It’s the…
• You remember this! It’s just:
•
 f ( x  h)  f ( x ) 
|
f ( x) 


•
h

h 0 
• Seems simple? It is!! Just remember that you have
to factor out “h” from the numerator so that it
cancels with the denominator, and you’re set.
lim
– Unfortunately, teachers/exam writers know it’s easy so
they make this factoring process really really (really!)hard.
Try an Example
f ( x)  x 2  1
Start with the function
2
2

((
x

h
)

1)

(
x
 1) 
|
f ( x)  lim 

h
h 0 

2
2
2

x

2
hx

h

1

x
 1
|
f ( x)  lim 

h
h 0 

Input it into the rule
2


2
hx

h
|
f ( x)  lim 

h
h 0 

 h(2 x  h) 
|
f ( x)  lim 

h

h 0 
Now you have to factor out
h…
f | ( x)  lim 2 x  h
And apply the limit!
h 0
f | ( x)  2 x
TADAAAA!!
Expand the factorial
Remember: h cannot =0 but
it gets SO CLOSE we just
can ignore it!
Need MORE PraCTiCE?
• Don’t be fooled – this powerpoint makes first
principles simple and, while the concept is, the
test questions can really do test your factoring
abilities.
• Pointers:
– You may be left with “h”s all over after factoring it out
– as long as none are denominators, you should be
ok!
• Practice:HL Math Book page 316 (example)
– Questions on page 321… try with first principles!
Power Rule, Chain Rule, &
Implicit Differentiation
Peter Tu
IB Math HL (C)
Power Rule
• The power rule comes
 f  x   x  f  x 
f '  x   lim 

x


into effect for
 a  x   x   ax 
f '  x   lim 

differentiation for
x


equations who are in the  x   x   x   1n  x  x   n2  x  x    n3  x  x     nn  x  x  ,
 

n
n
n
n
form, a being a constant,
  x    x  x    x  x     x  x      x  x   x  
1
 2
 3
n

 f '  x   lim  a 



x the variable, and n also
x
 

 
 
a constant:
 

n
n
f  x   ax n
 x 0
n
n
 x 0
n
n 1
n
n
n2
n 1
n 3
2
n2
n 3
2
nn
3
nn
3
 x 0
f '  x   lim  a  nx n 1    x n  2  x     x n 3  x  
 x 0
 2
 3
 
1
f  x   ax n
• From the method of first
principles, we find that:
2
  x 
n 1

 
f '  x   anx n 1
• Hence, the power rule
states:
f '  x   anxn1
n
n
n
Example
• Since it is possible to actually split up
different terms during differentiation,
• We can use this to differentiate the equation:
f  x 
5 4 3 3 8 2 2 1 1
x  x  x  x 
4
2
7
3
7
• In the following fashion:
5 4 3 3 8 2 2 1 1
x  x  x  x 
4
2
7
3
7
5

3

8

2

1

f '  x    4   x 41    3  x 31    2   x 21   1  x11    0   x 01 
4

2

7

3

7

9
16
2
f '  x   5 x3  x 2  x1  x 0  0 x 1
2
7
3
9
16
2
f '  x   5 x3  x 2  x 
2
7
3
f  x 
Chain Rule
• What should we do if we have to differentiate a
function respective to another function?
• Given x, y, and z, can we differentiate x relative
to z indirectly?
• If we write the differential as:
f ' x 
x
y
f ' y 
y
z
• Since a differential is defined as the small
change of one over the small change of the
other, we can apply a multiplication rule of
fractions for the chain rule:
x y x


y z z
Example
• The factory produces cups modeled by
function:
c  5t  2t
2
c
 10t  2
t
• These cups in turn generate revenue
modeled by the function:
r  2c  5c
2
r
 4c  5
c
• What is the rate of revenue generated?
c r r
 
t c t
r
 10t  2  4c  5   40tc  8c  50t  10
t
Further exploration
• Note, that the chain rule has no limits on
how many elements it can have, since it is
really based on multiplication of fractions,
so it is possible to have:
 a b c  d e  f  g  h i  j  k  a
 




   

b c  d e  f  g  h i  j  k l l
Implicit Differentiation
• What happens when we try to differentiate
a relationship between two variables that
are no given as explicit functions?
• Take:
y 2  2 x 2 y  3x  3xy
• Implicit functions are dealt with the same
way, by differentiating each term with
respect to one variable, in this case x:




2
2
y 
2x y  
 3x    3xy 


x
x
x
x
• Then, by using the chain rule and product
rules, we can simplify the equation:

y  2y

 y

y2  
  2x
 4 xy   3   3x
 3y 

y
x 
x

 x

y  2y

 y

2y
  2x
 4 xy   3   3x
 3y 
x 
x

 x

y
y
y
2y
 2 x2
 4 xy  3  3 x
 3y
x
x
x
• Then we move the terms with the
differential of y in respect to x to one side,
and factor:
2y
y
y
y
 2 x2
 3x
 3  3 y  4 xy
x
x
x
y
2 y  2 x 2  3 x   3  3 y  4 xy

x
 y 3  3 y  4 xy

 x 2 y  2 x 2  3x
Further Example
• Consider a simpler case:
 2x  13 y  5  7
• The concept is the same:
 2 x  1 3 y  5   7
3 y  6 xy  10 x  7




 3 y    6 xy   10 x    7 
x
x
x
x

y  y
3
y
    6 x  6 y   10  0
y
x  x

y
y
 6x
 6 y  10  0
x
x
y
y
3
 6x
 10  6 y
x
x
y
 3  6 x   10  6 y
x
 y 10  6 y

 x 3  6x
3
Product Rule
Function
Derivative
y  uv
dy du
dv

v u
dx dx
dx
y  f ( x)  g ( x)
dy
 f '( x)  g ( x)  f ( x)  g '( x)
dx
Differentiate the following:
1
log e ( x)
x
x 2 sin( x)
2
Let y = x sin( x) so that
u = x² and v = sin (x)
This means that du
dx
and
= 2x
This time set up a table:
T
h
function
derivative
i
du
1
s
1
u
dv
= cos (x).
dx
Using the product rule we have that
dy du
dv

v u
dx dx
dx
 2 x  sin( x)  x 2  cos( x)
 2 x sin( x)  x cos( x)
2
x
v  loge x
dx

x2
dv 1

dx x
Adding, we have:
dy
1
1 1
  2  log e x  
dx
x
x x

1
(1  log e x)
x2
Quotient Rule
Function
u
y
v
f ( x)
y
g ( x)
Derivative
du
dv
v u
dy dx
dx

dx
v2
dy f '( x)  g ( x)  f ( x)  g '( x)

