Transcendental Functions - Unit 6
Transcendental numbers & functions:

An algebraic number is a real number that is the root of a polynomial equation with rational coefficients. All
rational numbers are algebraic; some, but not all, irrational numbers are algebraic. Examples: The rational number 3.5
is algebraic because it is a root of the polynomial 2x  7  0 . The irrational number
the root of x 2  5  0 .

5 is also algebraic since it is
1.Prove that 1  3 is algebraic.
A transcendental number is a real number that is not algebraic. All transcendental numbers are irrational. We can
say that the set of algebraic numbers, along with the set of transcendental numbers, for a partition of the real numbers
since the union of these two sets is all the reals, and their intersection is empty. Examples of transcendental numbers
include , e, sin(1), 2 , and ln(n), provided n  1 is positive and rational. Proving numbers to be transcendental
is very difficult, and this type of number theory is not part of our calculus course.
2.There are other, nonmathematical, definitions for the term “transcendental.” What are they? Don’t just copy them
down; explain them in your own words!
An algebraic function is a function that can be expressed in terms of a finite number of sums, differences,
products, quotients, exponents, and radicals.
2

d 3 x  cos2 x  sin2 x
e
□ Sample Problem (by Raquel Roney, Nayeon Kang, & Ayush Dulguun): dx
cos x  sin x  cos2x by the Double Angle Identity
2
2
d (3x cos2x )
e
so, dx
d x
d (3x cos2 xsin 2 x) (3x cos2x ) d
e  ex
e
e
 3x  cos2x
dx
dx
so, dx

= 3 2  sin 2x  e 3x cos2x
= 6sin 2x  e 3x cos2x




Chap 6: □ Sample Problem (by Marina Mendoza & Beni Atibalentja):: (easy)

x3

4  x2
dx
SOLUTION: u  4  x 2 , du  2 xdx

x3
4  x2
dx =

x
4  x2
dx  
3
4  x2
dx
=
1 du
1
 3

2 u
4  x2
=
1
x
1
2
(
u
)
 3 sin 1
(Notation for arcsin=

2
2
a
2
1
)
 x2
2x  7  0
= 
1
1
x
(2u 2 )  3 sin 1
2
2
=  4  x 2  3 sin 1
x
2
Chap 6: □ Sample Problem (by Marina Mendoza & Beni Atibalentja):: (Easy)
Solve:
6 x 3  3x
 ( x 2  4) 2 dx
SOLUTION:
3 x(2 x 2  1)
 ( x 2  4) 2 dx
u – 4 = x2
Let u = x2 + 4
du = 2xdx
3 2(u  4)  1
du
2
u2
3 2u  7
du
= 
2
u2
3
1
= [2  du  7  u  2 du ]
2
u
21
= 3 ln | u |  u 1
2
21
= 3 ln |x2 + 4| +
2x 2  8
=
Chap 6: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is a basic-level problem for
students learning Euler’s method. It can be used as an example to test understanding.
Let y = f(x) be the solution to the differential equation
dy x  2 y
with the initial condition f(1) = 4. What is

