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Rates of Change Applications

CSEC Additional Mathematics Essential Knowledge: Differentiation isn’t just an abstract mathematical concept – it has powerful real-world applications. Rates of change problems connect calculus to physics, economics, biology, and engineering. In CSEC Additional Mathematics, you’ll encounter problems involving velocity, acceleration, related rates, optimization, and growth/decay. Mastering these applications shows you truly understand what derivatives mean beyond the formulas.

Key Concept: The derivative \(\frac{dy}{dx}\) represents the instantaneous rate of change of \(y\) with respect to \(x\). In practical terms, this could represent: velocity (rate of change of position), acceleration (rate of change of velocity), marginal cost (rate of change of cost), population growth rate, or how quickly a balloon’s radius is expanding.

Part 1: Motion Along a Straight Line

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Displacement, Velocity, and Acceleration

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The Motion Formulas

For an object moving along a straight line with displacement \(s\) (position) at time \(t\):

\[ \begin{align*} \text{Velocity: } & v = \frac{ds}{dt} \\ \text{Acceleration: } & a = \frac{dv}{dt} = \frac{d^2s}{dt^2} \\ \text{Speed: } & |v| \text{ (magnitude of velocity)} \end{align*} \]

Memory Aid: “Differentiate displacement to get velocity, differentiate velocity to get acceleration.” Velocity tells you how fast and in which direction; speed tells you only how fast (always positive).

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Interpreting Signs in Motion

The signs of displacement, velocity, and acceleration have physical meaning:

Quantity Positive Sign Negative Sign Zero
Displacement (\(s\)) Right/up of origin Left/down of origin At origin
Velocity (\(v\)) Moving in positive direction Moving in negative direction Stationary (at rest)
Acceleration (\(a\)) Speeding up if \(v > 0\)
Slowing down if \(v < 0\)		 Slowing down if \(v > 0\)
Speeding up if \(v < 0\)		 Constant velocity
📝 Example 1: Basic Motion Problem

A particle moves along a straight line such that its displacement \(s\) meters from a fixed point \(O\) at time \(t\) seconds is given by \(s = t^3 – 6t^2 + 9t + 2\). Find:

(a) The velocity after 2 seconds

(b) When the particle is at rest

(c) The acceleration when \(t = 3\)

1
Find velocity function: \(v = \frac{ds}{dt} = 3t^2 – 12t + 9\)
2
(a) Velocity at \(t=2\): \(v(2) = 3(2)^2 – 12(2) + 9 = 12 – 24 + 9 = -3 \text{ m/s}\)
3
(b) At rest means \(v=0\): \(3t^2 – 12t + 9 = 0\)
Divide by 3: \(t^2 – 4t + 3 = 0\)
Factor: \((t-1)(t-3) = 0\)
So \(t = 1\) or \(t = 3\) seconds
4
Find acceleration function: \(a = \frac{dv}{dt} = 6t – 12\)
5
(c) Acceleration at \(t=3\): \(a(3) = 6(3) – 12 = 18 – 12 = 6 \text{ m/s}^2\)
📝 Example 2: Maximum Height Problem

A ball is thrown vertically upward. Its height \(h\) meters after \(t\) seconds is given by \(h = 20t – 5t^2\). Find:

(a) When the ball reaches its maximum height

(b) The maximum height

(c) When the ball hits the ground

1
Velocity function: \(v = \frac{dh}{dt} = 20 – 10t\)
2
(a) At maximum height, \(v=0\): \(20 – 10t = 0 \Rightarrow 10t = 20 \Rightarrow t = 2\) seconds
3
(b) Maximum height at \(t=2\): \(h(2) = 20(2) – 5(2)^2 = 40 – 20 = 20\) meters
4
(c) Hits ground when \(h=0\): \(20t – 5t^2 = 0\)
\(5t(4 – t) = 0\)
So \(t=0\) (start) or \(t=4\) seconds

Part 2: Related Rates Problems

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Connecting Different Rates of Change

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Related Rates Strategy

Related rates problems involve finding how different quantities change together. The key is using the Chain Rule:

\[\text{If } y = f(x) \text{ and } x = g(t), \text{ then } \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\]
1
Identify given rate(s) and rate to find
2
Find an equation relating the variables
3
Differentiate both sides with respect to time \(t\)
4
Substitute known values and solve
📝 Example 3: Expanding Circle

A circular oil spill is expanding such that its radius increases at a constant rate of 0.5 m/s. How fast is the area increasing when the radius is 10 m?

