. in this case we have used I . But if we might change we might recommend one to use a . Why , because if you know that the value of a = 1, it would be easier to substitute the value of 1 from how the values of P( a: y) and (a+ h : y). The expression hiligted is not true to all values of P( x:y) and Q ( x: y ) if you placed a =1 or a or 1 against
the formula of a funtion of a parabola as long as they both fall into the same point of contact to the function , and this may also apply to any given function. To express distancees in terms of a , h or 1 and h or h or 1 , in this case we have used 1 : but you are not limited to take/ test a , as soon as you finish observation below. In point P ( 1: x ) , where x = 1, y = f(x) = x 2, so y = f( 1) = (1) 2 = 1, ao this demonstrates that there can be many ways of expressing y –co-ordinates of P . 1 st it can be P 12). Or P ( a: a 2). In this case we are using [( I : f(1) ¿
¿ , 2nd , P[( a : f (a)], 3rd P (1: ¿
or the actual values following the furethe
substitution P( 1: 12) or ( 1: 1) . In point Q ( x: y ) ,where Q is (1+h:y), then (1+h: y) where x = 1 + h , y = f(x) = x 2, so y = f ( 1+ h) 2= (1+h)2 = 1+ 2h + h2 . Or In point Q ( x: y ) , where Q is (a+h : y), then , where x = a+ h , y= f(x) = x 2, so y = (a+h)2 = a2 + 2ah +h2 , so Q ( x: y ) could be expressed asQ [(a+h : f( a+ h)], or Q ( a+h : a 2 + 2ah +h2) But How does this situation take place: OT // ST Distance :- OT = OS + ST -----But OT = OS + ST, but also OT = 1 + PM , since OS = a = 1 ( In this case we use 1) . So OS = 1 + PM --------- but given PM = h Therefore OS = 1+ h. ( hint you have moved from known to uknown Q ( x: y) to ( 3: 9) where you expect h to take a value of (2) . But keep this in mind . For the purpose of this exercise the values of Q ( x: y) were retrieved or rubbed off. Q (x;y) are coordinate of a graph of a parabola . The relationship between P and Q is mediated by a gradient of line PQ, Theorem of Pythagoras and y = x 2, the equation of a function of a parabola. The solution was to produce line from point P to run horizontally to meet a horizontal line also drawn from Q to M and T . M Is named after the successful meeting of vertical line QT and horizontal line PM . The horizontal line lie parallel to x- axis and vertical line lie parallel to y-axis . Then the gradient of line PQ is given by M pQ =
Y 2– Y 1 move from Q to P X 2−X 1
The key point lie in the Origin . Distance OT = 1+h
( where h is given to line PM) .
We can derive y- co-ordinates of Q ( 1+h) : y) using x –co-ordinates of Q = 1+h to subtitute it into y = f(x) = x2. So for P [ (1+ h) : y )] , this are co-ordinates of point Q and not a function itself , a function is y = x2. How since y = f(x) = x2 F( x+ h) = (1+h)2 = (1+h) (1+ h) = 1( 1+h) + h ( 1+h) . = 1+h+ h+ h2. = 1+ 2h + h2 So the co-ordinates of Q [ 1+h : f(1+h)] or Q [( 1+q) : 1+ 2h +h2)] Cosider that if you have chosen a , you would have Q [( a+ h) : y)]. Your answer is going to match by putting a in position of 1 or 1+h = a+ h . so this would look like Q [( a+h: f( a+h)] = Q [( a+h: ( a+h)2] = [ ( a+h: ( a2+2ah+h2 )] In other words a+ h is the x – co-ordinate of Q( x:y ) such that Q[( x+h ) : y], on which y is uknown . So the presence of y = f(x) = x2 , sothat students subtitute x+h into f(x) since these points belong together . To everyone , whether a , or I or a+ h or I+ h : we have to apply the gradiet of MQP =
MQP =
y 2− y 1 x 2=x 1
move from P to Q or apply M =
Yq−Yp Xq− Xp
=
y 2− y 1 x 2=x 1
Yq−Yp Xq− Xp
Now you any of the points which you have found to be Q ( x: y ) and P ( x : y ) . In this case we have two options of a+h : f( a+h)
and a : f( a)
or [1+h : f( 1 + h)]
and [ 1 : f(1) ] you may express
them fully but on the serious not their abbreviated expression trully arrives to anyform you . MQP =
y 2− y 1 x 2=x 1
since we are orientated to a= 1, so
[ 1 : f(1) ] and P [1+h : f( 1 + h)] ( x 1 : y1)
f (1+h) MQP = ¿−f (1) ¿ 1+h−1
MQP
f (1+h) = ¿−f (1) ¿ h
MQP =
f (a+ h) ¿−f ( a) ¿ h
(x2 : y2 )
-------------- (1)
if you took a path where x =a and x = a+h you would have the ff .
