Section 2.3 (cont.)

Lesson 16
Higher-Order
Derivatives
Real World
Examples
Acceleration
Section 2.3 (cont.)
February 19th, 2014
Lesson 16
Higher-Order
Derivatives
Real World
Examples
Acceleration
In this lesson, we will introduce the notion of higher-order
derivatives.
As a motivating example, let f (x) = 3x 2 − 7x + 8 and
g (x) = 6x − 7. Then
f 0 (x) = 6x − 7
and g 0 (x) = 6.
Note that f 0 (x) = g (x). Therefore g 0 (x) is the derivative of
f 0 (x). Put another way, g 0 (x) is the second derivative of f (x).
Lesson 16
Higher-Order
Derivatives
Definition
The second derivative of a function is the derivative of the
2
derivative. If y = f (x), we write ddxy2 or f 00 (x).
Real World
Examples
Acceleration
Example
Compute the second derivative of the given functions.
f (x) = x 4 + 10x 3 − 12 x 2 +
√
3 x
2
The first step in computing the second derivative of a
function is always computing the first derivative.
f 0 (x) = 4x 3 + 30x 2 − x +
f 00 (x) = 12x 2 + 60x − 1 +
3
√
4 x
√3
8 x3
Lesson 16
Higher-Order
Derivatives
Real World
Examples
Acceleration
Notice that there’s no reason to stop at two. If we differentiate
n
n times, we get the nth derivative. If y = f (x), we write ddxyn or
f (n) (x).
Example
Compute the second derivative of the given functions.
g (t) = (t 2 + 2)(3 − 4t 2 )
g 0 (t) = 2t(3 − 4t 2 ) + (t 2 + 2)(−8t)
We could take the second derivative now, but simplifying
first will make taking the derivative easier.
g 0 (t) = 6t − 8t 3 − 8t 3 − 16t = −16t 3 − 10t
g 00 (t) = −48t 2 − 10
Lesson 16
Example
Higher-Order
Derivatives
Real World
Examples
The manager of the Many Facets jewelry store models total
sales by the function
Acceleration
S(t) =
2000t
4 + 0.3t
where t is the time (years) since the year 2010 and S is
measured in thousands of dollars.
a. At what rate were sales changing in the year 2012?
b. What happens to sales in the long run (that is, as
t → +∞)?
Lesson 16
Higher-Order
Derivatives
Real World
Examples
a. Since S(t) gives the sales, the derivative S 0 (t) will give
the rate of change of sales.
S 0 (t) =
Acceleration
(4 + 0.3t)2000 − 2000t(0.3)
8000
=
2
(4 + 0.3t)
(4 + 0.3t)2
Since t is measured in years after 2010, the rate that sales
8000
are changing in 2012 is S 0 (2) = (4.6)
2 ≈ 378.07 thousand
dollars per year.
b. To determine what happens to sales in the long run, we
take the limit of S(t) as t goes to infinity.
2000t
2000
=
≈ $6, 666.67 thousand
t→∞ 4 + 0.3t
0.3
lim
Lesson 16
Before working the next example, we make a quick digression.
Higher-Order
Derivatives
Real World
Examples
Acceleration
The acceleration of an object tells how quickly that object is
speeding up or slowing down. In other words, it tells us the
rate of change of velocity. Therefore acceleration is the
derivative of velocity.
Similarly, the velocity of an object tells how quickly that object
is moving, i.e. changing position. Therefore velocity is the
derivative of position.
Combining these two points, we have that acceleration is the
second derivative of position.
Lesson 16
Example
After t hours of an 8-hour trip, a car has gone
Higher-Order
Derivatives
D(t) = 48t +
Real World
Examples
11 2 2 3
t − t
5
7
Acceleration
kilometers.
a. Derive a formula expressing the acceleration of the car as
a function of time.
b. At what rate is the velocity of the car changing with
respect to time at the end of 5 hours? Is the velocity
increasing or decreasing at this time?
c. By how much does the velocity of the car actually change
during the sixth hour?
Lesson 16
Higher-Order
Derivatives
a. Let V (t) and A(t) be the velocity and acceleration of the
car, respectively. Then
Real World
Examples
V (t) = D 0 (t) = 48 +
Acceleration
22
6
t − t 2,
5
7
and so
A(t) = V 0 (t) = D 00 (t) =
22 12
− t.
5
7
60
146
2
b. A(5) = 22
5 − 7 = − 35 ≈ −4.17 km/h . Since the value
is negative, the velocity is decreasing at this time.
c. V (6) − V (5) =
1524
35
−
340
7
= − 176
35 ≈ −5.03 km/h.

Download Report