1
Given: \(\frac{dr}{dt} = 0.5\) m/s, \(r = 10\) m, find \(\frac{dA}{dt}\)
2
Relating equation: Area of circle \(A = \pi r^2\)
3
Differentiate with respect to \(t\): \(\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 2\pi r \cdot \frac{dr}{dt}\)
4
Substitute values: \(\frac{dA}{dt} = 2\pi (10) \cdot (0.5) = 10\pi\)
5
Interpretation: The area is increasing at \(10\pi \approx 31.4\) m²/s when radius is 10 m
📝 Example 4: Ladder Sliding Down Wall

A 5 m ladder leans against a vertical wall. The bottom is pulled away from the wall at 0.5 m/s. How fast is the top sliding down when the bottom is 3 m from the wall?

1
Diagram: Let \(x\) = distance from wall to ladder bottom, \(y\) = height of ladder top
2
Given: \(\frac{dx}{dt} = 0.5\) m/s, ladder length = 5 m (constant), find \(\frac{dy}{dt}\) when \(x = 3\)
3
Relating equation (Pythagoras): \(x^2 + y^2 = 5^2 = 25\)
4
Differentiate with respect to \(t\): \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\)
5
Find \(y\) when \(x=3\): \(3^2 + y^2 = 25 \Rightarrow y^2 = 16 \Rightarrow y = 4\) (positive since height)
6
Substitute into differentiated equation: \(2(3)(0.5) + 2(4)\frac{dy}{dt} = 0\)
\(3 + 8\frac{dy}{dt} = 0\)
7
Solve: \(8\frac{dy}{dt} = -3 \Rightarrow \frac{dy}{dt} = -\frac{3}{8} = -0.375\) m/s
8
Interpretation: The top is sliding down at 0.375 m/s (negative indicates downward direction)

Part 3: Optimization Problems

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Finding Maximum and Minimum Values

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Optimization Strategy

To find maximum or minimum values of a quantity subject to constraints:

1
Identify quantity to optimize (maximize/minimize)
2
Write equation for this quantity in terms of one variable
3
Find derivative and set equal to zero
4
Verify it’s a maximum/minimum (second derivative or sign test)
5
Answer the question in context
📝 Example 5: Maximum Area of Rectangle

A farmer has 100 m of fencing to enclose a rectangular area adjacent to a river (so only 3 sides need fencing). Find the dimensions that maximize the area.

1
Variables: Let \(x\) = length perpendicular to river, \(y\) = length parallel to river
2
Constraint (fencing): \(2x + y = 100\) (only 3 sides: two \(x\)’s and one \(y\))
So \(y = 100 – 2x\)
3
Quantity to maximize: Area \(A = x \cdot y = x(100 – 2x) = 100x – 2x^2\)
4
Find derivative: \(\frac{dA}{dx} = 100 – 4x\)
5
Set to zero: \(100 – 4x = 0 \Rightarrow 4x = 100 \Rightarrow x = 25\) m
6
Find \(y\): \(y = 100 – 2(25) = 50\) m
7
Verify maximum: Second derivative: \(\frac{d^2A}{dx^2} = -4 < 0\), so maximum
8
Maximum area: \(A = 25 \times 50 = 1250\) m²
📝 Example 6: Minimum Cost Container

A rectangular box with a square base and open top must have volume 32,000 cm³. Find dimensions that minimize the material used (surface area).