2
dx
[ g ( x)]
Differentiate the following:
2
x 1
sin( x)
Let u = x² + 1 and v = sin (x)
This gives the following derivatives,
du
and
 2x
dx
dv
 cos( x)
dx
Using the quotient rule we have:
du
dv
v u
dy dx
dx

dx
v2
2 x  sin( x)  ( x 2  1)  cos( x)

[sin( x)]2
2 x sin( x)  ( x 2  1)cos( x)

sin 2 ( x)
Differentiation of Trig. Functions
Examples
DIFFERENTIATION
Log and Euler Functions
Differentiation of Log
Functions
1
f ( x)  ln x f '( x ) 
x
NOTE! The derivative of the natural
log of any value nx, example: ln 5x,
is equal to 1 over x!
y  5ln x
5
y'
x
g ( x)  4 ln 2 x
1
g '( x)  4(2 )
2x
4
g '( x) 
x
g ( x)  log a x
1
g '( x) 
x ln a
y  5log 2 6 x
1
y' 5
6
6 x ln 2
5
y'
x ln 2
Differentiation of Euler
Functions
The derivative of
ex
is
ex .
f ( x)  2 xe 2 x
f '( x)  2 x(2e 2 x )  2e 2 x
f '( x)  4e 2 x  2e 2 x
Jessie Chiang
Block C
Tangents and Normals
Finding Tangents: (the line that
touches a point on a function and
shares the slope of the point on the
function)
1.
FIND THE SLOPE OF THE
TANGENT (slope of tangent
at (x,y) will be f ’(x) at x)
2.
PLUG THE SLOPE (m)
INTO LINEAR EQUATION
y= mx+b
3.
PLUG ORIGINAL POINTS
(x,y) INTO LINEAR
EQUATION TO FIND b
4.
DONE!! 
Tangents and Normals
•
Finding Normals: (the perpendicular line to a point on a function)
1. FIND THE SLOPE OF THE TANGENT (slope of tangent at (x,y)
will be f ’(x) at x)
2. REMEMBER THAT LIKE ALL PERPENDICULAR LINES, THE
SLOPE OF THE NORMAL X THE SLOPE OF THE TANGENT
= -1.
3. USE SLOPE OF TANGENT TO GET SLOPE OF
PERPENDICULAR
4. PLUG THE SLOPE (m) INTO LINEAR EQUATION y= mx+b
5. PLUG ORIGINAL POINTS (x,y) INTO LINEAR EQUATION TO
FIND b
6. DONE!! 
Let’s try a problem!!
Given: y(x) = x^2
Let's find the tangent at x = 2. So what point is that? We know the x-coordinate of
the point. To find the y-coordinate, simply use the equation of the curve!
Step 1: Find the derivative of the curve. The equation of the curve is given in
4.5-2. You know how to take its derivative. It's
y'(x) = 2x
Step 2: Evaluate the derivative at the point given. The problem says do it at x
= 2. The derivative at that x is y'(2) = 4. That is your m. Write it down. m = 4.
Step 3: Solve for b. Remember, y = mx + b. We know what x is, what y is, and
what m is. If you plug them all in, you get
5 = 4×2 + b a And it's trivial algebra to go from there to b = -3
Step 4: Write the equation. You know m and b now. Simply substitute them
into y = mx + b. You get
y = 4x - 3 That's it. We're done. Once you know how to take the derivative, these
problems are easy. Just follow the four steps here.
Figure 4.5-1 shows a graph of this problem. Observe how the tangent line just
kisses the curve at (2, 5). The angle they meet at is, in fact, zero. Plotting the
curve and the tangent line is one way you can check your work to see if your
answer is right.
http://www.karlscalculus.org/calc4_5.html
Another problem!
Let's take our example from before where the curve is
y = x^2 + 1 and we want the normal line to pass through (2, 5).
Step 1 is the same as before. It is still the case that y' = 2x. In
step 2, though, after we evaluate y'(x) at x = 2 and find that it's
equal to 4, instead of setting m to that, we set m to its negative
reciprocal, which is -1/4.
The remaining steps are the same. Solve for b with the x and y
coordinates you have and the m you have just determined. In
this case we have:
5 = (-1/4)×2 + b and solving for b, we have b = 5.5 So the answer
here is y = (-1/4)x + 5.5
Figure 4.5-2 shows a plot of our curve and the line that is normal to
it at x = 2. Observe that the normal line crosses our curve in
two places. At (2, 5) is the crossing at right angles that we
expected to get. To the left there is another crossing at a point
we have not determined yet.
http://www.karlscalculus.org/calc4_5.html
Finding the turning point of a function.
The derivative of f(x) gives
the slope of f(x) at a point,
where x is the x-coordinate
of that point.
At the turning point of f(x),
slope=0.
A turning point can be a
local maximum or a local
minimum.
To find the maximum or
minimum, differentiate
f(x). Where f ’(x)=0, x
gives the x-value of the
turning point of f(x).
Vanessa Carnegie (C block)
There are two ways to determine whether a turning-point is a local
maximum or a local minimum.
1. Determine the slope on either side of the turning
point. On a graph of f ’(x), the turning is where the
derivative crosses the x-axis. If f ’(x) is negative to the
left of the turning point and is positive to the right, it is a
local minimum.
f ’(x) of a
local
minimum
If f ’(x) is positive to the left of the turning
point and is negative to right, it is a local
maximum.
f ’(x) of
local
maximum
2. Using the second derivative of f(x). The second
derivative is the slope of the first derivative. If the
second derivative of f(x) at its turning point is
positive, it is a local minimum. If f ”(x) < 0, it is a
local maximum.
And now for an example…
An open rectangular box with square base is to be made from 48 ft.2 of material.
What dimensions will result in a box with the largest possible volume ?
Find an equation for the surface area of the box.
2
4 xy  x  48
Rearrange this to solve for y.
48  x 2
y
4x
Now, find an equation for the volume of the box.
V  width  length  height  x 2 y
Replace y in so that x is the only variable.
2
3
48

x
x
V  x2 y  x2 
 12 x 
4x
4
To find the maximum volume for
the box, differentiate V and let it
equal zero. Solve for x.
dV
3
 12  x 2
dx
4
To verify our answer, find the
second derivative for the volume.
d 2V
3


x
2
dx
2
At x=4,
d 2V
3


(4)
dx 2
2
dV
0
dx
0
3
3
(16  x 2 )  ( x  4)( x  4)
4
4
x  4, 4
x4
2
dV
0
2
dx
This confirms that
there is a local
maximum at x=4.
(x cannot be a negative value
because it represents a length.)
Now we can find y.
48  x 2 48  (4)2
y

