Download Practice problems with rates of change

Section 3.4b
OTHER APPLICATIONS
WITH RATES OF CHANGE
The “Do Now” – p.130-131, #16
(a) How fast was the rocket climbing when the engine stopped?
v = 190 ft/sec
(b) For how many seconds did the engine burn?
2 seconds
(c) When did the rocket reach its highest point? What was its
velocity then?
After 8 seconds, and its velocity was 0 ft/sec
(d) When did the parachute pop out? How fast was the rocket
falling then?
After about 11 seconds, and it was falling at
about 90 ft/sec
The “Do Now” – p.130-131, #16
(e) How long did the rocket fall before the parachute opened?
About 3 seconds (from the rocket’s highest
point)
(f) When was the rocket’s acceleration greatest? When was
the acceleration constant?
The acceleration was greatest just before the
engine stopped. The acceleration was
constant from t = 2 to t = 11, while the rocket
was in free fall.
Derivatives in Economics
In manufacturing, the cost of production c(x) is a function of
x, the number of units produced. The marginal cost of
production is the rate of change of cost with respect to the
level of production, so it is called dc/dx.
Ex: Suppose c(x) represents the dollars needed to produce x
tons of steel in one week. It costs more to produce x + h tons
per week, and the cost difference divided by h is the average
cost of producing each additional ton:
c  x  h  c  x

h
The average cost of each of
the additional h tons produced
Derivatives in Economics
In manufacturing, the cost of production c(x) is a function of
x, the number of units produced. The marginal cost of
production is the rate of change of cost with respect to the
level of production, so it is called dc/dx.
The limit of this ratio as h  0 is the marginal cost of
producing more steel per week when the current production
is x tons.
Marginal cost of Production:
c  x  h  c  x
dc
 lim
dx h0
h
Guided Practice
Suppose it costs
c  x   x  6x  15x
3
2
dollars to produce x radiators when 8 to 10 radiators are
produced, and that r x  x3  3x 2  12 x
 
gives the dollar revenue from selling x radiators. Your shop
currently produces 10 radiators a day. Find the marginal cost
and marginal revenue.
The marginal cost of producing one more radiator a day when
10 are being produced is c 10
 
d 3
2
2

c  x    x  6 x  15 x   3x  12 x  15
dx
c 10  3 100 12 10  15  $195
Guided Practice
Suppose it costs
c  x   x  6x  15x
3
2
dollars to produce x radiators when 8 to 10 radiators are
produced, and that r x  x3  3x 2  12 x
 
gives the dollar revenue from selling x radiators. Your shop
currently produces 10 radiators a day. Find the marginal cost
and marginal revenue.
The marginal revenue:
d 3
2

r  x    x  3x  12 x   3x 2  6 x  12
dx
r 10  3 100  6 10  12  $252
Guided Practice
When a certain chemical was added to a nutrient broth in
which bacteria were growing, the bacterium population
continued to grow for a while but then stopped growing and
began to decline. The size of the population at time t (hours)
was b t  106  104 t  103 t 2

Find the growth rates at t = 0, t = 5, and t = 10 hours.
Bacteria growth rate:
b t   10,000  2000t
b  0  10,000 bacteria/hour
At t = 5: b  5  0 bacteria/hour
At t = 10: b 10  10,000 bacteria/hour
At t = 0:
Guided Practice
The position of a body at time t seconds is s  t  6t  9t
meters. Find the body’s acceleration each time the velocity
is zero.
2
Velocity: v t  s t  3t  12t  9
3
2
 
Acceleration: a  t   v  t   s  t   6t  12
Find when velocity is zero:
3t 2  12t  9  0
3  t  4t  3  0
2
3 t 1 t  3  0
 t  1,3
Find acceleration at these times:
a 1  6m sec
a  3  6m sec
2
2