Math 151, c Benjamin Aurispa
4.5 Exponential Growth and Decay
Suppose the rate of change of y with respect to t is proportional to y itself. So there is some constant k
such that
dy
= ky
dt
The only solution to this equation is an exponential function of the form y(t) = Aekt where y(0) = A.
Verify this:
Notice that the proportion constant k becomes the coefficient of t in the exponent and that A is the initial
value of y. This is often also written as
y(t) = y0 ekt
Suppose the population of a town is growing so that the rate of growth per year is always 3 times the current
population size. If the current population is 10,000, find a function that models the population of the town
t years after the current year.
Example: A bacteria culture starts with 5000 bacteria and triples every 2 hours. Assume the population
grows at a rate proportional to its size.
(a) Find a function that models how much bacteria is present after t hours.
(b) At what rate is the bacteria population growing after 30 minutes?
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Math 151, c Benjamin Aurispa
Suppose for a function that the slope of the tangent line at any point is always 7 times the x-coordinate. If
the curve passes through the point (1, 9), find an equation for the curve.
The half-life of a radioactive substance is the amount of time it takes for half of the substance to decay.
This means that the rate of decay is proportional to the amount present at that moment.
Example: (#8, 4.5) Polonium-210 has a half-life of 140 days.
(a) If a sample has a mass of 200 mg, find the mass after t days.
(b) When will the mass be 10 mg?
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Math 151, c Benjamin Aurispa
Example: Suppose that a radioactive substance decays to 48% of its original amount in 12 days. What is
the half life of this substance?
Continuous Compound Interest: If an amount A0 is invested at an interest rate i (as a decimal) and interest
is compounded continuously, then the amount of the investment after t years is given by
A = A0 eit
Example: How long will it take for an investment to triple in value if the interest rate is 4% compounded
continuously?
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Math 151, c Benjamin Aurispa
Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature
difference between the object and its surroundings.
dy
= k(y − T )
dt
where y is the temperature of the object and T is the room temperature (i.e. the temperature of the
surroundings). The solution to this equation is a function of the form:
y(t) = (y0 − T )ekt + T
where y0 is the initial temperature of the object.
Example: An object with a temperature of 190◦ F is taken out of an oven and into a room with temperature
72◦ F. After 10 minutes the object has cooled to 175◦ F. Find a function y(t) that models the temperature
of the object after t minutes.
Suppose the rate of change of atmospheric pressure P with respect to altitude h is proportional to P (at a
fixed temperature). Then...
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Math 151, c Benjamin Aurispa
4.6 Inverse Trigonometric Functions
Remember that only one-to-one functions have inverses. So, in order to find the inverse functions for sine,
cosine, and tangent, we must restrict their domains to intervals where they are one-to-one.
To find the inverse sine function, we restrict the domain of sine to [−π/2, π/2].
We define the inverse sine function, sin−1 x by sin−1 x = y ↔ sin y = x.
sin−1 x, or arcsin x, has domain
and range
.
sin−1 x is the ANGLE in the interval [−π/2, π/2] whose sine is x.
Remember that for inverse functions, we have “cancellation” laws. So, for the sine function
sin(sin−1 x) = x for −1 ≤ x ≤ 1
sin−1 (sin x) = x for − π2 ≤ x ≤
π
2
Examples:
sin−1
√
arcsin(− 21 )
3
2
sin(sin−1
9
10 )
sin−1 2
sin−1 (sin 4π
3 )
5
arcsin(sin π5 )
Math 151, c Benjamin Aurispa
In order to have an inverse for cosine, we restrict the domain of cosine to the interval [0, π].
The inverse cosine function cos−1 is defined by cos−1 x = y ↔ cos y = x.
cos−1 x, or arccos x, has domain
and range
.
cos−1 x is the ANGLE in the interval [0, π] whose cosine is x.
Cancellation Laws.
cos(cos−1 x) = x for −1 ≤ x ≤ 1
cos−1 (cos x) = x for 0 ≤ x ≤ π
Examples:
cos−1 (−
arccos(0)
cos(cos−1
12
11 )
√
2
2 )
cos(cos−1 (− 17 ))
cos−1 (cos 7π
6 )
6
arccos(cos 2π
3 )
Math 151, c Benjamin Aurispa
In order to have an inverse for tangent, we restrict the domain of tangent to the interval (−π/2, π/2).
