General Principles of Intellectual Property: Concepts of Intellectual Proper...
Higher Order Deriavatives
1. Composition of Functions:
The process of combining two or more functions in order to create
another function.
One function is evaluated at a value of the independent variable and
the result is substituted into the other function as the independent
variable.
The composition of functions f and g is written as:
𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥
1.7 – The Chain Rule
The composition of functions is a function inside another function.
8. 1.7 – The Chain Rule
The position of a particle moving along a coordinate line is, 𝑠 𝑡 = 12 + 4𝑡, with s in
meters and t in seconds. Find the rate of change of the particle's position at 𝑡 = 6
seconds.
𝑠 𝑡 = 12 + 4𝑡
𝑠 𝑡 = 12 + 4𝑡
1
2
𝑑𝑠
𝑑𝑡
= 𝑠′ 𝑡 =
1
2
12 + 4𝑡 −1
2 4
𝑑𝑠
𝑑𝑡
= 𝑠′ 𝑡 =
2
12 + 4𝑡
1
2
𝑎𝑡 𝑡 = 6,
𝑑𝑠
𝑑𝑡
= 𝑠′ 6 =
2
12 + 4 6
1
2
𝑑𝑠
𝑑𝑡
= 𝑠′ 6 =
1
3
𝑚𝑒𝑡𝑒𝑟𝑠/𝑠𝑒𝑐𝑜𝑛𝑑𝑠
9. 1.7 – The Chain Rule
The total outstanding consumer credit of a certain country can be modeled by 𝐶 𝑥 =
0.21𝑥4 − 5.98𝑥3 + 50.11𝑥2 − 18.29𝑥 + 1106.47 , where C is billion dollars and x is
the number of years since 2000.
a) Find
𝑑𝐶
𝑑𝑥
.
b) Using this model, predict how quickly outstanding consumer credit will be rising in
2010.
𝐶 𝑥 = 0.21𝑥4
− 5.98𝑥3
+ 50.11𝑥2
− 18.29𝑥 + 1106.47
𝑑𝐶
𝑑𝑥
= 0.84𝑥3 − 17.94𝑥2 + 100.22𝑥 − 18.29
a)
b) 𝑥 = 2010 − 2000 = 10 𝑦𝑒𝑎𝑟𝑠
𝑎𝑡 𝑥 = 10,
𝑑𝐶
𝑑𝑥
= 0.84 10 3
− 17.94 10 2
+ 100.22 10 − 18.29
𝑑𝐶
𝑑𝑥
= 29.91 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 𝑑𝑜𝑙𝑙𝑎𝑟𝑠/𝑦𝑒𝑎𝑟
12. 1.8 –Higher-Order Derivatives
Velocity: the change in position with respect to a change in time. It is a rate of
change with direction.
𝑣 𝑡 = 𝑠′ 𝑡 =
𝑑𝑠
𝑑𝑡
The velocity function, 𝑣 𝑡 , is obtain by differentiating the position function with
respect to time.
𝑠 𝑡 = 4𝑡2
+ 𝑡
𝑣 𝑡 = 𝑠′(𝑡) = 8𝑡 + 1
𝑠 𝑡 = 5𝑡3
− 6𝑡2
+ 6
𝑣 𝑡 = 𝑠′(𝑡) = 15𝑡2 − 12𝑡
Position, Velocity, and Acceleration
13. 1.8 –Higher-Order Derivatives
Velocity: the change in position with respect to a change in time. It is a rate of
change with direction.
𝑣 𝑡 = 𝑠′ 𝑡 =
𝑑𝑠
𝑑𝑡
The velocity function, 𝑣 𝑡 , is obtain by differentiating the position function with
respect to time.
