Calculus Notes: Differentiation Rules and Rates of Change
Section 2.2: Differentiation Rules and Rates of Change
Fundamental Differentiation Rules
Constant Rule: The derivative of a constant function is zero.
If f(x) = c, where c is a constant, then f'(x) = 0.
Power Rule: The derivative of x raised to a power n is n times x raised to the power of n-1.
If f(x) = x^n, then f'(x) = nx^{n-1} for any real number n.
Constant Multiple Rule: The derivative of a constant times a function is the constant times the derivative of the function.
If f(x) = c imes g(x), then f'(x) = c imes g'(x).
Sum/Difference Rule: The derivative of a sum or difference of functions is the sum or difference of their derivatives.
If h(x) = f(x) \pm g(x), then h'(x) = f'(x) \pm g'(x).
Derivative of Sine and Cosine:
If f(x) = \sin(x), then f'(x) = \cos(x).
If f(x) = \cos(x), then f'(x) = -\sin(x).
Derivative of the Natural Exponential:
If f(x) = e^x, then f'(x) = e^x.
Example 1: Finding Derivatives Using Basic Rules
A. f(x) = -2
Applying the Constant Rule: f'(x) = 0
B. g(x) = 3x - 1
Applying Constant Multiple, Power, Sum/Difference, and Constant Rules: g'(x) = 3(1)x^{1-1} - 0 = 3x^0 = 3
C. y = 2x^3 - x^2 + 3x
Applying Power, Constant Multiple, and Sum/Difference Rules: dy/dx = 2(3)x^{3-1} - 1(2)x^{2-1} + 3(1)x^{1-1} = 6x^2 - 2x + 3
D. y = 5 + \sin(x)
Applying Constant and Derivative of Sine Rules: dy/dx = 0 + \cos(x) = \cos(x).
E. g(t) = \pi \cos(t)
Applying Constant Multiple and Derivative of Cosine Rules: g'(t) = \pi(-\sin(t)) = -\pi \sin(t).
F. y = 3/x^2 = 3x^{-2}
Applying Power and Constant Multiple Rules: dy/dx = 3(-2)x^{-2-1} = -6x^{-3} = -6/x^3
G. y = 1/(\sqrt{x}) = x^{-1/2}
Applying Power Rule: dy/dx = (-1/2)x^{-1/2-1} = (-1/2)x^{-3/2} = -1/(2x^{3/2})
H. h(x) = x^{4/3}
Applying Power Rule: h'(x) = (4/3)x^{4/3-1} = (4/3)x^{1/3}
I. f(x) = \sqrt[3]{x} = x^{1/3}
Applying Power Rule: f'(x) = (1/3)x^{1/3-1} = (1/3)x^{-2/3} = 1/(3x^{2/3})
J. f(x) = 2\sin(x) + 3\cos(x) + 1
Applying Constant Multiple, Derivative of Sine/Cosine, and Constant Rules: f'(x) = 2\cos(x) + 3(-\sin(x)) + 0 = 2\cos(x) - 3\sin(x).
K. h(x) = e^x - 4x
Applying Derivative of Natural Exponential, Constant Multiple, and Power Rules: h'(x) = e^x - 4(1)x^{1-1} = e^x - 4.
L. g(x) = 7e^x - \cos(x)
Applying Constant Multiple, Derivative of Natural Exponential, and Derivative of Cosine Rules: g'(x) = 7e^x - (-\sin(x)) = 7e^x + \sin(x).
Example 2: Finding the Derivative at a Point
Find the derivative of y = 3x^5 + 10x^2 at the point (1, 13).
First, find the general derivative: dy/dx = 3(5)x^{5-1} + 10(2)x^{2-1} = 15x^4 + 20x
Then, evaluate the derivative at x = 1: dy/dx |_{x=1} = 15(1)^4 + 20(1) = 15 + 20 = 35.
Example 3: Finding Horizontal Tangents
A horizontal tangent occurs where the derivative of the function is equal to zero (i.e., the slope is zero).
A. y = x^2 + 1
Find the derivative: dy/dx = 2x
Set the derivative to zero: 2x = 0 \implies x = 0
Find the corresponding y-coordinate: y = (0)^2 + 1 = 1
The point with a horizontal tangent is (0, 1).
B. y = x^3 + x
Find the derivative: dy/dx = 3x^2 + 1
Set the derivative to zero: 3x^2 + 1 = 0 \implies 3x^2 = -1 \implies x^2 = -1/3
Since x^2 cannot be negative for real numbers, there are no real values of x for which the tangent is horizontal. Therefore, this function has no horizontal tangents.