1
Variables: Let \(x\) = side of square base, \(h\) = height
2
Volume constraint: \(x^2 h = 32000 \Rightarrow h = \frac{32000}{x^2}\)
3
Surface area (material used): Base + 4 sides = \(x^2 + 4xh\)
Substitute \(h\): \(S = x^2 + 4x\left(\frac{32000}{x^2}\right) = x^2 + \frac{128000}{x}\)
4
Find derivative: \(\frac{dS}{dx} = 2x – \frac{128000}{x^2}\)
5
Set to zero: \(2x – \frac{128000}{x^2} = 0 \Rightarrow 2x = \frac{128000}{x^2} \Rightarrow 2x^3 = 128000\)
\(x^3 = 64000 \Rightarrow x = 40\) cm
6
Find \(h\): \(h = \frac{32000}{40^2} = \frac{32000}{1600} = 20\) cm
7
Verify minimum: Second derivative: \(\frac{d^2S}{dx^2} = 2 + \frac{256000}{x^3} > 0\) for \(x>0\), so minimum
8
Minimum surface area: \(S = 40^2 + \frac{128000}{40} = 1600 + 3200 = 4800\) cm²

Part 4: Economics Applications

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Marginal Analysis in Business

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Key Economics Concepts

In economics, derivatives represent marginal quantities:

Function Meaning Marginal Concept Formula
Cost: \(C(x)\) Total cost to produce \(x\) units Marginal Cost \(MC = C'(x)\) ≈ cost to produce one more unit
Revenue: \(R(x)\) Total revenue from selling \(x\) units Marginal Revenue \(MR = R'(x)\) ≈ revenue from one more unit
Profit: \(P(x)\) Total profit from \(x\) units Marginal Profit \(MP = P'(x)\) ≈ profit from one more unit

Key Relationship: Profit = Revenue – Cost, so \(P(x) = R(x) – C(x)\), and therefore \(P'(x) = R'(x) – C'(x)\). Maximum profit occurs when \(P'(x) = 0\), i.e., when Marginal Revenue = Marginal Cost.

📝 Example 7: Maximizing Profit

A company’s total cost function is \(C(x) = 1000 + 20x + 0.1x^2\) and price per item is \(p = 100 – 0.2x\), where \(x\) is number of items produced and sold. Find:

(a) Revenue function

(b) Profit function

(c) Production level for maximum profit

(d) Maximum profit

1
(a) Revenue: \(R(x) = \text{price} \times \text{quantity} = (100 – 0.2x)x = 100x – 0.2x^2\)
2
(b) Profit: \(P(x) = R(x) – C(x) = (100x – 0.2x^2) – (1000 + 20x + 0.1x^2)\)
\(= 100x – 0.2x^2 – 1000 – 20x – 0.1x^2\)
\(= 80x – 0.3x^2 – 1000\)
3
(c) Maximize profit: Find derivative: \(P'(x) = 80 – 0.6x\)
Set to zero: \(80 – 0.6x = 0 \Rightarrow 0.6x = 80 \Rightarrow x = \frac{80}{0.6} = \frac{800}{6} = \frac{400}{3} \approx 133.33\)
4
Check maximum: Second derivative: \(P”(x) = -0.6 < 0\), so maximum
5
(d) Maximum profit: \(P\left(\frac{400}{3}\right) = 80\left(\frac{400}{3}\right) – 0.3\left(\frac{400}{3}\right)^2 – 1000\)
\(= \frac{32000}{3} – 0.3\left(\frac{160000}{9}\right) – 1000\)
\(= \frac{32000}{3} – \frac{48000}{9} – 1000\)
\(= \frac{96000}{9} – \frac{48000}{9} – \frac{9000}{9} = \frac{39000}{9} \approx 4333.33\)

Part 5: CSEC Past Paper Questions

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Exam-Style Questions and Solutions

📝 CSEC Additional Mathematics 2018 Question 9 (Adapted)

A particle moves in a straight line so that its displacement \(s\) meters from a fixed point at time \(t\) seconds is given by \(s = t^3 – 9t^2 + 24t\).

(a) Find expressions for velocity and acceleration.

(b) Find when the particle is at rest.

(c) Determine the intervals when the particle is moving in the positive direction.

