2.2: Basic Differentiation Rules and Rates of Change

Question 1

Constant Rule (for any real number, c)

Answer 1

d/dx [c] = 0

Question 2

Power Rule (n is a real number)

Answer 2

d/dx [x^n] = nx^n-1

Question 3

Constant Multiplier Rule (for any real number, c)

Answer 3

d/dx [cf(x)] = cf’(x)

Question 4

The Sum and Difference Rules (let f and g be differentiable)

Answer 4

1) d/dx[f(x) + g(x)] = f’(x) + g’(x)
2) d/dx[f(x) - g(x)] = f’(x) - g’(x)

Question 5

Derivatives of Sine and Cosine (Special Limits)

Answer 5

1) lim x->0 (sin(x) / x) = 1
2) lim x->0 (1 - cos(x) / x) = 0

Question 6

Trig Review

Answer 6

1) sin(A + B) = sin(A) cos(B) + cos(A) sin(B)
2) cos(A + B) = cos(A) cos(B) - sin(A) sin(B)

Question 7

Algebra Review: Formula: y - x = ?

Answer 7

• (x - y)

Question 8

Derivative of Sine (Formula)

Answer 8

d/dx(sin x) = cos x

Question 9

Derivative of Cosine (Formula)

Answer 9

d/dx(cos x) = - sin x

Question 10

Distance/Time (Formula)

Answer 10

1) D = RT
2) R = D/T

Question 11

Average Rate of Change of a Function, s(t), on the interval [a,b]

Answer 11

v(avg) = D/T = Δs/Δt = (s(b) - s(a) /b - a)

Question 12

Difference Between Instantaneous Velocity and Average Velocity

Answer 12

1) v(t) = s’(t)
2) s(b) - s(a) / b - a

Question 13

Suppose your grandmother lives 400 miles away and you decided to take a drive to visit her.

Questions:
1) What was the average velocity of your entire trip?
2) What was your instantaneous velocity at t = 1 hour, and at t = 3 hour?

Answer 13

1) v(avg) = 400/8 = 50 ~ 50 mph

2)
At t = 1
v(1) = s’(1) = 30 mphs

At t = 3
v(3) = s’(3) = 65 mphs

Question 14

Use the position function, s(t) = -12 cos (t), where s is measured in feet, and t is in seconds

Answer 14

1) Find the position of the object at t = 0 and t = π
2) Find the equation for the object’s instantaneous velocity, and evaluate it at t = 0 and t = π
3) Find the average velocity of the object on the interval [0, π]
4) When is the first time the position of the object is at 6 feet?
5) Find the equation for the object’s acceleration and evaluate it at t = 7pi/6

Question 15

Find the position of the object at t = 0 and t = π

Answer 15

1) s(0) = -12cos(0) = -12 = -12 ft
2) s(π) = -12cos(π) = 12 = 12 ft

Question 16

Find the equation for the object’s instantaneous velocity and evaluate it at t = 0 and t = π

Answer 16

v(t) = s’(t) = 12sin(t)
1) v(0) = 12 sin(0) = 0
2) v(π) = 12 sin(π) = 0

Question 17

Find the average velocity of the object on the interval [0, π]

Answer 17

v(avg) = s(π) - s(0) / π - 0 = 12 - (-12) / π = 24/π ft/sec

Question 18

When is the first time the position of the object is at 6 feet?

Answer 18

-12 cos t = 6 / -12
cost = -1/2
t = 2π/3 seconds

Question 19

Find the equation for the object’s acceleration and evaluate it at t = 7π/6

Answer 19

a(t) = v’(t) = d/dt[12 sint t] = 12 cos(t)
a(7π/6) = 12 cos (7π/6)
= 12( -cos (π/6))
= 12(-sqrt(3)/2)
= -6sqrt(3) ft/sec

Question 20

A bowling ball is dropped from a 200 ft tall platform

Answer 20

1) Find the position function
2) How long will it take for the bowling ball to hit the ground?
3) How fast is it traveling when it hits the ground?

Question 21

Find the position function

Answer 21

s(t) = 1/2(-32)t^2 + (0)t + (200)
= -16t^2 + 200

Question 22

How long will it take for the bowling ball to hit the ground?

Answer 22

-16t^2 + 200 = 0
200 = 16t^2
200/16 = t^2
sqrt(25/2) = sqrt(t^2)
t = 5/sqrt(2 (sqrt(2)/sqrt(2)) = 5sqrt(2)/2 sec

Question 23

How fast is it traveling when it hits the ground?

Answer 23

v(t) = s’(t)
= -32t
v(5sqrt(2)/2) = -32(5sqrt(2)/2)
= -80sqrt(2) ft/sec

Question 24

A particle is thrown upward with an initial velocity of 64 ft/sec from a height of 192 feet. Use the position function s(t) = -16t^2 + vot + so. Include appropriate units in all answers

Answer 24

1) Determine the position and velocity function of the particle
2) Find the average velocity of the particle over the interval [0,5]
3) What is the maximum height the object reaches?
4) What is its instantaneous velocity as it hits the grounds?