if you take it by inspection
But on the practical because MQP =
y 2− y 1 x 2=x 1
again move from P to Q , where is P
¿ ( a+h) : f( a + h) ] and Q [ a : f(a) ] ¿ xp
MQP =
MQP
Yp−Yp Xp−Xq
yp
xq
yq
move from p to q
f ( a+h )−f (a) a+h−a
=
f ( a+h )−f (a) h
----------------------------------------(2)
NB –crucial steps :- Now evaluate the values of 1 and 2 to the function of y =f(x) = x 2. So for this step we may say we are deriving the method for gradiet of line MQP directly from curve of y = x2 using the fomular of gradient via substiution of x-co-ordinate point into y = f(x) = x 2. Remember that you needed point of co-ordinates P and Q first , which in this case any two points from a graph of a straight line . Secondly you substituted those two points into a graph of a parabola , why because their P [(a: f(a)] and Q [( a+h : f( 1+ h)] Once the two set of Q ( x: y ) and P ( x: y ) emerged you then applied the formula of M PQ =
Yq−Yp Xq− Xp Or MPQ
y 2− y 1 x 2=x 1
on a serious warning that you ar moving from Q to P.
Then the new average gradient formula of line PQ emerged as M=
But this time around it is MPQ =
Yq−Yp Xq− Xp
=
f (1+h) ¿−f (1) ¿ h
Yq−Yp Xq− Xp f (a+ h) or MPQ = ¿−f ( a) ¿
h
If you derive a formula gradient from y= x 2, where one line simple go touches the points of a graph and s traight line is joined to mark their relation ( and rarely the intersection) you will need to express the two points in form of co-ordinate pints even before you apply a gradient formula . We take y = x2 for an example on which a curve graph is used together with co-ordinate poits of its
straight line and its gradient to derive the average gradient of a straight line . The reason being the fact that co-ordinates of a straight line that shares co-ordinate points with the graph of y = f(x) = x2 are unknown. That is where you produce vertical line and horizontal line to meet and produce a triangle with an aim of creating points that must be used to a gradient. A gradient of a triangle cannot be determined independent of x- and y- axis . So you need to check very well if points still start from the origin ( 0: 0) to their terminals where fixed points of ( x:y) and (x: y) are found . This is similar to the use of satellite in search of hidden metal substance with metal instead of digging of a gold without any direction of their golden rule of location , time and space. So , MPQ =
MPQ =
Yq−Yp Xq− Xp
f (1+h) ¿−f (1) ¿ h
=
f (1+h) ¿−f (1) ¿ h
Now apply M =
f (a+ h) or MPQ = ¿−f ( a) ¿
h
f (a+ h) or
MPQ = ¿−f ( a) ¿
h
f (1+h) f (a+ h) M ¿ PQ = ¿−f (1) ∧¿ ¿−f ( a) ¿ h h
So you will get Mpq = =
into y = f(x) = x2
***********
(1+h)2 – (1)2 h (1+h)(1+h) – (1)(1) h
= 1 + 2h + h2 -1 =
(1+h)(1+h) – (1)(1) h
=
(1+2 h+h 2 ) – 1 h
=
1+2 h+h 2 – 1 h
=
2 h+ h2 h
=
h(2+h) h
= 2+h This means that gradient of Mpq = 2+h Gradient does not apply when there is no steep distance for change of rise over run , so h
≠ 0.
