Announcements
Assignment 3 is now due November 3rd (instead of Oct 27).
Midterm rooms have been posted (Friday, Nov 1 from 6:00pm - 7:30pm).
Students registered with Student Accessibility Services please contact their
office ASAP to schedule your exam.
Midterm covers: 1.2-1.4, 2.1-2.7, 2.9-2.10, 3.1-3.3.
Official formula sheet to be posted on Sat (trig identities/basic integrals).
No calculators, no scrap paper, no electronic devices, closed book.
No labs next week.
To prepare review your notes, the midterm review sheet, weekly lab
problems, worksheet problems, and online assignments (in this order).
Study hard, and remember to sleep and eat.
For extra help you can visit the Devon Academic Resource Centre
(Engineering Drop-in Centre ENC 208). The tutoring schedule for AMAT
217 is posted on D2L. Or go to Mike’s office hours.
If you are going to miss the midterm then contact me as soon as possible
with official documentation.
AMAT 217 (University of Calgary)
Fall 2013
1 / 16
Inverse Trig Functions
The restricted trig functions are one-to-one:
π π
Restrict to − ,
2 2
h
i
Restrict to [0, π]
Restrict to
π π
− ,
2 2
The inverse trig functions are the inverses of restricted trig functions.
AMAT 217 (University of Calgary)
Fall 2013
2 / 16
Inverse Trig Functions
The last slide shows sin−1 x , cos−1 x and tan−1 x .
We now define the remaining ones as follows:
Inverse secant, inverse cosecant, inverse cotangent functions
1
sec−1 x = cos−1
, for |x | ≥ 1.
x
1
csc−1 x = sin−1
, for |x | ≥ 1.
x
1
cot−1 x = tan−1
, for x 6= 0.
x
Example
Evaluate csc−1 (1).
Solution: Using the definition we have:
csc−1 (1) = sin−1
1
π
=
1
2
from the graph of sin−1 x .
AMAT 217 (University of Calgary)
Fall 2013
3 / 16
Example
Evaluate cot(tan−1 (5)).
Solution: Using the definition of cot−1 x from the last slide we have:
1
cot−1 x = tan−1
x
Subbing in x = 1/5 we get:
cot−1
1
= tan−1 (5)
5
Thus, we have:
1
1
cot(tan−1 (5)) = cot cot−1
=
5
5
AMAT 217 (University of Calgary)
Fall 2013
4 / 16
Inverse Trig Derivatives
sin−1 x
0
=√
1
1 − x2
cos−1 x
0
= −√
1
1 − x2
tan−1 x
0
=
1
1 + x2
Some integral rules
Z
1
1
−1 x
dx
=
tan
+C
a2 + x 2
a
a
Z
√
x 1
+C
dx = sin−1
a
a2 − x 2
Example
Z
Evaluate
√
Z
3
dx .
2 − x2
3
√
dx = 3
2 − x2
AMAT 217 (University of Calgary)
Z
1
√
dx = 3 sin−1
2 − x2
x
√
2
+C
Fall 2013
5 / 16
Growth rates of various functions
Example
Compare the long term rate of growth of the following functions by plotting them:
√
e x , ln x , x , x 2 ,
x = x 1/2 .
Plot in your favourite program/software (e.g., Wolfram Alpha).
Observe that for large x we have:
e x > x 2 > x > x 1/2 > ln x
Higher powers of x grow faster than lower powers of x .
e x grows faster than any (positive) power of x !
ln x grows slower than any (positive) power of x !
http://abstrusegoose.com/218
xn
For any constant n > 0: lim x = 0
x →∞ e
AMAT 217 (University of Calgary)
and
lim
x →∞
ln x
=0
xn
Fall 2013
6 / 16
Real world problems: Exponential growth and decay
Many problems can be modelled with a simple differential equation.
(Half-life, Newton’s Law of Cooling, Bacterial Growth, Population Growth).
Suppose y represents a certain quantity at time t.
If the quantity changes at a rate proportional to its size then we have:
dy
= ky
dt
If the initial amount is y0 then we have an initial value problem:
dy
= ky ,
dt
y (0) = y0
The unique solution to this IVP can be proved to be:
y = y0 e kt
where y is quantity at time t, y0 is initial quantity, t is time and k is the
proportionality constant.
If k > 0, the quantity y exhibits exponential growth.
If k < 0, the quantity y exhibits exponential decay.
AMAT 217 (University of Calgary)
Fall 2013
7 / 16
Example
A cell culture grows at a rate proportional to the number of cells present. If the
culture contains 500 cells initially and 800 after 24 hours, how many cells will
there be after a further 12 hours?
Let y (t) be the quantity of cells present after t hours.
The growth follows an exponential decay process so can be described as:
y (t) = y0 e kt
Then y (0) = 500, so that y0 = 500:
y (t) = 500e kt
There are 800 cells after 24 hours, hence, y (24) = 800:
8
ln(1.6)
= 24k → k =
800 = 500e k(24) → ln
5
24
We need to find how many cells are present a further 12 hours later (t = 36):
y (36) = 500e
ln(1.6)
24 (36)
≈ 1012
Thus, cell count grew to about 1, 012 in the 12 hours after it was 800.