dx
2x
the approximation for f(1.4) if Euler’s method is used starting at x = 1 with an interval size of 0.2?
Solution:
To find the solution for this problem we need to take two iterations.
First:
f (1.2)  f (1)  0.2
dy
9
 4  0.2 * = 4.9
dx
2
Then:
f (1.4)  4.9  0.2
ans: 5.817
dy
11
 4.9  0.2 *
 5.817
dx
2.4
□ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): : This is a basic-level problem. It involves business
calc, covering how to calculate values that are compounded continually.
In the year 2005 you decide to start a time capsule in your backyard. You decide to start out with enough objects to have
a total value of $50. Compounding this property continuously, at what rate is it growing if in 2025 you have $1000?
Solution: From the question we know that t=20 years, the initial deposit is $50, and our end value is $1000
Because the interest is accumulated continuously, we will use the equation
Then we can plug in our known values, so
A(t )  A0 e rt
1000  50e r ( 20)
20= e
20 r
ln(20)=20r
r=.1498 *100=14.98%
Chap 6: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is a basic-level problem. It deals with
exponential growth. It should be used as an example when introducing this concept.
A population of bacteria given by y(t) grows according to the equation
dy
 ky , where k is a constant and t is measured
dx
in minutes. If y(10) = 10 and y(30) = 25, what is the value of k?
Solution: The general solution to the differential equation
dy
 ky is
dt
y  Ae kt
The two given conditions will allow us to solve for k.
Ce10k  10
Ce 30k  25
We can divide the two equations to get e 20k  2.5 , so 20k=ln2.5. So,
k
ln 2.5
 0.046
20
Chap 6: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): (inv trig) This is a basic-level problem for
students learning how to make u-substitutions. It can be used as a follow-up example to test understanding.
cos x
 1  sin
2
x
dx = ?
Solution:
To solve, we use a u-substitution u = sinx, du = cosxdx
Substituting, we now have:
1
 1 u
2
du
 arctan u  C
 arctan(sin x)  C
Chap 6: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is an intermediate-level logistic
growth problem. Students given this problem need to understand the concept and be familiar with patterns within logistic
growth, such as reaching a stable point.
The population of pandas in a particular location is given by the logistic differential equation
initial population at t=0 is 300, and t is in months, what is
p
dP
 P( 4 
) . If the
dt
200
lim P(t ) ?
t 
Solution: The easiest way to solve this problem is to understand that any solution to a logistic differential equation
eventually approaches one of its stable points, or a value of P for which
Since P ( 4 
P(4 
dP
0
dt
p
) is quadratic we know there are two zeroes, or stable points.
200
p
) , so P = 0 or P = 800.
200
The initial condition p(0) = 300 makes P ( 4 
p
dP
) positive, meaning
is positive and the population is increasing,
200
dt
which means that our answer is 800.
You could also derive or remember the solution to the logistic differential equation, but this is much more complicated.
Chap 6: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is an intermediate-level problem.
Students should understand differential equations, and know how to integrate well.
If
dy
 2xy 2 and y = 3 when x = 1, what is the value of y when x = 3?
dx
Solution: To find y as a function of x, we must solve the differential equation
dy
 2xy 2 . This equation is separable,
dx
which means that we can algebraically manipulate to move all the y’s to one side and all the x’s to the other side.

dy
 2 xdx
y2
Now, we can integrate both sides and express y as a function of x.

dy
 2 xdx
y2 
1
 x2  C
y
1
y 2
x C
Plugging in the initial condition x = 1 and y = 3 we get:
3
We now have y as a function of x,
y
1
,
1 C
C
2
3
1
x2 
2
3
So, solving for y when x = 3:
y
1
3

3  3 25
2
Chap 6: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is an advanced-level problem.
Students need to be comfortable with Euler’s method and differentiation.
dy x  y
dy
. If the solution, f, of
contains the point (0,2), approximate f(1) using Euler’s method with three

dx
2x
dx
steps of equal size.
Solution: When solving this problem, we will first split the interval into equal parts and then find the three points.
Step one: (1,2) to (4/3, ?)
m=
dy x  y 1  2 3
=


dx
2x
2
2
dy = m*dx
dy =
3 1 1
5
*  so ? = . Therefore, our first point is
2 3 2
2
4 5
3 2
5
3


Step two: ( , ) to  , ? 
dy x  y 23


dx
2x
16
23 1 23
dy 
* 
16 3 48
m
 5 143 

 3 48 
Point two:  ,
 5 143 
 to (2, ?)
 3 48 
Step three:  ,
4 5
 , 
3 2
m
dy x  y 223


dx
2x
160
dy 
223 1 223
* 
160 3 480


Point three:  2,
551 

160 
□ Sample problem (by Fan Huang & Fernanda Mendez): Chap 6: Show that  ln[csc( x)  cot( x)] =
ln[csc( x)  cot( x)] using the integral  csc( x) dx .
Solution:
 csc( x)
dx 
csc( x)  cot( x)
 csc( x) csc( x)  cot( x) dx
csc 2 ( x)  csc( x) cot( x)
 csc( x)  cot( x) dx 
 csc( x)
dx 