2
4x
4(4)
The dimensions that will result
in a box with the largest
possible volume are:
4 ft  4 ft  2 ft  32 ft 3
Example taken from
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/maxmindirectory/MaxMin.html
(full of applied maximum/minimum problems)
Maximum and
Minimum Word
Qu3stions
Jason L3i
Max / Minimum Word Questions
• Application of calculus to find maximums
or minimums
• Example
– The points PQR form the corner of a house,
where angle PQR is a right angle. Running
parallel to these walls is a garden patch. 20
meters of fencing is available to create an
enclosure PUTSRQ, in such a way that PU =
RS = x. Assume PQ = QR = y.
Example (continued):
How should the fencing be placed so that
area PQRSTU is a maximum?
We know total fencing = 20 m
2(y+x)+2x=20
2y+4x=20
1. 2y = 20 – 4x
Area = 2(x*y) + x^2
= x^2 + 2xy Sub 1. Into this equation
= x^2 + 20x - 4x^2
= 20x – 3x^2
Example(continued)
We know now A = 20x – 30x^2
Using the first derivative we can find the x value that will give a
maximum Area
f’(x) = 20 – 60x
60x = 20
x=1/3
Let f’(x) = 0
Still we need to check if this is a maximum, minimum or point of
inflexion
For x values lower than 1/3, f’(x) is +
For x values higher than 1/3, f’(x) is –
Therefore it is a maximum point!!!
What are they?
Points of inflection are points on a
graph where the concavity changes!
This means that the derivative of this graph
will have a a minimum at the x where the
point of inflection occurs, because points of
inflection create maxs or mins unless it is a
flat P.I.
The flat P.I. is where the point is an xintercept on the graph of the derivative.
Example: Find the points of inflection of the graph y = 5 x 2  2 x 3
Solution: In order to find the points of inflection of this graph you need to
differentiate the function twice, because maxs and mins always become zero
points on the differentiated graph.
 10 x  6 x
dy
dx
d2y
dx
2
2
 10  12 x
From this point we set the
value to zero to find x
-10=12x
-5/6=x
-.833=x
X is the point of inflection, the
point where concavity
changes!!!
AND THAT IS THAT
Related
Rates of
Change!
Related Rates of Change
What’s this all about?
• We use related rates to relate the rate of change of
one variable with another
• Everybody could have their own method of doing
these that they are comfortable with!
• Generally, start with this when you get a question:
– Given: Write what information you have.
– Want: What are you looking to find?
– When: When!
• You need to know your formulae for shapes!
– Ex// Volumes and surface areas of spheres, cones,
cylinders
• Try to visualize each problem!
Let’s do a problem together!
Get a pencil and pen!
• Air is leaking from a spherical balloon at a rate of 5 cm3 /min. Fin the rate
at which the surface area of the balloon is changing when the radius is 10
4 3
cm.
A  4 r 2
• Given:
• Want:
• When:
dV
cm3
 5
dt
min
dA
dt
r  10
V  r
3
dV
Good Job!
 4 r
2
dr
dr
1

dV 4 r 2
dr
dr dV
 ( )( )
dt
dV dt
dr
5

dt 4 r 2
dA
 8 r
dr
dA
dA dr
 ( )( )
dt
dr dt
5
(
)(8 r )
4 r 2
dA 10

dt
r
r  10
dA 10

dt
10
dA
cm 2
 1
dt
s
Integration of
LOG and EULER
functions
By: Cassandra Wee
Block G
So…we already know how to do basic integration.
Now, we’ll learn how to do it with log and Euler
functions! =D
For log functions, you
want to change the
integral into the following
form:

f ' ( x)
 dx
f ( x)
For Euler functions, you
want to change the integral
into the following form:

f ' ( x)  e
f ( x)
 dx
Let’s try to integrate with a log function…

2x

dx
2
2x  2
REMEMBER:
our goal is to get this

f ' ( x)
f ( x)
- let the denominator be f ( x)
- then f '( x) is 4 x
4x
 dx
2
2x  2
- BUT...because we changed the numerator from 2 x to 4 x, we need to multiply
1
the whole integral by :
2
1
4x
- this is what it should look like :  2
2
2 2x  2
d (ln( f ( x)) f '( x)
- because

, the answer is
dx
f ( x)
- the integral now looks like this : 
1
ln(2 x  2)  c
2
Now let’s try a Euler function…
e

2 x2  2
 dx
REMEMBER:
our goal is to get this
 f ' ( x)  e
f ( x)
- let 2 x 2  2 be f ( x)
- so f '( x) is 4 x
- the integral looks like this :  4 x  e
2 x2  2
 dx
- BUT...because we multiplied it by 4 x, we also need to multiply it by
1
to
4x
cancel it
- the integral now looks like this :
d (e f ( x ) )
- since
 f '( x)  e f ( x ) ,
dx
the answer to this problem is
1
2 x2  2
4
x

e
 dx (this is in the form we want)

4x
1 2 x2  2
e
c
4x
Integrating Trig
Functions
how exciting…
Elyn Tan
Integrating trig functions can be
tricky…
More specifically, let’s look at the cosine double angle rule…
cos 2
This tricky little thing can equate to other tricky little things that
can eventually help you solve complex trig functions!!
cos 2  cos   sin 
2
2
 2 cos   1
2
 1  2sin 
2
Example
This is a common type of situation where you would
need to use the double angle rule…
2
cos
 x.dx
Then, looking at the previous slide, choose the
appropriate rule to use. In this case, we should use
cos 2  2cos 2   1
1  cos 2
2
cos  
2
And so using that rule we can
now integrate the trig function…
1  cos 2 x
2
 cos x   2
1
  1  cos 2 x.dx
2
1
 ( x   cos 2 x.dx )
2
Here we must use the substitution
rule
Let
u  2x
du
2
dx
1
dx  du
2
 cos 2 x.dx
1
  cos u   .du
2
1
 (sin 2 x)  C
2
And so…
1
 ( x   cos 2 x.dx)
2
1
1
 ( x  sin 2 x)  C
2
2
Another example
 cos x.dx
3
Looks hard? No worries…
First begin with this:
3
cos
 x.dx
  cos x cos 2 x.dx
  cos x(1  sin 2 x).dx
  cos x  cos x sin 2 x.dx
 sin x   cos x sin x sin x.dx
Again, we must use the
substitution rule
du
 cos x
Let u  sin x
dx
du
dx 
cos x
 sin x   cos x(u 2 ).
 sin x   u 2 .du
1 3 
 sin x   u   C
3 
1 3
 sin x  sin x  C
3
du
cos x
Differentiation and Integration of Inverse
Trigonometric Functions
Differentiation of inverse sine:
x  tan y  y  arctan x.
sin 2 y  cos 2 y  1,
Differentiate both sides with respect to x
1  cos y 
Differentiation of inverse tangent:
dy
dy
1