The inverse tangent function tan−1 is defined by tan−1 x = y ↔ tan y = x.
tan−1 x, or arctan x, has domain
lim arctan x =
x→∞
and range
.
lim arctan x =
x→−∞
tan−1 x is the ANGLE in the interval (−π/2, π/2) whose tangent is x.
Cancellation Laws.
tan(tan−1 x) = x for all x
tan−1 (tan x) = x for − π2 < x <
π
2
Examples:
tan−1 ( √13 )
arctan(−1)
π
))
tan−1 (tan(− 12
arctan(tan( 5π
4 ))
7
tan(arctan 1000)
Math 151, c Benjamin Aurispa
Examples: Evaluate the following expressions.
tan(sin−1 45 )
sin(2 cos−1 (− 32 ))
sin(tan−1 x)
cot(cos−1 x)
Find the domain of f (x) = arccos(3x − 8).
Find the domain of f (x) = tan−1 (x2 − 9).
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Math 151, c Benjamin Aurispa
Derivatives of Inverse Trig Functions
1
d
sin−1 x = √
dx
1 − x2
d
1
cos−1 x = − √
dx
1 − x2
d
1
tan−1 x =
dx
1 + x2
Show that
d
1
sin−1 x = √
.
dx
1 − x2
Find the derivatives of the following functions.
f (x) = arcsin(3x2 + ln x)
y = arctan(5x2 + 9x) arccos(5x)
Find the equation of the tangent line to the graph of f (x) = arctan(−2x) at the point where x =
9
√
3
2 .
Math 151, c Benjamin Aurispa
4.8 Indeterminate Forms and L’Hospital’s Rule
We have seen limits in the past that take the form 00 , ∞
∞ , and ∞ − ∞. When we encountered these, we had
to do something else...algebra, simplification, factoring...to be able to find the limit. These types of limits
are examples of indeterminate forms.
0
∞
f (x)
= or
, then we can use L’Hospital’s Rule to find the limit.
x→a g(x)
0
∞
If lim
L’Hospital’s Rule: Suppose f and g are differentiable functions. If lim
x→a
f (x)
0
∞
= or
, then
g(x)
0
∞
f (x)
f ′ (x)
= lim ′
x→a g(x)
x→a g (x)
lim
Notes: The limit could also be of the form
−∞ −∞
−∞ , ∞ ,
or
∞
−∞ .
f (x)
x→a g(x)
=
0
∞,
this is NOT indeterminate: The limit is 0.
f (x)
x→a g(x)
=
∞
0 ,
this is NOT indeterminate: The limit will be ∞ or −∞.
If lim
If lim
ex − 1
x→0 sin 3x
(1) (#5, 4.8) lim
(ln x)3
x→∞
x2
(2) (#14, 4.8) lim
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Math 151, c Benjamin Aurispa
Indeterminate Products: If lim f (x)g(x) = 0 · ∞, this limit is indeterminate. Why?
x→a
1 2
·x
x→∞ x
3
·x
x→∞ x2
lim
5
· x2
x→∞ x2
lim
lim
To find the limit, the goal is to write the indeterminate product in the form
Rule.
(1) lim+ csc x ln(1 + sin 7x)
x→0
(2) (#40, 4.8) lim xex
x→−∞
11
0
0
or
∞
∞
and use L’Hospital’s
Math 151, c Benjamin Aurispa
Indeterminate Difference: If lim [f (x) − g(x)] = ∞ − ∞, this limit is indeterminate. To find the limit, the
x→a
goal is once again to convert this difference into a quotient that we can use L’Hospital’s Rule on if necessary.
lim
x→0+
2x + 1 1
−
sin x
x
Indeterminate Powers: If lim [f (x)]g(x) is of the form 00 , ∞0 , or 1∞ , these are indeterminate. These cases
x→a
are treated by first taking the natural logarithm, which will make the limit of the form 0 · ∞. Then, proceed
as we did with indeterminate products. However, we must remember to “undo” the natural logarithm to
find our final answer. (Note that 0∞ is NOT an indeterminate form. A limit of this form will be 0.)
(1) lim
x→∞
4
1+ 2
x
x2
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Math 151, c Benjamin Aurispa
(2) lim xtan x
x→0+
(3) (#62, 4.8) lim (ex + x)1/x
x→∞
Summary: There are 7 basic indeterminate forms:
0 ±∞
,
, ∞ − ∞, 0 · ∞, 00 , ∞0 , 1∞ .
0 ±∞
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