Acording to the knoweldge of a book : Mpq = 2+h Remember a graph of y = x2, where points were not retrieved , where Q ( 3: 9) and P ( 1: 1) , with a straight line function of y =f(x) = 4x -3. So the gradient = 4, since y = mx+c .
The boks argue that 3 = h + 2 = 1+ 2 = 3. So for 3 = h+ 2 , h =2 . I do not see a connection between 3= 1+2 , so to conclude that h=2 becaise gradient = 4 and not 3 Instead I could argue , MQP = h+2 , but since Mpq from non retrieved section ( A) is 4.
≠ a so to call it h = 1.
So 4 = h+ 2, so h = 4-2 = 2. Since h
But let see if we apply co-ordinates of Q (x: y) and P(x:y) into a gradient , where ‘a’ is used to express the cordiantes of Q (x: y) and P(x:y s follows . P [(a: f(a)] and Q [( a+h : f( a+ h)] Similary you shall be required to move from Q to P .
Y 2−Y 1 X 2−X 1
MQP =
P [(a: f(a)] and Q [( a+h : f( a+ h)] (a 1 : y1 )
( x2 : y2)
f (a+h) a+ h−a : ¿ ¿−f ( a) ¿ ¿ f (a+h) a+ h−a : ¿ ¿−f (a) ¿ ¿
MQP =
=
(a+ h)( a+h)−axa (a+h−a)
=
MQP = a2 + 2ah+ h2 – a2 /
=
(a+ h)( a+h)−axa (a+h−a)
a+ h−a
= 2ah + h2/ h =
h ( 2 a+ h ) h
MQP = 2a + h In exercise B we know that MQP = 4. This means that MQP 4 = 2a + h 4 = 2a + h, where a = 1( given ) you can make substitution to 4 = 2a + h 4 = 2a + h 4= 2(1) + h 4=2 +h 4-2 = h 2 =h So h = 2 , where h = zero We acknowledge the slite mistake that a book aithor made to call h by a …. We strongly encourage
careful mapping of (x: y ) co-ordinates in future. Please notice that a gradient of point P or a slope of QP is also an average gradient of a TANGENT to a curve ( y = x 2) . This takes point P exclusively but it may use the other points of a seecant for culculation of a the gradient oa a slope. You have to be very careful what is a tangent and what is a secant. So if you were given P ( a: f(a) ) as co-ordinates of P , required to culculate gradient . You will be required to determine the average gradient of a tangent , because the gradient = of a secant as long as they maintain their contact at P . Even so do they touche , the gradient culculation must be an average gradient of a curve . Then the gradient of a curve = gradient of tangent . If you know how to culculate or express the gradient of a curve through M = Y1-Y2 / X2 –X2 --- then you can determine the gradient of the tangent. So for P ( a : f(a) ) , the gradient of a slope of a tangent is given by f( a+h ) – f(a) / h …. And it is said this apply for all valuse of h closer to zero. The next lesson plan , for applications will question the status of average gradient of a tangent which is defined with respect to its gradient of a curve that is traced from secant , where the nature of secants can differ from how the one stands against the y-axis.
Planned Questions; What is a formula of a gradient What is a secant What is a parabola What is a difference between a tanget and a secant What do you think is In your own opinion why do you think the average gradient of a curve , traced through secant and curve points tend to be equal to a tangent to the cuve.
Classwork :
Student’s Textbook’s Applications : Teacher’s Homework : - explanation and applcations of limits and differentiation .