.
dx cos y
dx
Differentiate both sides with respect to x
1  sec2 y 
dy
dy
1


.
2
dx
dx sec y
But, from the identity: sin y  cos y  1,
But, from the identity: 1  tan y  sec y,
We have:
We have:
2
cos y  1  sin 2 y  1  x 2 ,

d
1
.
 arcsin x  
2
dx
1 x
2
2
dy
1
1


,
dx 1  tan 2 y 1  x 2

d
1
.
 arctan x  
dx
1  x2
2
Differentiation and Integration of Inverse
Trigonometric Functions
Integration to inverse sine:
Integration to inverse tangent:
d
1
.
 arcsin x  
2
dx
1 x
Express in terms of differentials:
d  arcsin x  
1
1  x2
dx.
Integrate both sides:

1
1  x2
dx  arcsin x  C.
d
1
.
 arctan x  
dx
1  x2
Express in terms of differentials:
d  arctan x  
1
dx.
1  x2
Integrate both sides:

1
dx  arctan x  C.
2
1 x
Differentiation and Integration of Inverse
Trigonometric Functions
Example 1:
Example 2:
1
1
dx

 16  x2
 42  x2 dx
d
arcsin  ln x  
dx
By the chain rule:
Knowing that:
d
1
1
arcsin  ln x  

2
dx
1   ln x  x

1
x 1   ln x 
2
.
1
x
dx

arctan
   C,
 a2  x2
a
We find that:
1
 x
dx

arctan
   C.
 42  x 2
4
Integration by Substitution
Normal…
Integration by Substitution is
pretty self-explanatory.
Basically, we let something
in the original function equal
to something new to make
the integration easier or
work.
Consider:
f ( x)   sin  cos  d
First step, we would look for any
derivatives of a term in the integral
function.
We can spot two. sin ' s derivative is
cos ' s derivative is  sin  .
cos
.
Then, we choose a term in the integral to be
substituted by another easier variable, say u.
In this particular function, we can substitute
either term in the integral ( sin and cos) 
I will substitute sin for u.
Therefore, u  sin 
f ( x)   sin  cos  d And… substitute to get
f ( x)   u cos  d
REMINDER: IF INTEGRAL IS
DEFINITE, THEN DON’T
FORGET TO CHANGE THE
BOUNDARIES WITH THE
SUBSTITUTED VALUE!!!!!
Since
we
know
derivative sin  cos .
Therefore, u  sin 
du
 cos 
d
Rearrange it and we get
f ( x)   u cos  d
du  cos .d
Look at the original equation, there is also
cos .d
We can substitute it for du, forming the equation…
f ( x)   u.du
Which is much easier than it was at first.
Now… You try a question.

3
f ( x)   tan  sec 2  d

4

d  sec  
 sec  tan 
d
3
f ( x)   tan  sec 2  d

d  tan  
 sec2 
d
4
Substitute:
1.732
f ( x) 

u sec2  d
Let u=tan
1
1.732
 udu
 u2 
f ( x)  

2
 
du
 sec 2 
d
du  sec2  d
f ( x) 
1
1.732
Change boundaries:
Lower boundary:

u  tan
1
1.732 1 
f ( x)  
 
2
 2
2
f ( x)  0.999912  1
2
4
1
Upper boundary:

u  tan
3
 1.732
Unusual and Special
Substitutions
Julian Springer
Integration by Substitution
Integration by substitution is used to make
the integration process easier. Usually,
what needs to be substituted is easy to
see, but there are cases where it is not
obvious what part of the function needs to
be substituted to make the integration
easier to solve.
Examples


There is a way to solve
2
these problems, but it is
x 9
dx not immediately obvious. In
x
the next slide, the
substitution will be shown.
2
9x  16
dx
3
x
2
27

4
x
dx

The Substitution
Function Contains: Substitute:
a b x
2
2
a
x  sin 
b
2
Ex: 16  9x 2
a b x
2


2
2
Ex: 16  9x
b x a
2
2

2
2
Ex: 9x 2 16



x
4
sin 
3
x
a
tan 
b
x
4
tan 
3
a
x  sec 
b
x
4
sec 
3
Examples Solved
Here, the first example is solved:
This is the substitution:
 x 2  9 
  x dx
x  3sec 
dx
 3sec  tan 
d
dx  3sec  tan  d
 9sec 2   9 
  
3sec  tan  d
3sec



 9tan  tan  d

2
 3 tan 2  d

 sec d   1d 
3
2
 3tan   3  C

x 2  9  3sec 1
x 
C
3 
x 9
2
3

x

The final step is done using triangles, and finding

what angle theta holds.
Integration by Parts
Lena Shen
So what is Integration by Parts?
I
• Used when the integral contains the product of
two functions of x
• We know from product rule that:
d (  )
     
dx
• Therefore:



.
dx


.
dx




So what is Integration by Parts?II
• Changing the sides, we get:
  .dx    .dx
• WHICH IS OUR RULE~
（＊＾Ｕ＾）人（≧Ｖ≦＊）/
• Written in rule book as:
d
d
  dx dx     dx dx
Sample Problem!
• Integrate  ln x 
2
  ln x  dx
2
let
  ln x    ln x
 
1
x
  x ln x  x
1
  x ln x  x  ln x      x ln x  x  .dx
 x
  x ln x  x  ln x   ln x  1.dx
  x ln x  x  ln x   x ln x  x  x   c
 x  ln x   x ln x   x ln x  2 x   c
2
 x  ln x   2 x ln x  2 x  c
2
 ln x.dx
let
  ln x    1
 