 u  csc( x)  cot( x)
 du
  ln( u )  C1   ln[csc( x)  cot( x)]  C1
u
csc( x)  cot( x)
 csc( x) csc( x)  cot( x)
csc 2 ( x)  csc( x) cot( x)
 csc( x)  cot( x) dx 

du  [ csc( x) cot( x)  csc 2 ( x)] dx
 u  csc( x)  cot( x)
du  [ csc( x) cot( x)  csc 2 ( x)] dx
du
 ln( u )  C 2  ln[csc( x)  cot( x)]  C 2
u
Therefore  ln[csc( x)  cot( x)]  C1  ln[csc( x)  cot( x)]  C 2
To show that  ln[csc( x)  cot( x)]  ln[csc( x)  cot( x)] , we must show that C1 = C2.
C1 = C2
 ln[csc( x)  cot( x)] 1  ln[csc( x)  cot( x)]  [csc( x)  cot( x)] 1  csc( x)  cot( x)
1
1
1
cos( x)
1
1  cos( x)

 csc( x)  cot( x) 




1
cos( x) sin( x) sin( x)
1  cos( x)
csc( x)  cot( x)
sin( x)

sin( x) sin( x)
sin( x)

sin( x)  1  cos( x)  1  cos( x)
sin( x)[1  cos( x)] 1  cos( x)
1  cos( x) 1  cos( x)

 




2
1  cos( x)  1  cos( x) 
sin( x)
sin( x)
sin( x)
sin( x)
sin ( x)
Chap 6: □ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu):
Find the equation of the curve having the following conditions: (1)At each point (x,y) on the curve, the slope of the curve
is 
x
. (2) The curve passes through the point (2,1).
2y
Solution: Since the slope of the curve is given, we have the differential equation
y (2)  1 . Separating the variables and integrating, 2 y dy   x dx 
 2y
dy
x

with the initial condition
dx
2y
dy    x dx
x2
22
 c . Using the initial condition, y (2)  1 12  
 c So, c = 3. Therefore, the equation of the
2
2
x2
x2
2
2
2
2
specified curve is y  
 3 or x  2 y  6 . y  
c.
2
2

y2  
Chap 6: □ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu): a) Find the derivative of sin 1 ( x 3 )
(Use the inverse trigonometric functions and the chain rule).
Solution:
d
sin-1 x3 =
dx
1
1  (x3 )2
(3 x 2 ) =
3x 2
1 x6
.
b) Find the derivative of tan-1 (x3)
Solution:
d
1
3x 2
2
tan-1 x3 =
.
(
3
x
)

dx
1  (x3 )2
1  x6
d x ln x
(e
tan x )
Chap 6: □ Sample problem (by Morgan Holbrook & Justin Parks): Find the derivative: dx
.
d x ln x
(e
tan x )
x ln x
tan( x)  e x ln x sec 2 ( x)
Solution: dx
= (ln x  1)e
Chap 6: □ Sample problem (by Liz King & Katherine Wallig): (logs, deriv rvw)
This is a complicated derivative problem involving the majority of derivative-taking techniques (chain rule, power rule,
3 cos x  ( x s  3x) 4 (ln x)
product rule, natural logs, quotient rule, trig). Find the derivative of
4x 2
Solution: It looks scary, we know. But just remember to work from the inside out, and split things up to make it easier
for yourself whenever possible. :
d 3 cos x  ( x s  3x) 4 (ln x)
d 3 cos x
d ( x 2  3x) 4 (ln x)
(
(
)

(
)
)
=
dx
dx 4 x 2
dx
4x 2
4x 2
Phew, now that we have two separate parts to the problem, let’s just do one at a time:
First part: Just use the quotient rule:
4 x 2 (3 sin x)  3 cos x(8 x)  12 x 2 sin x  24 cos x