The gradient of a slope of our tangent to a curve ( any curve ) at point P = gradient of a secant PQ. . We say the average gradient PQ gets closer and closer as Q moves closer and closrer to P In the diagram above we have been estimating the gradient or slope of a tangent to a curver at a : f(a) by expanding the points from Q to P . We let distance PM = h. Initially there is no point M but we produce vertical and horizontal lines videlicent QM and PM respectively to meet at M. Where these lines run parallel to their corresponding y-axis and x-axis , we discovered that some sisdes take common values. Then the power of equalizing and subtituting the values to their appropriate values gave us confidence to invent uknown co-ordinates of Q as ( x: y) where [( x+h) : f( a+h) ] . Soon as these valuse are produced or invented we have to apply the gradient formula M PQ =
Then we’ve moved from Q to P with our values so that M PQ =
y 2− y 1 x 2−x 1
f ( a+h )−f (a) a+h−a
=
f ( a+h )−f (a) h This is a gradient or slope of a tangent to a curve at P , where p adopts a role of making its coordinates equals to
¿ P(a : f(a)]. This rule takes any form of a curver to define the value of the ¿
average gradient of a tangent in the point P . So the answer is alwas a gradient or slope of a tangent .
What gives the value of a gradient of a slope of a tangent at point a: f (a) if x = 1 Solution : y = x2, and
P ( x: y )
is
P
[(a : f(a)] is given by M =
f ( a+h) – f ( a) h
=
f (1+h) – f (1) h M=
f (1+h) – f (1) h
But most pupil can choose to proceed with a and introduce the actual value of a at the end So
M=
f (a+h) – f (a) h
Where y = f(x) = x2 M=
f ( a+h) – f (a) h
=( a+h)2- a2 / a+ h –a
M = a2 + 2ah +h2- a2 / h M = 2ah +h2/h = h(2a + h) /h = 2a + h M = 2a + h M = 2(1) + h = 2 + h M = 2+ h but M = 2 , so 2 = 2+ h . In this graph of
f ( a+h) – f ( a) h
we can see that whenever we express the gradient of a tangent
to a curve , h is likely to approach the value of zero ( 0). Teacher illustrate as follows : This draw us back to the gradient and ( or slope ) of a curve at a point (P) or any point at a curve mainly the point of a TANGENT. If P ( a: f(a) ) is a point on the curve , y = f(x ) , we call this the limit . You have seen above that h was likely to approach 0, since 2+ h = 2, h = 2 -2 = 0. Thso this means that the gradient of the curve y = f(x) = x 2, (where ?) at x = 1 is 2. This value becomes a gradient of a tangent as far as the law applies and you can also reflect that from a graph below.
Consider if the gradient of the curve y = f(x) = x 2 at x = 2, using the rule of diffeentiating of (a: f(a) ) we can see that P ( 2: Y ) becomes (2: f(2) ) as if you have ( a : f(a) ). We have seen that the values of h are approacing zero though illusration of a gradient of y = x 2, with y = 2x -1 becoming a tangent of the curve graph of y = x 2, since m = 2(a) + h , , where m = 2 ( hint straight line gradient ) and a = 1 hint x = 1) , so 2 = 2(a) + h , is 2 = 2(1) + h , is 2= 2+h , is 2-2 = h , is 0= h or h = 0. This implies m = 2 a+ h = 2(1) + o , therefore for any gradient of the curve given y = f(x) , where x = a , you should have ( a: f(a) ) you shall be reminded that h approaches 0. This means that :Y = f(x ) in this case take x2 as an exmple. For y = f(x ) , where f(x) = x2, where x = 2 say in point P ( you will be reminded that : a: f(a) becomes ( 2: f(2) . . Now you trully know that h approaches zero (0) . The gradient is not y = f(x ) = x 2, nore it is (a: f(a) or 2: f(2) where x = 2. The gradient is fixed to rely on M =
y 2− y 1 x 2−x 1
.
