2.2: Basic Differentiation Rules and Rates of Change Flashcards
Constant Rule (for any real number, c)
d/dx [c] = 0
Power Rule (n is a real number)
d/dx [x^n] = nx^n-1
Constant Multiplier Rule (for any real number, c)
d/dx [cf(x)] = cf’(x)
The Sum and Difference Rules (let f and g be differentiable)
1) d/dx[f(x) + g(x)] = f’(x) + g’(x)
2) d/dx[f(x) - g(x)] = f’(x) - g’(x)
Derivatives of Sine and Cosine (Special Limits)
1) lim x->0 (sin(x) / x) = 1
2) lim x->0 (1 - cos(x) / x) = 0
Trig Review
1) sin(A + B) = sin(A) cos(B) + cos(A) sin(B)
2) cos(A + B) = cos(A) cos(B) - sin(A) sin(B)
Algebra Review: Formula: y - x = ?
- (x - y)
Derivative of Sine (Formula)
d/dx(sin x) = cos x
Derivative of Cosine (Formula)
d/dx(cos x) = - sin x
Distance/Time (Formula)
1) D = RT
2) R = D/T
Average Rate of Change of a Function, s(t), on the interval [a,b]
v(avg) = D/T = Δs/Δt = (s(b) - s(a) /b - a)
Difference Between Instantaneous Velocity and Average Velocity
1) v(t) = s’(t)
2) s(b) - s(a) / b - a
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?
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
Use the position function, s(t) = -12 cos (t), where s is measured in feet, and t is in seconds
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
Find the position of the object at t = 0 and t = π
1) s(0) = -12cos(0) = -12 = -12 ft
2) s(π) = -12cos(π) = 12 = 12 ft
Find the equation for the object’s instantaneous velocity and evaluate it at t = 0 and t = π
v(t) = s’(t) = 12sin(t)
1) v(0) = 12 sin(0) = 0
2) v(π) = 12 sin(π) = 0
Find the average velocity of the object on the interval [0, π]
v(avg) = s(π) - s(0) / π - 0 = 12 - (-12) / π = 24/π ft/sec
When is the first time the position of the object is at 6 feet?
-12 cos t = 6 / -12
cost = -1/2
t = 2π/3 seconds
Find the equation for the object’s acceleration and evaluate it at t = 7π/6
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
A bowling ball is dropped from a 200 ft tall platform
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?
Find the position function
s(t) = 1/2(-32)t^2 + (0)t + (200)
= -16t^2 + 200
How long will it take for the bowling ball to hit the ground?
-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
How fast is it traveling when it hits the ground?
v(t) = s’(t)
= -32t
v(5sqrt(2)/2) = -32(5sqrt(2)/2)
= -80sqrt(2) ft/sec
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
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?
Determine the position and velocity function of the particle
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
Find the average velocity of the particle over the interval [0,5] or [a,b]
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
What is the maximum height the object reaches?
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
What is its instantaneous velocity as it hits the grounds?
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
Suppose the function (in cm) of a function is given by: s(t) = 1 + 2 sin(t). Time is in minutes
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]
What is the initial position of the particle
At t = 0, given: s(t) = 1 + 2 sin(t)
s(0) = 1 + 2 sin (0)
s(0) = 1 + 2 (0) = 1 cm
When is the first time particle’s position is 0?
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
Find the acceleration of the particle at t = 5pi/3
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
When is the first time the particle’s velocity is 0?
The particle’s velocity is 0, when:
v(t) = 2 cos (t) = 0, when cos (t) = 0
at t = pi/2
FInd the average velocity of the particle over the interval [2pi/3, 11pi/6]
avg velocity = s(b) - s(a) / b - a (formula)
Substitute 2pi/3, 11pi/6 to the function
Find s(11pi/6) = ?
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
Find s(2pi/3) = ?
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
Compute the average velocity
(avg velocity = s(b) - s(a) / b - a
s(11pi/6) - s(2pi/3) / 11pi/6 - 2pi/3
Simplify the denominator
11pi/6 - 2pi/3 = 11pi/6 - 4pi/6 = 7pi/6