(4 x 2 ) 2
16 x 4
Okay, that wasn’t so bad. Now let’s do the second part:
d x 2  3x 4 (ln x)
(
)
dx
4x 2
1
4 x 2 [( 4( x 2  3x) 3 (2 x  3)(ln x))  ( ( x 2  3 x) 4 )]  8 x( x 2  3 x) 4 (ln x)
x
(4 x 2 ) 2
16 x 2 ( x 2  3x) 3 (2 x  3) ln x  4 x( x 2  3x) 4  8 x( x 2  3x) 4 ln x
16 x 4
2
2
 12 x sin x  24 cos x  16 x ( x 2  3x) 3 (2 x  3) ln x  4 x( x 2  3x) 4  8 x( x 2  3x) 4
=
16 x 4

Chap 6: □ Sample problem (by Liz King & Katherine Wallig): This one covers the basics of logs in calculus. Pretty
much all you need to know is the derivative of lnu, and substitute whatever is inside the log with u. Differentiate:
d
[ln(cos x 2 )]
dx
d
1 du
dy
(ln u ) 
 2 x sin x 2 . So for the final answer we get
. Let y = cos(x2) , so
dx
u dx
dx
d
1 dy y '  2 x sin x 2
.
(ln y ) 
 
dx
y dx y
cos x 2
Solution: Remember that
Chap 6: □ Sample problem (by Liz King & Katherine Wallig): This is a simple problem to test whether students
understand the basic principles of taking the derivative of an equation involving e.
3x
d 5 4
e
dx 7
3x
3x
3x
d 5 4 5 3 4 15 4
e  ( )e 
e
Solution:
dx 7
7 4
28
Chap 6: □ Sample problem (by Liz King & Katherine Wallig): This problem deals with Newton’s law of cooling. You
will need to find the cooling constant k first, then use that constant to figure out the right amount of time it takes to cool to
the given temperature. For this problem you will also need to know how to work with the number e and natural logs.
A pie is taken out of the oven and is 238ºF and set in a room that’s temperature is 76ºF. After 20 minutes the pie was
184ºF, at what time is the pie 123ºF?
Solution: Use the formula T-Ts= (T0-Ts)e-kt to find k
184  76  (238  76)e 20k
108  162e  20k
2
 e  20k
3
2
ln  20k
3
3
 ln  20k
2
1.5
k
20
123  76  (238  76)e
t (
ln 1.5
)
20
ln 1.5
t (
)
47
 e 20
162
47
ln 1.5
ln
 t (
)
162
20
47
 20 ln(
)
162  t  61.04 min utes
ln 1.5
Chap 6: □ Sample problem (by Liz King & Katherine Wallig): For this one you need to know the derivatives of
inverse trig functions, do not attempt to do this problem with merely a basic thing such as u substitutions, but try to figure
out which inverse trig function to use (hint, you may have to use two functions).

4
6x
  ( x  2)2  16  25  x4

dx

Solution: First let’s divide it up into two separate antiderivatives.
4
 ( x  2)
2
 16
dx  
6x
25  x 4
dx
4
dx
2x
 3
dx
2
( x  2)  16
52  ( x 2 ) 2
recognize this form? You should! Now all you need to do s make a few u
substitutions, u = x + 2 for the first and u = x2 for the second
du
du
1
u
u
 3
 4( ) tan 1  3 sin 1  C 
2
2
2
4
4
5
u 4
5 u
( x  2)
x2
tan 1
 3 sin 1
C
4
5
4
2
chap 6: □ Sample problem (by Merla Hübler & Lisa Portis): Basic inverse trigonometric integration:

Solution:
dx
a2  x2
 sin 1
x
c
a
 a5

dx
25  x 2
 sin 1

dx
25  x 2
x
c
5
chap 6: □ Sample problem (by Merla Hübler & Lisa Portis): Basic hyperbolic inverse functions: Find the
antiderivative:

dx
9  9x 2
d
1
In this particular problem you can factor out the 9 to get it into the general
sinh 1 x 
dx
1 x2
1
1
dx
dx
=
= sinh 1 x  c

3
3 1 x2
9  9x 2
Solution: 
form.