𝑠 𝑡 = 4𝑡2
+ 𝑡
𝑣 𝑡 = 𝑠′(𝑡) = 8𝑡 + 1
𝑠 𝑡 = 5𝑡3
− 6𝑡2
+ 6
𝑣 𝑡 = 𝑠′(𝑡) = 15𝑡2 − 12𝑡
Position, Velocity, and Acceleration
14. Position, Velocity, and Acceleration
Acceleration: the change in velocity with respect to a change in time. It is a rate
of change with direction.
The acceleration function, 𝑎 𝑡 , is obtain by differentiating the velocity function
with respect to time. It is also the 2nd derivative of the position function.
𝑎 𝑡 = 𝑣′
𝑡 =
𝑑𝑣
𝑑𝑡
= 𝑠′′
𝑡 =
𝑑2 𝑠
𝑑𝑡2
𝑠 𝑡 = 4𝑡2 + 𝑡
𝑣 𝑡 = 𝑠′(𝑡) = 8𝑡 + 1
𝑠 𝑡 = 5𝑡3
− 6𝑡2
+ 6
𝑣 𝑡 = 𝑠′(𝑡) = 15𝑡2 − 12𝑡
𝑎 𝑡 = 𝑣′
𝑡 = 𝑠′′
𝑡 = 8 𝑎 𝑡 = 𝑣′ 𝑡 = 𝑠′′(𝑡) = 30𝑡 − 12
1.8 –Higher-Order Derivatives
15. The position of an object is given by 𝑠 𝑡 = 2𝑡2 + 8𝑡 , where s is measured in feet and
t is measured in seconds.
a) Find the velocity
𝑑𝑠
𝑑𝑡
and acceleration
𝑑𝑣
𝑑𝑡
functions.
b) What are the position, velocity, and acceleration of the object at 5 seconds?
𝑣 𝑡 =
𝑑𝑠
𝑑𝑡
= 4𝑡 + 8a)
b)
1.8 –Higher-Order Derivatives
𝑎 𝑡 =
𝑑𝑣
𝑑𝑡
= 4
𝑠 5 = 2 5 2
+ 8 5 = 90 𝑓𝑒𝑒𝑡
𝑣 5 = 4 5 + 8 = 28 𝑓𝑒𝑒𝑡/𝑠𝑒𝑐
𝑎 5 = 4 feet/sec/sec or 𝑓𝑒𝑒𝑡/𝑠𝑒𝑐 2
16. 1.8 –Higher-Order Derivatives
The position of a particle (in inches) moving along the x-axis after t seconds have
elapsed is given by the following equation: s(t) = t4 – 2t3 – 4t2 + 12t.
(a) Calculate the velocity of the particle at time t.
(b) Compute the particle's velocity at t = 1, 2, and 4 seconds.
(c) Calculate the acceleration of the particle after 4 seconds.
(d) When is the particle at rest?
𝑣 𝑡 =
𝑑𝑠
𝑑𝑡
= 4𝑡3 − 6𝑡2 − 8𝑡 + 12a)
b)
c)
d)
𝑣 1 = 2 𝑖𝑛𝑐ℎ𝑒𝑠/𝑠𝑒𝑐
𝑣 2 = 4 𝑖𝑛𝑐ℎ𝑒𝑠/𝑠𝑒𝑐
𝑣 4 = 140 𝑖𝑛𝑐ℎ𝑒𝑠/𝑠𝑒𝑐
𝑎 𝑡 =
𝑑𝑣
𝑑𝑡
= 12𝑡2 − 12𝑡 − 8
𝑎 4 = 136 𝑓𝑒𝑒𝑡/𝑠𝑒𝑐2
𝑣 𝑡 = 0 𝑎𝑡 𝑟𝑒𝑠𝑡
0 = 4𝑡3
− 6𝑡2
− 8𝑡 + 12
0 = 2𝑡2 2𝑡 − 3 − 4 2𝑡 − 3
0 = 2𝑡 − 3 2𝑡2 − 4
𝑡 =
3
2
, 1.414 𝑠𝑒𝑐.