1
x
 x
1
 x ln x      x  .dx
 x
 x ln x  x  c
INTEGRATION BY
PARTS (double)
Yung Lam Ho
Integration by parts (double) is
when you have to integrate by
parts TWICE!
Example:
e
x
sin x.dx
PART 1:
Let u=sinx
u  cos x
v  e x
v  ex


uv
dx

uv

u

 v.dx
x
x
x
e
sin
x
.
dx

sin
xe

cos
xe
.dx


Now we need to use integration by parts one more
time to integrate cosxex!
PART 2:
Now let's integrate:
x
cosxe
.dx

v  e x
v  ex
Let u= cosx
u= -sinx
From the previous slide, we already found that:
x
x
x
e
sin
x
.
dx

sin
xe

cos
xe
.dx


Now, we can integrate cosxe x directly into the above equation

x
x
x
x
e
sin
x
.
dx

sin
xe

e
cos
x


sin
xe
.dx


x
x
x
x
e
sin
x
.
dx

sin
xe

e
cos
x

sin
xe
.dx


2 sin xe x .dx  sin xe x  cos xe x
ANSWER:
x
e
x
sin
xe
.dx   sin x  cos x 

2

Differential Equations:
•
•
•
•
Used to solve problems in the form: dy
2e x

dx
y
Strategy to solve such problems:
Bring all the y-terms and x-terms such that: f ( y ) dy  g ( x)
dx
Example:
dy
2e x

dx
y
dy
y
  2e x
dx
dy
x
 y dx .dx   2e .dx
 y.dy  2e
x
c
The dx cancels here so left side becomes
only in terms of y and the right side
only in terms of x
y2
  2e x  c
2
y  4e x  2c
Note: do NOT forget the constant c and simply
add a “+ c” at the end of the expression
because this would be WRONG!
I.e.
y  4e x  c
Would be wrong.
Another example! Step by step:
2
dy
 3 x 2 ( y  1)
dx
2 dy
 3x 2
( y  1) dx
2
dy
Then integrate both sides with respect to x.
 ( y  1) dx .dx   3x .dx
2
2ln y  1  x3  c
x3
2
y 1  e e
x3
2
c
2
y  Ae  1
First bring the expression with y on the left side.
Finish the integration.
Re-arrange the equation to make it in terms of y
(don’t forget the constant c because this is an
indefinite integral)
Where A  e
c
2
which is a constant.
Last example this time with definite integral
equations!
•
•
dy
Consider the following expression:  2e  y ( x  2)
= 1, find the definite expression fordxy.
The working out to solve this is shown below:
and given that for y = 0, x
dy
 2e  y ( x  2)
dx
1 y dy
e
 x2
2 dx
1 y dy
e
.dx   ( x  2).dx
2  dx
1 y x2
e   2x  c
2
2
y
2
e  x  4 x  2c
y  ln x 2  4 x  2c
•
Now, at this point, sub in the values given above and solve for c:
0  ln 12  4(1)  2c
e0  5  2c
2c  4
c  2
therefore the y expression is:
y  ln x 2  4 x  4
Finding Area bound by x-axis
Indefinite integral:
Definite integral:
2
(
x
  3)dx

2
1
( x  3)dx
2
The graph shows the function f(x) = x2+3. The
shaded area represents the definite integral

2
1
( x 2  3)dx
The procedure for
calculating area
under x-axis:
(1) Integrate the
function, omitting
the constant of
integration
(2) Put the result in
square brackets
with the limits
outside
(3) Substitute the
limits into the
integrated function
(upper limit first)
and subtract the
two numbers
Let’s try an example!
Find the area bound by the curve
y=12x2(1-x) and the x-axis
(1)find where the graph cuts the xaxis by graphing
We see that the limits are from 0
to 1 (zeros of the curve)
(2) Integrate. Do not forget to
multiply out before integrating.
1
 12 x
0
2
 12 x 3dx
1
1
0
0
  12 x 2 dx   12 x 3dx
1
12 x  12 x 

 

 3   4 
3
0
4
1
(3) Substitute limits in
0
12(1)3 12(0)3
12(1)1 12(0)1
(

)(

)
3
3
4
4
 4 3 1
Another example!
Find the area bound by f(x) = (x2-1)(x-3) = x3-3x2-x+3 and the
x-axis. Graph it!
We can see from the graph that the limits are from -1 to 1, and 1
to 3.
We must be careful here. There is a negative area. You cannot
integrate separately and add the two areas. You must
subtract the negative area from the positive because we want
a positive area!

1
1
3
( x  3x  x  3)dx   (x 3  3x 2  x  3)dx
3
2
1
 x 4 3x 3 x 2



  3x 
4 3 2

 4  (4)
 8units
1
1
 x 4 3x3 x 2



  3x 
4 3 2

3
1
Area bound by y axis
1
Ex/ Find the exact value of the area enclosed by the curve with equation: y  1  ,
x
the y-axis and the lines y=4/3 and y=2.
First, sketch the graph of the function.
The working is the same, except that we
must first write the function in terms of
1
y.
y  1
x
1
y 1 
x
yx  x  1
x( y  1)  1
1
x
y 1
Now we can work out the integral.
2

4
3
2
1
dy   ln( y  1) 
4
y 1
3
  4 
  ln(2  1)    ln   1 
  3 
1
 ln1  ln
3
 ln 3
Jenny Wong
Area Between Two Curves
Boundaries
• First, the intersection points between the two
graphs needs to be found. To do this, set the
two functions to be equal to each other and
then solve for x.
• These intersection points are the boundaries
for the integration. Generally there will only
be 2 intersection points, but if there are more,
consider the functions to be in different
sections. For example, if there are
intersection points at
x = -2, 2 and 4, then integrate between -2 and
2 and then 2 and 4.
What next?
• Before you integrate the functions, first find
which function is will have a greater area
beneath it. To do this, you can simply graph
the two functions and see which is further
from the axis which you will be integrating
from (this will almost always be the x-axis)
Finally integrate
• The area bound by the two curves is the
difference in the areas beneath each of the
curves. This can be done by integrating the
difference of the curves. Eg.
or
f ( y )  g ( y ).dy

dx
 f ( x)  gof( x).how
• For good examples
to do it, go to
http://archives.math.utk.edu/visual.calculus/5/a
rea2curves.3/index.html
Volume Around X Axis
By: Patrick Bai
Volume Around the X Axis
• Consider the region between the graph of a
y  fand
( x) the x axis
continuous function
from where x = a to x = b
• Take for example where
f ( x ) andx
1 x  4
2-D image
Revolved around the x axis
Volume Around X Axis
b
• Volume =

• V=
4
1
• V=
2

y
 .dx
where
y  f ( x)
a
 ( x )2 .dx

4
1
x.dx
4

x
• V=

2 1
2
15
• V= 2 
Taken from:
http://curvebank.calstatela.edu/volrev/volrev.htm
Volume Around Y Axis
By: Michael Chang
Volume Around the X Axis
• Consider the region between the graph of a
continuous function y  f ( x)
• To find the volume of revolution around the y
axis, express the equation in terms of x
using y. x  f ( y )
• Find the limits in terms of y; y = a to y = b
• Take for example where f ( x)  x and 1  x  4
y x
y2  x
1 y  2
Volume Around X Axis
b
• Volume =
• V=
• V=
  x .dy
2
2
a