Rates of Change and Motion Applications
Average Rate of Change: Represents the slope of the secant line between two given endpoints.
Formula: (f(x2) - f(x1)) / (x2 - x1)
Instantaneous Rate of Change: Represents the derivative of the function evaluated at a specific point. This is the slope of the tangent line at that point.
Motion Terminology (for position function s(t)):
s(t): Position of an object at time t.
v(t) = s'(t): Velocity of the object at time t. It is the first derivative of the position function.
a(t) = s''(t) = v'(t): Acceleration of the object at time t. It is the second derivative of the position function or the first derivative of the velocity function.
Projectile Motion Formula: Describes the height of an object under constant gravitational acceleration.
s(t) = (1/2)gt^2 + v0t + s0
s_0: Initial height (position at t=0).
v_0: Initial velocity (velocity at t=0).
g: Acceleration due to gravity.
g = -32 \text{ ft/sec}^2 (for feet per second units)
g = -9.8 \text{ m/sec}^2 (for meters per second units)
Speed: The magnitude of velocity.
\text{Speed} = |v(t)|
Speed Increasing/Decreasing:
Increasing: If velocity v(t) and acceleration a(t) have the same sign (both positive or both negative), the speed is increasing.
Decreasing: If velocity v(t) and acceleration a(t) have opposite signs (one positive, one negative), the speed is decreasing.
Worksheet Problems
Problem 1: Ball Dropped from CN Tower
A ball is dropped from an upper observation deck of the CN Tower in Toronto, 450 m above the ground.
Initial conditions: s0 = 450 \text{ m}, v0 = 0 \text{ m/s} (since it's dropped), g = -9.8 \text{ m/s}^2.
a. Write the equation of the distance of the ball after t seconds.
Using s(t) = (1/2)gt^2 + v0t + s0:
s(t) = (1/2)(-9.8)t^2 + (0)t + 450
s(t) = -4.9t^2 + 450
b. Find the average velocity of the ball from the time period of 3 seconds to 5 seconds.
s(3) = -4.9(3)^2 + 450 = -4.9(9) + 450 = -44.1 + 450 = 405.9 \text{ m}
s(5) = -4.9(5)^2 + 450 = -4.9(25) + 450 = -122.5 + 450 = 327.5 \text{ m}
Average Velocity = (s(5) - s(3)) / (5 - 3) = (327.5 - 405.9) / 2 = -78.4 / 2 = -39.2 \text{ m/s}
c. Find the velocity of the ball after 5 seconds.
First, find the velocity function: v(t) = s'(t) = d/dt (-4.9t^2 + 450) = -4.9(2)t^{2-1} + 0 = -9.8t
Evaluate at t = 5 seconds: v(5) = -9.8(5) = -49 \text{ m/s}
d. At what time will the ball hit the ground?
The ball hits the ground when s(t) = 0
-4.9t^2 + 450 = 0
4.9t^2 = 450
t^2 = 450 / 4.9 \approx 91.8367
t = \sqrt{91.8367} \approx 9.583 \text{ seconds} (only positive time is physically relevant)
e. With what velocity will the ball hit the ground?
Use the velocity function v(t) = -9.8t and the time t \approx 9.583 \text{ s} when it hits the ground.
v(9.583) = -9.8(9.583) \approx -93.9134 \text{ m/s}
Problem 2: Ball Thrown into the Air
If a ball is thrown into the air with a velocity of 40 ft/s, its height in feet after t seconds is given by y = -16t^2 + 40t.
Initial conditions match the formula: g = -32/2 = -16 \text{ ft/s}^2, v0 = 40 \text{ ft/s}, s0 = 0 \text{ ft}.
a. Find the average velocity for the time period beginning when t = 2 lasting:
The height function is y(t) = -16t^2 + 40t
y(2) = -16(2)^2 + 40(2) = -16(4) + 80 = -64 + 80 = 16 \text{ ft}.