AMAT 217 (University of Calgary)
Fall 2013
8 / 16
Half-life is the amount of time required for a quantity to fall to half its value as
measured at the beginning of the time period.
The Half-Life Formula
t/T
1
, where
Formula: y (t) = y0
2
y (t) = amount at time t,
y0 = original amount,
t = time (in years),
T = half-life (in years), that is, the time it takes to cut the quantity in half.
Proof: Half-life follows an exponential decay process so can be described as
y = y0 e kt
Let T be the half-life so that: y (T ) = 12 y0 .
Then substituting t = T and y (T ) = 12 y0 we have:
1
1
y0 = y0 e kT
→
= e kT
2
2
Thus, half-life can be described as:
y = y0 e
ln(1/2)
t
T
AMAT 217 (University of Calgary)
→
→
y = y0 e
kT = ln(1/2)
ln( 12 )t/T
→
→
k=
y = y0
ln(1/2)
T
t
1
2
T
Fall 2013
9 / 16
Example
Find the half-life of a radioactive substance if after 300 years only 12.5% of the
original amount remains.
t/T
1
, where
The formula is y (t) = y0
2
I
I
I
I
y (t) = amount at time t,
y0 = original amount,
t = time (in years),
T = half-life (in years).
We are given that y (300) is 12.5% of the original amount y0 .
That is, y (300) = 0.125 y0 .
Thus, substituting t = 300 into the formula gives:
300/T
1
0.125 y0 = y0
2
Cancelling y0 ’s gives:
300/T
300 ln(1/2)
300
1
→ T =
ln(1/2) = 100
0.125 =
→ ln(0.125) =
T
ln(0.125)
2
AMAT 217 (University of Calgary)
Fall 2013
10 / 16
Newton’s Law of Cooling says that a hot object introduced into a cooler
environment will cool at a rate proportional to the excess of its temperature above
that of its environment.
Newton’s Law of Cooling Formula
Formula: T (t) = Te + (T0 − Te )e kt , where
T (t) = temperature of object at time t,
T0 = original temperature of object,
Te = temperature of environment (medium/room temp),
t = time,
k = a constant dependent on the material properties of the object.
Proof: Newton’s law (as a formula) says that:
dT
= k(T − Te )
dt
where y = T − Te in the original equation y 0 = ky .
The solution is then y = y0 e kt where y = T − Te and y0 = T0 − Te :
T − Te = (T0 − Te )e kt
AMAT 217 (University of Calgary)
Fall 2013
11 / 16
Example
An object of temperature 60◦ C is placed in a room with temperature 20◦ C .
At the end of 5 minutes the object has cooled to a temperature of 40◦ C .
What is the temperature of the object at the end of 15 minutes?
The formula is T (t) = Te + (T0 − Te )e kt , where
I
I
I
I
T (t) = temperature of object at time t,
T0 = original temperature of object,
Te = temperature of environment (medium/room temp),
k = a constant dependent on the material properties of the object.
We are given that T0 = 60 and Te = 20:
T (t) = 20 + 40e kt
We are also given that T (5) = 40 which will allow us to find k:
40 = 20 + 40e k(5)
→
20 = 40e 5k
→
k=
ln(1/2)
5
We need to find T (15):
T (15) = 20 + 40e
ln(1/2)
(15)
5
= 20 + 40e ln(1/2)
3
= 20 + 40
3
1
= 25
2
Therefore, the temperature at the end of 15 minutes is 25◦ C .
AMAT 217 (University of Calgary)
Fall 2013
12 / 16
Hyperbolic Functions
We define two new functions called hyperbolic sine and hyperbolic cosine:
sinh x =
e x − e −x
2
cosh x =
e x + e −x
2
We define hyperbolic functions for the others as you would expect:
tanh x =
sinh x
cosh x
cschx =
1
sinh x
sechx =
1
cosh x
cothx =
1
tanh x
The graphs of sinh x and cosh x are:
Note: cosh x ≥ 1.
AMAT 217 (University of Calgary)
Fall 2013
13 / 16
Example
Evaluate tanh t given that sinh t = 2.
Recall that we have the following formula:
cosh2 t − sinh2 t = 1
Then solving for cosh t we get:
cosh2 t = 1 + sinh2 t
→
cosh t =
p
1 + sinh2 t
We took the positive square root since cosh x ≥ 1.
As sinh t = 2 we get:
p
√
cosh t = 1 + 22 = 5
Now by definition of tanh t we have:
tanh t =
sinh t
2
=√
cosh t
5
Side Note: By dividing the formula above by cosh2 t we get the formula:
1
1 − tanh2 t =
→
sech2 t = 1 − tanh2 t
cosh2 t
AMAT 217 (University of Calgary)
Fall 2013
14 / 16
An Application of Hyperbolic Functions
The most famous application of hyperbolic functions is when describing the shape of a
hanging chain. It can be proved that if a heavy flexible chain is suspended between two
points at the same height, then it takes the shape of a curve with equation
y = c + a cosh(x /a).
This is what we call a catenary (the Latin word catena means chain).
AMAT 217 (University of Calgary)
Fall 2013
15 / 16
A silly joke?
AMAT 217 (University of Calgary)
Fall 2013
16 / 16