2
4
  y .dy
1
1
 ( y 2 )2 .dy
• V=
y 
 
5 1
• V=
32   31

5
5
5
2
Volumes of revolution – bound
by 2 curves!
Equation of Volume of
Revolution
The equation for obtaining the volume of Revolution is:
Suppose two equations are as follows:
Calculations
First find the intersection points (a,b) by equating the two curves:
After calculating the intersection you will find:
,
)
Second find plug the functions into the equation:
It’s g(x) – h(x) because g(x) is on top!
*Note* Remember to graph the functions so you can see which is on top
The equation can then be worked out:
We can then use the fnInt (f(x), x, a, b) function to figure out the volume
= π ( 15.46875 – 5.0520833)
= π (10.4167)
= 125 π/12 units3
Calculus Test#1
7. Find the total area enclosed by the function and the
x-axis.
Graph the function on the calculator.
It shows that area between x=1 and x=1 has a positive yvalue and areas between x=1 and x=2, and between x=1 and x=-2 have negative y-values.
While integrating, set up the equation so that there’s a
negative sign in front of areas with boundaries that have
negative y-values.
:
Area
1
1
2
1
2
1
  x 4  5 x 2  4.dx   x 4  5 x 2  4.dx   x 4  5 x 2  4.dx
1
1
2
x5 5 x3
x5 5 x3
x5 5x3
 
 4 x.dx   
 4 x.dx   
 4 x.dx
5
3
5
3
5
3
1
2
1
1
1
2
 x5 5 x3

 x5 5 x3

 x5 5 x3

 
 4x   
 4x   
 4x
3
3
3
5
 1  5
 2  5
1
1 5
32 40
1 5
 1 5
  1 5
  32 40

 (   4)  ( 
 4)   ( 
 4)  (

 8)   (   8)  (   4) 
5
3
3
5
3
3
5 3
 5 3
  5
  5

8
Ans: 8 square units
8. The shaded area shown below the function is 6a
units. Find a.
Integrate the given equation using a and -a as
boundaries
a
  x 2  2.dx 
0
a
0

x 2  2.dx
a
0
x

x

   2x    2x
3
0  3
a
3
3
 a3
  03

03
(  a )3
  (  2a )  (  0)   (  0)  (
 2( a)) 
3
3
 3
  3

a3
a3

 2a 
 2a
3
3
Set equation of area under the curve = 6a
3
2a
 4a  6a
3
3
2a
 2a
3
2
a 3
a 3
By Robert Xie
The beginning…
The maximum is a stationary point.
Stationary points can be found by finding the value of x
when first derivative is equal to zero.
ln x
f  x 
x
f '  x    ln x    x 2    x 1  x 1 
ln x 1
f ' x   2  2
x
x
1  ln x
f ' x 
x2
Cont..
Now find x when the first derivative is equal to zero.
1  ln x
0
x2
0  1  ln x
ln x  1
xe
This gives the stationary point.
It cannot yet be determined whether it is a maximum point.
Thus, we use the second derivative. If the value is
greater than zero, it is a minimum; if it is less than zero, it
is a maximum.
f ''  x   1  ln x   2 x 3    x 2   x 1 
ln x  3
2x2
Now find the second derivative when x  e
f ''  x  
ln e  3
f ''  e  
2e 2
1
f ''  e   2  0
e
Since the second derivative is less than zero,
the stationary point is a minimum
ii) To find the coordinates iii) Inflection occurs when
of the maximum, we
the second derivative of
substitute the x value
f(x) is equal to zero
into the original f(x)
ln x  3
f ''  x  
function.
2x2
let x  e
0  ln x  3
ln e
f e 
ln x  3
e
3
x

e
1
f e 
e
 1
 The coordinates are at  e, 
 e
c) To find the area bound by the x axis, function f and x=5,
the intersection between the function and the x axis
must be found
ln x
f  x 
x
0  ln x
Therefore the limits are x=1 and x=5.
1 x
The next step is integration.
5 lnx
1 x .dx

5
1
ln x x 1.dx
To integrate, the substitution rule will be used.
let u  ln x
du
1
x
dx
5
  ln x x
1
1
du
 u
dx
1
dx
5
5
  u du
1
5
u 
 
 2 1
2
  ln x 

 2
2
5


1
ln 5    ln1


2
2
2
d) Finding the volume can be done using the
calculator since the question does not ask for an
exact answer.

5
1
  ln x x
 .dx  1.38
1 2
Semester 2 Exam #3
a)
Area of trapezium:
1
(OA  CB )  h
2
h
 h  r sin
sin  
r
OA  r
CD  r cos *CB  2CD
 CB  2r cos
T




h(OA  CB )

2
r sin  ( r  2r cos  )
2
2
r sin  (1  2 cos  )
2
2
r (sin   2 sin  cos  )
2
2
r
(sin   sin 2 )
2
b)
r2
T  (sin   sin 2 )
2
differentiate
dT
r 2 (cos   2 cos 2 )

d
2
r 2 (cos   2(cos 2   sin 2  )

2
2
2
2
r (cos   2(cos   (1  cos  ))

2
2
2
r (cos   2(2 cos   1)

2
r2
2

(4 cos   cos   2)
2
To calculate the value of

when T is maximum, let T=0
2
r
(4 cos 2   cos   2)  0
2
 4 cos 2   cos   2  0
let x  cos
4 x2  x  2  0
x  0.59307, 0.84307
  53.6,147.5
BUT  cannot be greater than 90º, so  is 53.6º
Verify!
x
x<53.6
x>53.6
f(x)
f’(x)>0
f‘(x)<0
 Maximum when x=53.6º
c)
BOC  180  2
KOA  
AOB  180  (180  2 )  
AOB  
Using Cosine Rule:
AB  r  r  2r cos 
2
2
AB  0.813r
2
2
2
 AB  0.902r
2
Parameter = OA+OC+CB+AB
= r + r + 1.18 r + 0.902 r
75cm = 4.08 r
 r = 18.4 cm
T is maximum when   53.6, r  18.4
2
18.4
T
(sin 53.6  sin 2(53.6))
2
 297.961943...
T  298
Question 6
from 2006 Calculus Exam #2
Given the two functions, f ( x)  sin 2 x and
g ( x)  cos x where x  0 and is an angle in
radians, find
(a) The exact area bound by the two curves
and the y axis. (6 marks)
(b) The exact volume of the solid generated
when the area found in part (a) is rotated
fully around the x axis. (6marks)
Yujin Park
Area bound by the two functions,
f ( x)  sin 2 x and g ( x)  cos x, and the y-axis.
Find the intersection of the two curves by letting f(x) = g(x)
Intersection
(
g (x)

6
,
3)
2
Intersection
(0.5 , 0)
f (x)
Area that
we want!
(a) The exact area bound by the curves and the y axis.
b
Area under a curve =
 f ( x) dx,
a
where the boundaries (limits) are x-values,
from a to b.
We already know the boundaries, 0 to
and thus, a=0 and b=

.