So gradient derived between curve and secant becomes gradient of a TANGENT This has a strict rule that h approaches 0 So the actual gradient of a tangent to a curve is limited to h approaching 0 and co-ordinates of P defining the gradient in form of a: f(a ) , where a value of x is concerned as a . ( hint P to Q , is P(a : f(a) )andQ (a+h ; f( a+h)) . This is oberserved by producing verical and horizontal parallel lines to their perspective y-axis and x-axis respectively so that their points meet at their right angle and let us observe the rise over run as well as the valuse of a and a+ h , as shown below. The change of distance and run gives rise to
f( a+ h ) – f (a)
f (a+h) – f ( a) a+h−a
=
f ( a+h) – f ( a) h This becomes the fixed value of any gradient of a tangent to a circle needless to explore the operation of points P and Q and y = x2 , as done above. Mathematians then adopt that :
lim h appr 0
f ( a+h) – f ( a) = f(‘a) h
This expression is called the derivative of a . Because most books uses x instead of a , they will argue that this is aderivative of x. The derivative of x is f(‘x ) =
lim h appr 0
f ( a+h) – f ( a) h
The f(‘x ) =
lim h appr 0
f ( a+h) – f ( a) h
= a value of a tangent of a curve in any form of a
curved graph . This is true for all values of f(x ) since if y = x 2, and the value of x =1 ( a: f(a)) becomes (1: f(1) which takes f(‘x ) =
lim h appr 0
If y = f(x) =x2,
then f(‘x ) =
f ( a+h) – f ( a) = h
lim h appr 0
lim h appr 0
f (1+h) – f (1) h
f ( a+h) – f ( a) h
=
lim h appr 0
(1+h)(1+h) – (1)(1) h
Therefore f(‘x ) =
1+2 h+ hxh – 1 lim ¿ = ¿ h appr 0 h
lim 2 h+ hxh h appr 0 h
=
h(2+h) h
lim h appr 0
=
2+
h Therefore f(‘x ) =2+ h = 2+ 0, since h
0.
Therefore the derivative of y = f(x) is y’ = f(‘x) = 2. Similary in a curve y = x2 , where x = 2 takes the value of (a: f(a) ) . You can adopt P ( 2: f(2) ) Y’ = f ( x') =
lim f (a+h)– f (a) = h appr 0 h
lim f (2+ h) – f (2) h appr 0 h
Now apply this to y = x2 . Y’ = f ( x') =
lim h appr 0
Therefore f(‘x )
lim h appr 0
Therefore
=
lim f ( a+h) – f (a) = h appr 0 h 4 +2 h+2 h+hxh – 4 lim ¿ ¿ h appr 0 h
=
=
lim 4 h+ hxh h appr 0 h
4+ h
lim f (2+ h) – f (2) = h appr 0 h = 4+ h = 4+ (0) = 4
lim h appr 0
(2+h)(2+h)– (2)(2) h
4+ h
=
h(4 +h) h
=
Therefore Y’ = f ( x') =
lim f (2+ h) – f (2) h appr 0 h
=4
This means that the gradient of a tannget that to a curve of y = x 2 is Y’ = f ( x') = 4. The very same Y’ = f ( x') = 4 is equal to a gradient of a tangent on a curve graph . Threre is a difference between y = f(x) and y’ = f(x’) y = f(x) is an equation of a curve, where y = x 2 ,
f ( a+h) – f ( a) or y= h Therefore
lim h appr 0
f (x +h) – f ( x) h
but y’ = f(x’) =
lim h appr 0
.
lim f ( a+h)– f (a) = f(‘x) = y’ is a derivative of y = f(x) = x 2. h appr 0 h
The entire process that you have been dealing with is called differenctiation. But the word differentiation begin to apply if you have
lim f (a+h) – f ( a) = f(‘x) = y’, where y h appr 0 h
0.