In chapter 8 we’ll learn how to do this integral without hyperbolic functions.
chap 6: □ Sample problem (by Merla Hübler & Lisa Portis): Intermediate problem for practice with different
x2  7
dx .
integration techniques. Integrate: 
x5
2
7
Solution: Noticing that a u-substitution would not work in this case, long division should be tried since the numerator is
larger than the denominator, and this method might be effective in making the term easier to integrate.
x-5+ x185
x+5 x 2  7
x2+5x
-5x-7
-5x-25
18
7
 (x  5 
18
x 5
)dx
2
 [ x2  5 x  18 ln( x  5)] 72
2
 [ 492  35  18 ln( 12)]  [2  10  18 ln( 7)]
  52  18ln 127
chap 6: □ Sample problem (by Merla Hübler & Lisa Portis): Basic problem for practicing slope fields. Calculate and
draw the slope field for f ( x)  x 3  c at increments from -3 to 3 on both coordinates. Then draw f(x) with the initial
condition of: f (0)  1 .
Solution: The slope field is a graphical representation of a function created with
the slope of the graph at multiple points. To do this, find the derivative of the
function and evaluate it at the various x-coordinates:
f ' ( x)  3x 2
 f ' (3)  27
 f ' (2)  12
 f ' (1)  3
 f ' ( 0)  0
 f ' (1)  3
 f ' (2)  12
 f ' (3)  27
The initial condition helps to find where to draw the line because the given f(x) has a “+ c”, and the initial condition
allows for this to be solved for.
The function to be drawn is: f ( x)  x 3  1 .
chap 6: □ Sample problem (by Merla Hübler & Lisa Portis): Intermediate problem for practicing Euler’s method. lem:
Given that
dy
 x 3 y 2 and given the point (1, 2) , find y at 1.5, using Euler’s method. Use x  0.1 . Check the accuracy
dx
of the result.
Solution: Euler’s method: Given f ( x) 
y0  2
(1, 2)
y1  2  [ f (1, 2)]( 0.1)
y1  2  4(0.1)
y1  2.4
(1.1, 2.4)
y 2  2.4  [ f (1.1, 2.4)]( 0.1)
y 2  2.4  (7.67)(0.1)
y 2  3.17
(1.2, 3.17)
y 3  3.17  [ f (1.2, 3.17)]( 0.1)
y 3  4.9
(1.3, 4.9)
dy
, y n 1  y n  f ( x n , y n )x
dx
y 4  4.9  [ f (1.3, 4.9)]( 0.1)
y 4  10.17
(1.4,10.17)
y 5  10.17  [ f (1.4, 10.17)]( 0.1)
y 5  38.57
The final coordinate is (1.5, 38.57), therefore the solution given by the
approximation is 38.57.
Check:
dy
 x3 y 2
dx
dy
 x 3 dx
y2
y

2
dy   x 3 dx
1 x4

c
y
4
→ Solve for c by using the given point (1, 2):
x4 1
c

4 y
(1) 4 1
c

4
2


c
3
4
1
3 x4
→   
y
4 4

y (1.5) :
1
3 (1.5) 4
 
y
4
4
y  -1.94
→ It can be seen that the actual solution to the problem is very different than what was found using Euler’s method for
approximation. This demonstrates that the method is not the most useful for this case. The inaccuracy is a result of the
extremely exponential increase in the slope, and because Euler’s method approximates this sloping only roughly.
chap 6: □ Sample problem (by Merla Hübler & Lisa Portis): (part 2 of a chap 9 prob): Solve the differential equation
dy
2
dx  ( 4 x  5 x  3)( y  7) , if one condition involves the following values of x and y: x is the smallest positive integer
for the convergence of the previous series and y is the largest negative integer for the convergence of the previous series.
Solution: First, get the x’s and y’s on opposite sides of the equals sign to take the integral of both sides and solve the
equation for y. To get the constant, plug in the given condition.
dy
dx
 (4 x 2  5 x  3)( y  7) 
ln y  7 
4 x3
3
dy
y 7
 52x  3x  C  e
2
 (4 x 2  5x  3)dx 
4 x3 5 x 2
 2
3
3 x C