6
6
Also, since the area is under g ( x) but above f ( x),
the area would be Area under g ( x) - Area under f ( x)
Area bound by two curves and the y-axis


6
6
0
0
=  cos( x)dx   sin(2 x)dx


6
6
0
0
Area ,  cos( x)dx   sin(2 x) dx


1
 [sin( x)]06  [- (cos(2 x))]06
2

1

 [sin( )  sin(0)]  [(cos(2  )  cos(0)]
6
2
6

1

 [sin( )  0]  [(cos( )  1]
6
2
3
1
1 1
 [  0]  [  1]
2
2 2
1 1 1
1 1
  (- )  
2 2 2
2 4
1
 units 2
4
(b) The exact volume of the solid generated when the
area found in part (a) is rotated fully around the x axis.
Volume generated when part of the curves are rotated through 360o around the x-axis,
b
V=   { f ( x)}2 dx,
a
where the boundaries (limits) are x-values, from a to b.
From part (a), a=0 and b=

, and
6
V= V from g ( x) is rotated around, excluding volume from f ( x) rotated around,
 Volume would be Volume under g ( x) - Volume under f ( x)


6
6
0
0
V    {cos( x)}2 dx    {sin(2 x)}2 dx
using cos2 = 2cos 2  1 =1  2sin 2 

6
1
1
   (cos(2 x)  1)  (1  cos(4 x))  dx
2
2
0

6

  cos(2 x)  cos(4 x)  dx
20

1
 [ sin(2 x)  sin(4 x)]06
2 2
4
 1
 1

1
2
1
1
 [ sin( )  sin( )  sin(0)  sin(0)]
2 2
3 4
3
2
2
 1
3 1
3
 [ 
  ]
2 2 2 4 2
 3
3
 [

]
2 4
8
3 3

units 3
16
• Question 1.
• (Hint: a calculator may be used for (b) and (c) but give some
explanation as to how you solved it)
Solution!
(a) Use the chain rule:
f '( x)  3 x 2 cos x  x 3 sin x
(b) Find the point at which f '( x)  0
0  3 x 2 cos x  x 3 sin x
Graph in Calculator and find zeros within 0  x 

2
y  0 at x  0;1.19245...
Graph the original equation
Use the trace function to find the aforementioned zeros
This will tell you if it is a maximum on the graph [highest point]
We find it is a maximum at x  1.19245...
Cont…
(c) To find the point of inflection find the second derivative
f '  x   3x 2 cos x  x 3 sin x
f ''  x   3x 2 cos x  6 x sin x  3 x 2 sin x  x 3 cos x
Graph this function to locate its zeros between 0 and
we find y  0 at x  0
If this point is traced on the original graph,
it is found this is the original point of inflection

2
Question 2.
Air is leaking from a spherical
balloon at a rate of 5cm3/min. Find
the rate at which the surface area
of the balloon is changing when the
radius is 10cm.
Solution
dV
Given
 5
dt
Find
dA
when r = 10
dt
4
V  ( pie)r 3 -------->Differentiate both sides
3
dV
2 dr
 4( pie) r
dt
dt
When r = 10,
-5  4( pie)(10) 2
dr
dt

1
80( pie )
dr
dt
Cont…
2
A  4( pie ) r        Differentiate both sides
dA
 8( pie ) r
dt
dr
dt
When r = 10,
dA
1
 8( pie )(10)
80( pie )
dt
dA
 1
dt
Thus area is changing at -1 cm^2/sec when r = 10
Calculus Test #1 Q9
f ( x)  2 x 2
To make it easier, we’ll rename the functions
f(x) and g(x).
x2
g ( x)  1 
2
(a) On the same set of axis, sketch the
curves and find the intersection points.
5 y
(-0.632, 0.8)
-2
0
x2
g ( x)  1 
2 -5
f ( x)  2 x 2
(0.632, 0.8)
0
2
x
That’s easy,
the question
doesn’t ask for
exact values;
just find them
on your
calculator:
(-0.632, 0.8)
(0.632, 0.8)
(b) Find the exact volume of revolution formed when the region enclosed
by the curves in (a) is rotated about the y axis.
Ok, so we can’t really find the volume around the y axis…
We have to rewrite the functions in terms of y so that we can find the
volume of revolution around the x axis:
x2
y  1
2
x  2  2y
y  2x2
x
y
2
f ( x) 
x
2
g ( x)  2  2 x
The second graph shows the functions in terms of y. Notice that the
volume of revolution around the y-axis for the first graph will be the same
as the volume of revolution around the x-axis for the second graph.
5 y
2
y
1
-2
0
0
2
x
-0.5
-5
0
-1
x
0
0.5
1
1.5
We will now integrate the functions, being
careful to use the right boundary points.
1.5
y
To find the volume of
revolution for the area A, we
will integrate f(x) from x=0 to
x=0.8.
x  0.8
2
x
VA    ( ) .dx
0
2
0.8 x

.dx
0 2
0.8
g ( x)  2  2 x
f ( x) 
1
x
2
0.5
-0.5
0
-0.5
A
0
0.5
B
x
1
1.5
0.8
x 
  
 4 0
4

units3
25
2
To find the volume of revolution for the area B, we will integrate g(x) from
x=0.8 to x=1.
1.5
1
VB    ( 2  2 y ) 2 .dx
x  0.8
y
0.8
1
g ( x)  2  2 x
   2  2 y.dx
0.8
x
f ( x) 
2
1
2 1
   2 y  y 
0.8

0.5
-0.5
0
-0.5
A
0
0.5
B

2
units3
x
1
1.5
Total volume:
V  VA  VB


5
units3
(c)
Write an integration equation involving only a single integral that
could be used to find the volume of revolution if the region
described in (a) is rotated about the x axis.
5 y
f ( x)  2 x 2
(-0.632, 0.8)
-2
0
0
The exact value of the intersection
point 0.632 from (a) can be stored on
your calculator and used as the
boundary points for the integration.
Note the area revolved is g(x) – f(x).
(0.632, 0.8) x
2
V 
0.632
0.632
g ( x) .dx   
2
0.632
0.632
f ( x) 2 .dx
0.632
x2 2