NB it does not mean that students are going to do all the exercise below, the selection is subject to a teacher.
Assessment:This assessment not only tests but also it examine student’s listening skills and their independent effort towards handling limits, differentiation and understanding of application of differential calculus in general. Section A: Question 1
Given f (x’) =
lim h appr 0
f ( x +h) – f ( x) h
1) If y = 3x2, Determine y’ or f’(x) from first principles 2) Prove that f’(2) = 12 3) What can you conclude about finding a derivative point of f’(x) at a specific point where x=a. 4) Look at your caps book, what are other standards used to express differentiation? lim 5) If the shortest version of a rule of derivative or rather f’ (x’) = y’ = h appr 0 f ( x +h) – f (x) h
is now
dy dx
= nn-1
consider that y = xn. ( this will be discussed
throoughly in the next lesson plan) . But for now , consider that n ∈ R . , this rule entails that n is an interger . It would be tricky to deal with n – rational . This acordingly becomes a major undertaking of n an irrational number. Determine derivative of a) y = x-4, b) y = x3, and c) y = x2 using the shortest method dy = nxn-1 ) , on which y = xn initially. dx 6) A student applied the rule of differentiation from first principles in which y’ = f(x’) He followed method of P (a: f (a)) paying attention that the tangent to y = f(x) is going ( of
to be y’ = f (x’) =
f ( a+h) – f ( a) . h f (x +h)– f ( x) h
This is not different from saying y’ = f (x’) =
- this is what she
called her gradient. a) If y’ = f (x’) = x2, show that a gradient of y’ = f (x’) = 2x, using her system of the '
‘rule from first principles’ on which f ( x )=
lim h→0
f ( x +h) – f (x) h
.
b) If you have point of x=1 at a tangent point , also show that the total gradient of a tangent at curve of y = x2 = 2 Hint apply f (x’) from the answer that you got above and substitute x value from the gradient owed to the application of first principles). c) Hence determine the equation of a tangent to y = x2, provided that the tangent has the gradient of f (x’) = 2, passing through points (1: 1). d) Find the derivative of the following i) f(x ) = 2x-3 , at x = -2, x=0, and x = 4 ii) g(x ) = -3x2, at x = -3, x=0 and x = 3 iii) f(t) = t2 -3 , at t = -1, t = 0 and t = 2 7) The following sums require no application of any rule , solve for the LHS by equating it to RHS lim ¿ x 2−4 a) If x →2 x−2 , ¿
prove that the entire expression = 4
2
b) If
lim ¿ 4 x −9 3 2 x +3 x→ 2 ¿
2
, ,
prove that
lim ¿ 4 x −9 3 2 x +3 x→ 2 ¿
=0
Section B: i)
The equation of the tangent to y = 3x2 – 2x + 1 at x = 1
Prove that the equation of the tangent of the above equation is y = 4x-2. ii)
The equation of the tangent to y = x3 –x2 + 1 at x = 2. Prove that the equation of the tangent of the above equation is y = 8x- 11.