dy
y 7
  (4 x 2  5 x  3)dx 
 y  7  y  C1e
4 x3 5 x 2
 2
3
3 x
 7 . The smallest positive integer for
the convergence of the previous series is 5, and the largest negative integer for the convergence of the previous series is
-6. So x = 5 and y = -6.
 6  C1e
y e
4x3
3
920
3
 7  1  C1e
2
 5x2
920
3
 C1  e
920
3
. So, the equation is y  (e
920
3
) (e
4x
3
3
2
 5x
3x
2
)7 
 3x 920
3
chap 6: □ Sample problem (by Kevin Stanford & Mike Mitchner): derivative of natural log functions. Find the
derivative of the following function: y  x 2 ln 2 x .
Solution:


dy d 2
dy
dy
dy 2 x 2
 1 

x ln 2 x 
 x  2 ln 2 x .
 x 2  2  ln 2 x2 x  

 2 x ln 2 x 
dx dx
dx
dx
dx
2x
 2x 
dy
 4 xy, y 2  9 ,
dx
chap 6: Kevin Stanford & Mike Mitchner Integration & Euler method Given
Solution Intro
In order to solve the problem we first find the analytical solution by moving like variables to the same side of the
equation. Then, we proceed to take the antiderivative of both sides of the equation in order to get rid of any derivates and
get a solution in the form of y=x. Next Euler’s method is used to approximate the answer to the same problem. The most
important thing to remember here is the pronunciation of the name Euler. It is you-ler, not oil-er, no matter what
Beuschlein may tell you. After finding both the exact answer and the approximation all that is needed is to plug both
solutions into the percent error formula in order to find the percent error of the Euler approximation.
a. Find y(2.04) by solving the differential equation.
dy
dy
 4 xy 
Solution:
 4 xdx 
dx
y
y  e2x
9  C2 e
2
C
2( 2) 2

y  e 2 x  eC
2
 C2 e8  C 2 
dy
2
 y   4 xdx  ln y  2x  C 
y  C1e 2 x

2

y  C2 e 2 x
9
 0.0030191637 . Therefore,
e8
y  e2x  C
2

2
. When y(2) = 9,
y  0.003017637e 2 x
2
and
y2.04  12.436 . In part one of this problem the use of different constants can be tricky. The original C shown is of
c
course simply from finding the antiderivative of both sides of the equation. C sub 1 is created because
is simply
another constant, which we rename C sub 1. C sub 2 is created in order to get rid of the absolute value sign. By creating
a constant value that is always positive, the solution will always be positive, so by creating C sub 2 and making it
always positive we are able to remove the absolute value signs from y. After this is done all that is needed is to plug in
the given values of x and y to find the value of this C sub 2. After finding C sub 2 the x value we are looking for (2.04)
e
is plugged into this new equation of
y  C2 e 2 x
2
. This gives us the exact value of y(2.04).
b. Approximate y(2.04) using Euler’s method with x  0.01 .


Solution: Euler’s formula: y n  y n1  f xn1, y n1 x
This formula simply says in order to find y sub n you first need to know the last value of y found, first from that which
is given, in this case y(2)=9, then from that which has been found in the last step of the equation, i.e.: plug in y(1) when
finding y(2). After plugging in this y for y n 1 in Euler’s equation, the value of x n 1 can be found first from the given,
when x =2, then by multiplying the final sum of the last step of the equation by the x value from the last step of the
equation. This value is derived by adding the change in x, in this case 0.01, to the initial value of x progressively for
each step of the equation, starting with the given x, in this case 2, and ending after we have reached the value of x one
step before the value we are looking for, in this case 2.03. This means the values of x used will be progressively 2, 2.01,
2.02, and 2.03. After finding these values of x and y they are plugged into the initial derivative equation in order to find
f xn1 , y n1  The change in x is simply given and for this problem is always 0.01.
y1  9  180.01  9.18
y 2  9.18  73.80720.01  9.918072
y 3  9.918072  80.138021760.01  10.71945222
y 4  10.71945222  87.041952010.01  11.58987174
c. Find the % error of your approximation
Solution: %error = [(actual answer – approximation)/ (actual answer)] x 100
12.436  11.58987174
100  6.8%
12.436
% error =
chap 6: Kevin Stanford & Mike Mitchner basic log integral with u-sub Evaluate
Solution: Let
u  ln x And
du 
1
x
dx
 x ln x .
dx
1