Question 25

Determine the position and velocity function of the particle

Answer 25

1) Position Function:
s(t) = -16t^2 + vot + so
(substitute vo = 64 ft/sec and so = 192 ft)
s(t) = -16t^2 + 64t + 192

2) Velocity Function:
v(t) = d/dt (-16t^2 + 64t + 192)
v(t) = -32t + 64

Question 26

Find the average velocity of the particle over the interval [0,5] or [a,b]

Answer 26

1) Formula: avg velocity = s(b) - s(a) / b - a (if given a = 0 and b = 5)

2) Calculate s(0) and s(5)
s(0) = -16(0)^2 + 64(0) + 192 = 192 feet (a)
s(5) = -16(5)^2 + 64(5) + 192 = 112 feet (b)

3) avg velocity:
112 - 192 / 5 - 0 = -80/5 = -16 ft/sec

Question 27

What is the maximum height the object reaches?

Answer 27

1) To find max height, need to find the velocity is 0: v(t) = -32t + 64
2) Set the velocity equal to 0, and solve for t
-32t + 64 = 0
32t/32 = 64/32
t = 2 seconds
3) Substitute t = 2 to the function
s(2) = -16(2)^2 +64(2) + 192
= -16(4) + 128 + 192
= -64 + 128 + 192 = 256 feet

Question 28

What is its instantaneous velocity as it hits the grounds?

Answer 28

1) Set the position function equal to 0, s(t) = 0
-16t^2 + 64t + 192 = 0
2) Divide by -16 to simplify
t^2 - 4t - 12 = 0
Use quadratic to solve: -b +/- sqrt(b^2 -4ac) / 2a
3) Find the instantaneous velocity using velocity function
v(6) = -32(6) + 64 = -192 + 64 = -128 ft/sec

Question 29

Suppose the function (in cm) of a function is given by: s(t) = 1 + 2 sin(t). Time is in minutes

Answer 29

1) What is the initial position of the particle
2) When is the first time particle’s position is 0?
3) Find the acceleration of the particle at t = 5pi/3
4) When is the first time the particle’s velocity is 0?
5) Find the average velocity of the particle over the interval [2pi/3, 11pi/6]

Question 30

What is the initial position of the particle

Answer 30

At t = 0, given: s(t) = 1 + 2 sin(t)
s(0) = 1 + 2 sin (0)
s(0) = 1 + 2 (0) = 1 cm

Question 31

When is the first time particle’s position is 0?

Answer 31

1) Find time (t) when the particle’s position is zero. Set it to 0
1 + 2 sin(t) = 0
-1/2 = 2 sin(t)
-1/2 = sin (t)

2) Determine when s(t) = -1/2. The general solution for it: t = 7pi/6

Question 32

Find the acceleration of the particle at t = 5pi/3

Answer 32

Step 1: Find the velocity function v(t) => s(t) = 1 + 2 sin (t)
v(t) = d/dt [1 + 2 sin (t)] = 2 cos (t)
Step 2: Find the acceleration function a(t) => v(t) = 2 cos (t)
a(t) = d/dt [2 cos(t)] = -2 sin(t)
Step 3: Find the acceleration at t = 5pi/3
a(5pi/3) = -2 sin (5pi/3), since sin(5pi/3) = -sqrt(3)/2
a(5pi/3) = -2(-sqrt(3)/2) = sqrt(3) cm/min^2

Question 33

When is the first time the particle’s velocity is 0?

Answer 33

The particle’s velocity is 0, when:
v(t) = 2 cos (t) = 0, when cos (t) = 0
at t = pi/2

Question 34

FInd the average velocity of the particle over the interval [2pi/3, 11pi/6]

Answer 34

avg velocity = s(b) - s(a) / b - a (formula)
Substitute 2pi/3, 11pi/6 to the function

Question 35

Find s(11pi/6) = ?

Answer 35

s(11pi/6) = 1 + 2 sin (11pi/6), since sin(11pi/6) = -1/2, we have:
s(11pi/6) = 1 + 2 (-1/2) = 1 - 1 = 0

Question 36

Find s(2pi/3) = ?

Answer 36

s(2pi/3) = 1 + 2 sin (2pi/3), since sin(2pi/3) = sqrt(3)/2, we have:
s(2pi/3) = 1 + 2 (sqrt(3)/2) = 1+ sqrt(3) cm

Question 37

Compute the average velocity
(avg velocity = s(b) - s(a) / b - a

Answer 37

s(11pi/6) - s(2pi/3) / 11pi/6 - 2pi/3

Question 38

Simplify the denominator

Answer 38

11pi/6 - 2pi/3 = 11pi/6 - 4pi/6 = 7pi/6