Section C) Given f(x) = -x2 + 2x + 3 a) If x= 2 is a point of tangent to the curve above, needles to draw a graph -----just show that the gradient of a tangent = -2x + 2. (hint use short method to derive make derivative of y’ = f (x’), provided y = -x2 + 2x + 3. b) Now show that the equation of the tangent, the straight line that passes the graph at point on which x = 2 is given by y = -2x + 7. c) Given y = - x2 + 2x + 3 If the gradient of y = f(x) at any point (x: y) = f (x’). Use the above information to:i) Find the values of x at which the gradient of y = f(x) is
ii)
2 and also to Find the values of x at which the gradient of y = f(x) is
A memorandum is available very soon Section A
1) Given f (x’) =
lim h→ 0
f ( x +h) – f (x) h
lim h→ 0
f ( x +h) – f ( x) h
Given: - y= 3x2 ∴ f (x’) =
lim
= h→ 0
f (x +h) – f (x) h
=
lim h→ 0
x +h ¿ ¿ ¿ 2 – 3 x2 3¿ ¿
=
lim h→ 0
3 ( x +h ) ( x+ h )−(3 x 2) h
=
lim h→ 0
3 ( x 2 +2 x h+h 2) −( 3 x 2 ) h
=
lim h→ 0
3 ( x 2 +2 xh+h2 )−(3 x 2) h
=
lim h→ 0
3 x 2 +6 xh+3 h2−(3 x 2) h
=
lim h→ 0
+6 xh+ 3 h2 h
=
lim h→ 0
3 h (2 x +h) h
=
lim h→ 0
3 (2 x +h) 1
∴ f (x’) =
lim h→ 0
=
lim h→ 0
3( 2x+h)
=
lim h→ 0
3( 2x+0)
lim h→ 0
3( 2x)
f ( x +h) – f ( x) h
= 6x
2) f(‘x) = 6x
∴ f (‘2) = 6(2) = 12. 3) The value of a derivative at a point:
To calculate the value of a derivative point of f(x) at a specific point x = a, we can either determine f (x’) =
lim h→ 0
f ( x +h) – f (x) h
and then substitute f(a’) , this is to replace x
by a in f(x’). Others can choose to convert f (x’) =
lim h→ 0
f ( x +h) – f (x) h
into f (a’) =
f ( a+h) – f ( a) h Remember for P and Q, (a: f (a) or ((x: f(x)) 4) Most books prefer y’ = f (x’) =
Or
f (x’) =
lim h→ 0 lim h→ 0
or (a+ h: f(x+h)) or (x+ h: f(x+ h)). f ( x +h) – f ( x) h f ( a+h) – f ( a) h
lim h→ 0
It will always follow that :- of y = xn, is y’ =nxn-1. In most mathematics books y’ = f(x’) is known as
dy . dx Students are requred to know that the shortest method for deriving y = x n is y’ = nxn-1. Usually y’ deffer from y . if y= x2, then y’ = 2x. In this case y’ = f (x’) = derivative of y = x 2. In other words the derivative of y =f(x) = x2 is y’ = f(x’) = 2x2.
5) If the shortest version of a rule of derivative or rather f’ (x’) = y’ = f ( x +h) – f (x) h
is now
dy dx
= nxn-1 , if and only if y = xn.
lim h appr . 0
(Please note that the
expression of the shortened version for that for y = xn will discussed throoughly in the next lesson plan) . But for now , consider that n ∈ R . , this rule entails that n is an interger . It would be tricky to deal with n – rational . This acordingly becomes a major undertaking of n an irrational number.( most of the basic illustration will be demonstrated in the next lesson plan ). Determine derivative of a) y = x-4, b) y = x3, and c) y = x2 using the shortest method ( of dy dx
= nxn-1 ) , on which y = xn initially.
a) y = –x-4 y’=f(x’) =
dy dx
= -x-4 , since ----------(
dy dx
= nxn-1 )***
dy = -4(x)-4-1 = -4x-5 = -4x-5. dx So when ever you are dealing with differentiation and culculus sums of expressions in which you have not been requwted to determine f(x’) from first principles you will dy = nxn-1 )***. dx The above golden rule is used as the shortest method of deriving the value of y = xn kindly save your time and use (for y = xn, from first principles. Similarly For y = x3.
dy dx
y’=f(x’) =
Please not that f(x’) ≠ x3 from earliest convenience
= x3
since y ≠ y nor f ( x )=f ( x ' ) . However educators and professional as well as students often appear equating f(x’) to f(x) in the fashion shown above . Then we must not penelise students for doing so but they must remember that this serves the interest of euating LHS to RHS . In reality f(x’)
≠
f ( x ) , since y f ( x ) is a curve , with any point ( for say point P ) on whichthere is atangent
of a grap h givenby y=¿ f(x) = x3, whose gradient of a tangent is ggiven by y’ = f(x’) = 3x3-1, otherwise expressed as both y’ = f (x’) = dy dx
lim h→ 0
f (x +h)– f ( x) h
and y’ =
= nxn-1.