 x ln x  u du  ln u  ln ln x   C
chap 6: Kevin Stanford & Mike Mitchner Integration by Parts: Find the anti-derivative of the following function using
integration by parts: f ( x ) 
Solution:
ln x
.
x4
 udv  uv   vdu . The substitutions are u  ln x and
dv 
1
 x 4
4
x
. When I take the derivative of u
1
1
dx
v   x 3
x . When I take the anti-derivative of du I get:
3
I get:
. Plug the variables back into the integration
1 3
1 4
 x ln x    x
3
by parts formula 3
. Now the equation is in a form that we can use the power rule to solve
1
1
 x 3 ln x  x 3  C
3
9
.
du 
chap 6: Kevin Stanford & Mike Mitchner Integration using definition of the Natural logarithm and the fundamental
theorem of Calculus. Find the anti-derivative of the following function: f ( x) 
x
Solution: ln x 
1
F ( x)   udu 
1
 t dt
so, the derivative of
1
 x(3  ln x) dx .
1
d
1
ln x  . Let u  3  ln x . Then du  dx .
x
dx
x
1 2
1
u  C  (3  ln x) 2  C
2
2
4
dx . Incomplete
 6 x  11
u  x3
dx
dx
4
1
dx  4  2
 4 2
 4
Solution:  2
2
x  6 x  11
u  2 du  dx
( x  6 x  9)  11  9
( x  3)  2
chap 6: Completing the Square Kevin Stanford & Mike Mitchner Find
x
2
chap 6: Kevin Stanford & Mike Mitchner Newton’s law of cooling A turkey is taken out of a 375 oF oven where you set
it outside on the table for serving. An unfortunate disruption in your Thanksgiving Day activities breaks out as your
Uncle Henry and Grandpa Joe begin to fight. Meanwhile the bird sits in the 60o weather where it cools to 150o in the half
hour it takes for the police to arrive. How much longer before the bird reaches 100o?
Solution:
dT
 k (T  60) 
dt
dT
 (T  60)   kdt

e
ln T  60
 e kt C1
kt
 T  60  Ce  Ce kt  60 
375  Ce k ( 0 )  60  315  C . So, 150  315e k ( 30) and k  0.0247312448 . The equation we now have for
0.0247312448
the cooling of the bird is: T  315e
. Since we want to know how long it will take to cool from 150 to 100
degrees we plug in 100 in to the equation for T and find how long it takes to cool from 375 to 100 and subtract the
thirty minutes we already know it took to cool from 375 to 150.
So it will take 46.39 minutes to cool from 150 to 100 degrees.
100  315e 0.0247312448t
 t  46.39 minutes.
Chap 6: hard Patrick McCall & Nathan Dornfeld: (logs, hyperbolic functions) Let f (x) =
ln 3 cos 21x sin 21x  x 1 
e  x 
3e 3 cos 42 x
e 

ln 6 sin 42 x  ln 2 sin 42 x
12 e 3 cos 42 x
. Find f ’ (x) and write it in terms of a hyperbolic function.
1 

ln 3 cos 21x sin 21x e x  x 
e 

Solution: Simplifying, we have f (x) =
, with the qualification that cos 42x  0
(ln 6  ln 2) sin 42 x
4
(since in the original function this would be undefined). Because ln 6 - ln 2 = ln 3 and using the double angle
1 
1

4  sin 42 x  e x  x 
1 
e 

2

formula for sine, we can simplify further: f (x) =
= 2  e x  x  , cos 42x  0. Thus,
sin 42 x
e 

x
x
d x
e e
e  e  x = 2 (ex + e-x) = 4 
 4 cosh x , cos 42x  0. (If a function is undefined
f ’ (x) = 2
dx
2
for certain values of x, then so is its derivative.) Alternatively, we could have made use of a hyperbolic
d x
d
d  e x  ex 

 = 4
e  e  x = 4
sinh x
function before differentiating: f ’ (x) = 2
dx
dx
dx 
2 




= 4 cosh x, cos 42x  0.
chap 6: Raquel Roney, Nayeon Kang, & Ayush Dulguun basic u-sub, log integral Find
x
x2
dx .
2
 4x
Solution: Use u-substitution: u  x  4 x du = 2x + 4 . Next you want to change the numerator so that it looks
similar to du so, multiply everything by 2 , but then also multiply everything by 1/2 to cancel each other out.
2