i) .5 s (interval [2, 2.5])
y(2.5) = -16(2.5)^2 + 40(2.5) = -16(6.25) + 100 = -100 + 100 = 0 \text{ ft}
Average Velocity = (y(2.5) - y(2)) / (2.5 - 2) = (0 - 16) / 0.5 = -16 / 0.5 = -32 \text{ ft/s}
ii) .1 s (interval [2, 2.1])
y(2.1) = -16(2.1)^2 + 40(2.1) = -16(4.41) + 84 = -70.56 + 84 = 13.44 \text{ ft}
Average Velocity = (y(2.1) - y(2)) / (2.1 - 2) = (13.44 - 16) / 0.1 = -2.56 / 0.1 = -25.6 \text{ ft/s}
iii) .05 s (interval [2, 2.05])
y(2.05) = -16(2.05)^2 + 40(2.05) = -16(4.2025) + 82 = -67.24 + 82 = 14.76 \text{ ft}
Average Velocity = (y(2.05) - y(2)) / (2.05 - 2) = (14.76 - 16) / 0.05 = -1.24 / 0.05 = -24.8 \text{ ft/s}
iv) .01 s (interval [2, 2.01])
y(2.01) = -16(2.01)^2 + 40(2.01) = -16(4.0401) + 80.4 = -64.6416 + 80.4 = 15.7584 \text{ ft}
Average Velocity = (y(2.01) - y(2)) / (2.01 - 2) = (15.7584 - 16) / 0.01 = -0.2416 / 0.01 = -24.16 \text{ ft/s}
b) Find the instantaneous velocity when t=2.
Find the velocity function: v(t) = y'(t) = d/dt (-16t^2 + 40t) = -16(2)t + 40(1) = -32t + 40
Evaluate at t = 2: v(2) = -32(2) + 40 = -64 + 40 = -24 \text{ ft/s}.
(Observation: The average velocities approach -24 ft/s as the time interval shrinks, confirming the instantaneous velocity calculation.)
Problem 3: Arrow Shot Upward on the Moon
If an arrow is shot upward on the moon with a velocity of 58 m/s, its height in meters after t seconds is given by y = 58t - 0.83t^2.
Note: The gravitational acceleration on the moon is approximately -1.66 m/s^2, so (1/2)g = -0.83.
a. Find the average velocity over the given time intervals:
y(1) = 58(1) - 0.83(1)^2 = 58 - 0.83 = 57.17 \text{ m}.
i) [1, 2]
y(2) = 58(2) - 0.83(2)^2 = 116 - 0.83(4) = 116 - 3.32 = 112.68 \text{ m}
Average Velocity = (y(2) - y(1)) / (2 - 1) = (112.68 - 57.17) / 1 = 55.51 \text{ m/s}
ii) [1, 1.5]
y(1.5) = 58(1.5) - 0.83(1.5)^2 = 87 - 0.83(2.25) = 87 - 1.8675 = 85.1325 \text{ m}
Average Velocity = (y(1.5) - y(1)) / (1.5 - 1) = (85.1325 - 57.17) / 0.5 = 27.9625 / 0.5 = 55.925 \text{ m/s}
iii) [1, 1.1]
y(1.1) = 58(1.1) - 0.83(1.1)^2 = 63.8 - 0.83(1.21) = 63.8 - 1.0043 = 62.7957 \text{ m}
Average Velocity = (y(1.1) - y(1)) / (1.1 - 1) = (62.7957 - 57.17) / 0.1 = 5.6257 / 0.1 = 56.257 \text{ m/s}
iv) [1, 1.01]
y(1.01) = 58(1.01) - 0.83(1.01)^2 = 58.58 - 0.83(1.0201) = 58.58 - 0.846683 = 57.733317 \text{ m}
Average Velocity = (y(1.01) - y(1)) / (1.01 - 1) = (57.733317 - 57.17) / 0.01 = 0.563317 / 0.01 = 56.3317 \text{ m/s}
v) [1, 1.001]
y(1.001) = 58(1.001) - 0.83(1.001)^2 = 58.058 - 0.83(1.002001) = 58.058 - 0.83166083 = 57.22633917 \text{ m}
Average Velocity = (y(1.001) - y(1)) / (1.001 - 1) = (57.22633917 - 57.17) / 0.001 = 0.05633917 / 0.001 = 56.33917 \text{ m/s}
b. Find the instantaneous velocity after 1s.
Find the velocity function: v(t) = y'(t) = d/dt (58t - 0.83t^2) = 58(1) - 0.83(2)t = 58 - 1.66t
Evaluate at t = 1: v(1) = 58 - 1.66(1) = 56.34 \text{ m/s}.
(Observation: The average velocities are approaching 56.34 m/s, confirming the instantaneous velocity calculation.)
c. Is the arrow rising or falling at this time?
Since v(1) = 56.34 \text{ m/s} is positive, the arrow is rising at t = 1 second.