Introduction
Math 1210
Calculus 1 section 3.7
Julian Chan
Department of Mathematics
Weber State University
2013
Introduction
Objectives
Objectives:
Review of rates of change
Applications of derivatives
Physics
Biology
Chemistry
Economics
Introduction
Rates of change
1
The chagne in x is denoted by ∆x = x2 − x1 , and it has an associated
change in y which is ∆y = y2 − y1 = f (x2 ) − f (x1 ).
2
The average rate of change is given by
∆y
.
∆x
0
3
The relationship to the derivative f (x) is given by the essential idea of a
limit:
0
∆y
.
f (x) = lim
∆x→0 ∆x
4
Another intrepretation is that for ∆x close to zero we have an
approximation:
0
f (x) ≈ lim
∆x→0
∆y
.
∆x
Introduction
Physics
If a stone is thrown vertically upward from the surface of the moon with a
velocity of 10 meters a second it’s height in meters after t seconds is
f (t) = 10t − .83t 2 .
What is the velocity of the stone after 3 seconds?
Introduction
Physics
If a stone is thrown vertically upward from the surface of the moon with a
velocity of 10 meters a second it’s height in meters after t seconds is
f (t) = 10t − .83t 2 .
What is the velocity of the stone after 3 seconds?
We recall tjhat velocity is distance divided by time or we can use the
equation D = R × T to see this fact.
The only gives average velocity so to find instantaneous velocity we
compute
0
∆f
= f (t) = 10 − 1.66x.
v (t) = lim
∆x→0 ∆t
Introduction
Example
After 3 seconds we compute v (3) = 5.02.
What is the velocity of the object when it hits the ground?
Introduction
Example
After 3 seconds we compute v (3) = 5.02.
What is the velocity of the object when it hits the ground?
We solve for where f (t) = 0 and find that t = 0 or t = 10/.83 = 12.04.
Introduction
Example
After 3 seconds we compute v (3) = 5.02.
What is the velocity of the object when it hits the ground?
We solve for where f (t) = 0 and find that t = 0 or t = 10/.83 = 12.04.
We compute that v (0) = 10 and v (12.04) = −9.9.
Notice that these two values are not equal!
Introduction
Example
The fact that these two values are not equal means that the velocity has
changed.
We have studeied with something that deals witht rate of change ...
Introduction
Example
The fact that these two values are not equal means that the velocity has
changed.
We have studeied with something that deals witht rate of change ...
The derivative.
In this instance the derivative is the rate of change of velocity know as
acceleration, and a(t) = −1.66 a constant!
Note that in this instance we have a relationship between a(t), v (t), and
p(t). Later will will discover that knowing a(t) is enough to determine
v (t) and p(t) given some information!!!
This says that the derivative of a function determines the function!!!
Introduction
Chemistry
Consider the chemical reaction:
2H2 + O2 −→ 2H2 O
A natural question is how much 2H2 O will there be at any given time?
Introduction
Chemistry
Consider the chemical reaction:
2H2 + O2 −→ 2H2 O
A natural question is how much 2H2 O will there be at any given time?
We need a mathematical model!
Introduction
Chemisty
In the chemical reaction:
2H2 + O2 −→ 2H2 O
we call A = 2H2 and B = O2 the reactants and C = 2H2 O the product.
We can then model any chemical reaction as:
A + B −→ C
In chemistry one often studies concentation, and are taught that
concentration of a reaction is given in moles per liter.
Recall 1mole = 6.022 × 1023 molecules per liter.
Introduction
Chemistry
We denote the concentations of A, B, and C respectively as [A], [B], [C ] which
are all functions of time since the concentrations are changing over time due to
the reactions.
If these quantities are changing over time they are a function and we can study
how they CHANGE.