But we know that y = f(x’) = gradient of curve and y = f(x) = the curge itself. Listening skills are tested in this assessment exam. ( we do that just to make an equation , but keep in mind what is difference between a tangent and its curve expressio). If y’=f(x’) = ∴
dy dx
= x3
If y’=f(x’) =
y’=f(x’) =
dy dx
= x2
, then f(x’) = 3x3-1 = 3x2. dy dx
= 3x2, hint ( y’ =
where y = x3. dy dx
= nxn-1 )***.
Similary b) a derivative of the shoterned version of y = x2, is y’ = f(x’) =
dy dx
= nxn-1= 2x2-1-2x
Therefore ( ∴ ) f(x’) = -2x e) If y’ = f (x’) = x2, show that a gradient of y’ = f (x’) = 2x, using his rule from first principles. 6) a) If y’ = f (x’) = x2, show that a gradient of y’ = f (x’) = 2x, using her system of the ' ‘rule from first principles’ on which f ( x )=
f ( x ' )=
lim h→0
f ( x +h) – f (x) h
lim h→0
f ( x +h) – f (x) h
lim h→ 0
=
h x +¿ (¿ ¿ 2 – x 2) = h
2 lim ( x +h ) ( x+ h )−x h→ 0 h
=
2 2 2 lim ( x +2 xh+ h )−x h→ 0 h
=
lim h→ 0
( x2 +2 xh+h2 ) −x2
=
lim h→ 0
h(2 x +h) +2 xh+h 2 = h h
=
lim h→ 0
h
=
lim h→ 0
x 2 +2 xh+h2−x 2 h
= (2 x +h)
(2 x +h) = 2x + 0
( hint x-> 0, or x – approaches 0) c) f(x’) = 2x given the x-co-ordinate of the tangent to the curve of y = x2 is x = 1 a) If you have point of x=1 at a tangent point, also show that the total gradient of a tangent at curve of y = x2 is equal to 2. Hint is given that f (x’) = derivative of a gradient of a tangent to the curve. ∴ f (x’) = 2x ∴ f (1’) = 2(1) = 2 T h Is implies that in a point where the tangent and the curve share their point of x = 1, there would be a tangent with a gradient of 2. This is a gradient of a straight line that touches the curve of the equation of y = x2 and not the equation of a tangent. If you want an equation of a tangent, firstly you will need to determine the corresponding y – coordinate of a point at which x= 1 as informed above. So to determine the co-ordinate point of a tangent in which x= 1, let y = f(x) = x2. Since the tangent and a straight line that touches a curve of y = x2 has common point x: y, so we can substitute the value of x= 1 in y =f(x) = x2. So x =1, where f(x) = y = x2. f (1) = (1)2 = 1. This means that a point for say P in which a graph of y = x2 and straight line tangent share their common points have the (x: y) co-ordinates of (1:1) or P (1:1). The equation of a tangent of a curve is a straight line Y = mx + c Y= 2x + c ….since f (x’) = 2x or simple m = 2, where f (x’) = 2x, and where x= 1 is a point at the tangent to the curve of y = x2.
y = 2x +c
by calculations points at tangent are (1: 1) (x: y)
1
substitute (1:1) into y = mx+ c.
= 2(1) +c c= 1-2= -1 c = -1 m=
y2 – y1 X2 – X1
This is often trasnlated to
Y2 – Y1 = M (X2 – X1) Y – 1 = 2 (X – 1) Y -1 = 2x-2 Y= 2x – 2+1 Y = 2x -1.
y2 – y1 X2 – X1
=m
substitute (1: 1)