2x  4
du
dx  (1/2) 
 (1/2)ln u  c  (1/2)ln x 2  4 x  c In the second step du and u were substituted in
2
 4x
u
and then the integral is evaluated. At the end x 2  4 x is subsituted back in for u.
( 1/2)
x


chap 6: basic derivative of an exponential Given y  23 , find y’ (x) and show all steps that lead to your answer.
x
Solution: y ' ( x) 






d
d
d
d
d
2(3 x )  2  (3 x )  2 
(e ln 3 ) x  2  (e x ln 3 )  2  (e x ln 3 ) ( x ln 3)  2  (e x ln 3 )(ln 3)
dx
dx
dx
dx
dx
x
 2(ln 3) (e ln 3 )  2(ln 3)(3) x .
chap 6: Medium (by Raquel Roney, Nayeon Kang, & Ayush Dulguun): derivative of natural log Find
d
ln(sec x  e x ln 3 ) .
dx
Solution: We will make use of the following theorems:


d
f (x) d
d x
ln( f (x)) 
sec x  sec x tan x ; e x ln a  a x ; and
a  (ln a ) a x .
;
dx
f (x) dx
dx
Thus,
d
sec x tan x  (ln 3) 3 x
d
ln(sec x  e x ln 3 ) = ln(sec x  3 x ) 
.
dx
dx
sec x  3 x
Chap 6: (by Shaofeng Sun & Artem Rogachev): basic exponential growth problem, differential equations Once you
 built your piggy pen, you put in 25 pigs and then let them reproduce. The population of pigs can be described by the
differential equation:
dp
P
 7(1 
) You want 127 pigs for sale. If it is possible to have that many pigs calculate how
dt
125
much time it will take, if not say why?
Solution: No, the number of pigs in the pen can never reach 127. First, in the differential equation, it shows that the
carrying capacity which means max population in this case is only 125, and 127 is bigger than 125. If P 0 is large at the
beginning, the population may reach 127 at some point in time, but P0 is 25. Another reason is that the pigpen can only
hold up to 125 pigs as you calculated from number 2, so you can’t ever get 127!
Chap 6: (by Shaofeng Sun & Artem Rogachev): intermediate related rates and some simple differential eqns A right
cone with height 15 dm and radius 15 dm has two openings (as you see in the picture below). Below the cone is a cylinder
with height 6 dm and radius 4dm. Water is pouring into the cone from the larger opening at 10dm3/s, but it’s also coming
out from the bottom opening at 3dm3/s into the cylinder, for what time interval is
dh
of the water in the cone larger than
dt
dh
of the cylinder, if water starts to pour in at t=0?
dt
Solution: All right here is the plan. First we are going to find the equation for change in height in
terms of height in the cone. We konow that height in the cylinder is rising at a constant rate, since
cross sectional area is a constant. Then we find out for what H are the water levels rising at the same
rate. Later find out how much time it takes to get to that height.
1 2
r H , set up ratio between radius and
3
r 15
dV
1
height:
=10-3=7dm3/s Since 3dm is poured out. Substitute r for H we get: V = H 3 ,

 1 , so r =H.
H 15
dt
3
Let H be the height of water in cone, and h height in cylinder. V of cone =
so
dV
dt
dv
dH dH
7
,
 H 2
 dt2 = 2 .
dt
dt
 H dt H
7
2
≥0.0596831037,  H ≤117.2861257
2
dv
:
dt H
dh
dv
of cylinder =
/ ( 4 2 ) ≈ 0.0596831037 plug in 3 for
dt
dt
dH
so H≤6.11
H 2
 7 , → H 2 dH  7dt , bring t
dt
1
H 3  7t +C plug in known values of t and h like (0,0) so C is 0
3
1
dH dh
>
.
 611
. 3  7t , t = 34.125s so for time interval (0, 34.125)
3
dt dt
over Integrate both sides:
 H
2
dH   7dt , →