The AVERAGE rate of change of the concentration [C ] is
∆[C ]
[C ](t2 ) − [C ](t2 )
=
δt
t2 − t1
We are interested in studying how the concentration changes at any given
instant so let ∆t → 0 and obtain the reaction rate:
d[C ]
dt
Introduction
Chemistry
d[C ]
dt
1
Since [C ] is increasing we have
2
Since [A] and [B] are producing [C ] we have
3
We thus have:
4
is positive.
d[A]
dt
and
d[B]
dt
d[C ]
d[A]
d[B]
=−
=−
dt
dt
dt
More generally we have that
aA + bB −→ cC + dD
and
d[C ]
d[A]
d[B]
d[D]
=
=−
=−
ddt
cdt
adt
bdt
are negative.
Introduction
Chemistry
Consider an instance in which two chemicals combine to form their product.
1
. Find the reaction rate of
Each chemical A and B combinate at a rate of [C ]+5
chemical [C ].
Introduction
Chemistry
Consider an instance in which two chemicals combine to form their product.
1
. Find the reaction rate of
Each chemical A and B combinate at a rate of [C ]+5
chemical [C ].
We find that
d[C ]
dt
= − d[A]
=
dt
1
5+[C ]
Determine the concentration
of chemical C by verifying that a solution to the
√
equation is [C ] = −5+ 225+8t is a solution if there is no chemical present at time
zero.
Introduction
Chemistry
1
Boyle’s law states that when a sample of gas is compressed a constant
temperature the product of the pressure and the volume remains constant:
PV = C
2
Find the rate of change of volume with respect to pressure.
3
A sample of gas is in a container at low pressure and is steadily compressed at
constant temperature for 10 minutes, Is the volume decreasing more rapidly at
the beginning or at the end of 10 minutes?
Introduction
Chemistry
1
The isothermal compressibility is defined as:
β=−
1 dV
V dP
−(1/(k/P)) × (−k/P 2 ) = 1/P.
Introduction
Biology
1
When we consider the flow of blood through a blood vessel such as a vein or
artery we model this with a cylinderical tube. For any segment of the vein we
let the cylinder have length l and radius R.
2
The velocity of the blood v is greatest along the central axis and decreases as
the distance r from the center to the wall becomes zero (at the wall).
3
The relationship between v and r is given by Poiseuille:
v=
p
(R 2 − r 2 )
4nl
4
Where n is the viscosity of blood.
5
and p is the pressure difference between the ends of the tubes.
Introduction
Biology
1
We assume that n and p are constant so that v is a function of r alone.
2
The rate of change of the velocity if
3
The rate of change of the velocity of blood is called the velocity gradient given
0
by v (t).
4
For human arteries we take n = .027 and R = .008 cm, l = 2, and p = 4000
dynes per cubic cm.
∆v
δr
.
Introduction
Biology
We find that v (r ) = 1.85 × 104 (.000064 − r 2 ) and
0
v (.002) = 1.11 cm/sec, and v (.002) = −74.
Introduction
Economics
1
The price of a new computer for a specific company is modeled by
p(x) = 2000 − 25x where is the number of compters sold in thousands.
2
Find the revenue of the product?
Introduction
Economics
1
The price of a new computer for a specific company is modeled by
p(x) = 2000 − 25x where is the number of compters sold in thousands.
2
Find the revenue of the product?
3
It’s R(x) = x × P(x) = 2000x − 25x 2 .
4
Intrepret
dR
dx
and compute it when x = 5.
Introduction
Economics
1
The company has a fixed cost of 2000 from the costs of renting the machinery
and fixed cost of research for the development of this computer. In addition for
each unit of x produced in thousands the company incurs a cost of production
from hiring the workers (and other stuff) of 125.
2
Find that total cost function of the company, and compute how much it costs
to produce x = 5.
3
Intrepret
dc
dt
and determine marginal cost when x = 5.
Introduction
Economics
1
Find the revenue of the company.
2
Find and intrepret
dR
dt
and compute it’s value when x = 5.