(1  ( )) .dx   
(2 x 2 ) 2 .dx
0.632
0.632
2
0.632
x4
2

(1  x  )  4 x 4 .dx
0.632
4
0.632
x2
g ( x)  1 
-5
2

0.632
0.632
3.75 x 4  x 2  1 .dx
Madeleine Ong
Block C; Math HL II
Armstrong
A curve has equation
Find the equation of the
tangent to this curve at the point (1,1)
Step 1: Find the slope of the tangent
 formula is the derivative of the given function
dxy 3 d 2 x 2 d 3


dx
dx
dx
dy
dy
y 3  3 y 2 x  4 xy  2 x 2
0
dx
dx
dy
3 y 2 x  2 x 2   4 xy  y 3

dx
dy 4 xy  y 3
 2
dx 3 y x  2 x 2
 equation for the tangent's slope
Plug in given point: (1,1)
dy 4 11  1

dx 3 12 1  2 12
3
dy 4  1 5


 1
dx 3  2
5
Slope of the tangent at 1,1 is -1
Cont…
Step 2: Use the slope-intercept form to find the
equation of the tangent
slope  m   1
y  mx  b
y  x  b
plug in given point (1,1) to find y-intercept (b)
1  1  b
b2
 y   x  2  tangent equation
Step 1: Establish what rate(s) is known and needs to be found
Know :
dr
d
dA
2
 0.1 radians Find :
dt
dt
dt
When : r  3 and  

4
Cont…
Step 2: Manipulate equations to find the rate
Substep 1: Known: Asector
1 2
 r
2
dA 1  d r 2 

 

dt 2  dt 
Sub-step 2: Substitue into previous rate equation:
dA 1  2 d
dr 
 r
 2 r 
dt 2  dt
dt 
dA r 2  d
 
dt
2  dt

 dr 


r

 

 dt 
Cont…
Sub-step 3: Substitute in known values for rates:
dA 32

 .1   3 2 
dt
2
4
dA 9 3
cm 2


 5.16
dt 20 2
sec
Test #2
Short Answers 5 and 6
Richard Yeung
Short Answer #5
Part a
Part b
Short Answer #6
Graphing the problem makes it a lot easier!
Black is
Sin(2x)
Red is
Cos(x)
We are looking for the blue area
The graph
shows us that
even though we
are looking for
the area bound
by the two
curves and the y
axis, we can
find this in
terms of x!
The blue area is found by
p

0
Sin(2 x)  Cos( x)
Where, point p is marked by the
dotted line and is found by finding
the intersection of the two curves
Finding Intersect
• Let
– Sin(2x) = Cos(2x)
– 2Sin(x)Cos(x) = Cos(x)
– Sin(x) = 0.5
– Think of the 30-60-90 triangle
–X=

6
Finding area
• We know that p =

6
• Now we must find the area using
p

0
Sin(2 x)  Cos( x)
Solve
• We know that



6 u 2 .du
0

6 Sin(2 x)Cos ( x).dx
0
=
= 2

0
Then du = -Sin(x) . dx

1 3 6
= 2[ u ]
0
3
6 2Sin( x)Cos 2 ( x).dx
Let u = Cos(x)


1
3

2[
cos
( x)] 6
=
0
3
2 3 3
=
(
)
3 2
= 0.433
Calc test #1 Q5: Given that the curve y  x3  px 2  qx  r passes through (1,1)
and has turning points where x  1 and x  3, find the values of p, q, and r.
Step 1: Sub. in (1,1) into the eq. of the curve.
1 1 p  q  r
0 pqr
Step 2: Use the fact that turning pts occur when
dy
 0:
dx
Step 3: Solving the two equations simultaneously...
p  -3
q  -9
r... using the equation found in step 1, we get r  12
 0  3x 2  px  q
Step 3: Sub. in the given x values at which the
turning pts occur
Let x  1
0  3- 2p  q
-3  -2 p  q
Let x  3
0  27  6 p  q
-27  6 p  q
Answer: p  -3, q  -9, r  12
Calc test #1 Q6: An atheletics track has two straights of length g m (where g  0)
and two semi circular ends of radius x m. The perimeter of the track is 400 m.
(a) Show that g  200 -  x, and hence write down the possible values that x may have.
Step 1: DRAW.
Step 2: Set up the equation linking g and x.
g
x
x
1

s  400  2 g  2   2 x 
2

200  g   x
 g  200   x (shown)
Hence,
200  g
x
m, where x  0.

(b) What values of g and x produce the largest area inside the track
Step 1: Set up an equation of the area.
Step 3: But you must justify your reasoning...
Area A   x 2  2 gx
"This value of x gives the maximum area as A   x 2  2 gx
is a concave down parabola."
But, g  200   x :
A   x 2  2 x  200   x 
  x 2  400 x  2 x 2
 400 x   x 2
Step 4: Sub. in...
 200 
g  200   
0 m
  
Step 2:
For the largest area,
A '( x)  0  400  2 x
400 200
x

m
2

Answer: g  0 m, x 
200

m
Calculus Test #1
Numbers 3 and 4
Solutions!!
3. The tangent to the curve y = 2x²+ax+b at the
point (-2, 11) is perpendicular to the line
2y = x + 7. Find the value of a and b
• DIFFERENTIATE y = 2x²+ax+b
• y’ = 4x+a (This is now the slope)
• The slope
• 2y = x+7
• y= X 7
2
2
• Here we can see that the slope equals
1
2
• ATTENTION! The question asked for.. Perpendicular,
thus meaning that the slops is a negative reciprocal
• 4x+a = -2, where x = -2
• a=6
Number 3 continued
• Now that we’ve found (a), substitute a=6
into the original equation
• y = 2x²+6x+b
• Now substitute in the point given (-2, 11) to find b
• 11 = 2(-2)²+6(-2)+b
• 11 = 8-12+b
• b = 15
• So yeay!! Now we’ve found a AND b
• a=6
• b = 15
4. Given the sketch below
(a) If the line represents y = f(x) give the approximate x
values for when
(i) f’(x) = 0 ……………………………………………..
(ii) f’’(x) = 0 …………………………………………......
•
When the first derivative equals zero,
you look for the maximum and minimum
of the graph
–
•
Thus -5.5, -1/4, and 4(1/4)
When the second derivative equals zero,
you look for the x intercepts
–
Thus -1 and 2
(b) If the line represents y = f’’(x) write the approximate x values
where
(i) y = f(x) has minimum points …………………………………….
(ii) y = f(x) has maximum points …………………………………
(iii) y = f(x) has points of inflection ……………………………
• Since the second derivative is zero, the steps have to be taken
backwards to solve this problem.
• Use the x intercepts of y = f’’(x) to construct the first derivative graph
since they will be the maximum and minimum points
• Now use the x intercepts of y = f’(x) to construct the original graph of
y = f(x).
• Answers
– (i) x = -1
– (ii) x = 2
– (iii) x = -5.5, -1/4, 4