1
(a) Velocity: \(v = \frac{ds}{dt} = 3t^2 – 18t + 24\)
Acceleration: \(a = \frac{dv}{dt} = 6t – 18\)
2
(b) At rest when \(v=0\): \(3t^2 – 18t + 24 = 0\)
Divide by 3: \(t^2 – 6t + 8 = 0\)
Factor: \((t-2)(t-4) = 0\)
So \(t = 2\) or \(t = 4\) seconds
3
(c) Moving in positive direction when \(v>0\):
Test intervals: \(t<2\), \(24\)
For \(t=1\): \(v=3(1)^2-18(1)+24=3-18+24=9>0\)
For \(t=3\): \(v=3(9)-18(3)+24=27-54+24=-3<0\)
For \(t=5\): \(v=3(25)-18(5)+24=75-90+24=9>0\)
So moving positive when \(t<2\) or \(t>4\)
📝 CSEC Additional Mathematics 2016 Question 10 (Adapted)

A water tank has a rectangular base. The volume of the tank is 8 m³. The length of the base is twice the width. Material for the base costs $10 per m², and for the sides $6 per m². Find the cost of materials for the cheapest such tank.

1
Let width = \(x\) m, length = \(2x\) m, height = \(h\) m
2
Volume constraint: \(V = \text{length} \times \text{width} \times \text{height} = 2x \times x \times h = 2x^2h = 8\)
So \(h = \frac{8}{2x^2} = \frac{4}{x^2}\)
3
Surface areas:
Base: \(2x \times x = 2x^2\) (cost: \(10 \times 2x^2 = 20x^2\))
Sides: 2 of size \(2x \times h\) = \(4xh\), 2 of size \(x \times h\) = \(2xh\)
Total side area = \(4xh + 2xh = 6xh\) (cost: \(6 \times 6xh = 36xh\))
4
Total cost: \(C = 20x^2 + 36xh\)
Substitute \(h = \frac{4}{x^2}\): \(C = 20x^2 + 36x\left(\frac{4}{x^2}\right) = 20x^2 + \frac{144}{x}\)
5
Minimize cost: \(\frac{dC}{dx} = 40x – \frac{144}{x^2}\)
Set to zero: \(40x = \frac{144}{x^2} \Rightarrow 40x^3 = 144 \Rightarrow x^3 = \frac{144}{40} = 3.6\)
\(x = \sqrt[3]{3.6} \approx 1.53\) m
6
Minimum cost: \(C = 20(1.53)^2 + \frac{144}{1.53} \approx 20(2.34) + 94.12 \approx 46.80 + 94.12 = 140.92\) dollars

Quiz: Test Your Understanding

Rates of Change Applications Quiz
Question 1: A particle’s displacement is given by \(s = 2t^3 – 15t^2 + 36t\). Find when the acceleration is zero.
Answer:
Velocity: \(v = \frac{ds}{dt} = 6t^2 – 30t + 36\)
Acceleration: \(a = \frac{dv}{dt} = 12t – 30\)
Set \(a=0\): \(12t – 30 = 0 \Rightarrow 12t = 30 \Rightarrow t = 2.5\) seconds
Question 2: A spherical balloon is being inflated at 100 cm³/s. How fast is the radius increasing when the radius is 10 cm? (\(V = \frac{4}{3}\pi r^3\))
Answer:
Given: \(\frac{dV}{dt} = 100\), \(r = 10\), find \(\frac{dr}{dt}\)
\(V = \frac{4}{3}\pi r^3\)
\(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)
\(100 = 4\pi (10)^2 \frac{dr}{dt} = 400\pi \frac{dr}{dt}\)
\(\frac{dr}{dt} = \frac{100}{400\pi} = \frac{1}{4\pi} \approx 0.0796\) cm/s
Question 3: Find two positive numbers whose sum is 20 and whose product is maximum.
Answer:
Let numbers be \(x\) and \(20-x\)
Product: \(P = x(20-x) = 20x – x^2\)
\(\frac{dP}{dx} = 20 – 2x\)
Set to zero: \(20 – 2x = 0 \Rightarrow x = 10\)
Other number: \(20-10 = 10\)
Maximum product: \(10 \times 10 = 100\)
Question 4: The cost function is \(C(x) = 50 + 10x + 0.1x^2\) and price per item is \(p = 30 – 0.2x\). Find the production level that maximizes profit.
Answer:
Revenue: \(R(x) = (30-0.2x)x = 30x – 0.2x^2\)
Profit: \(P(x) = R(x) – C(x) = (30x-0.2x^2) – (50+10x+0.1x^2)\)
\(= 20x – 0.3x^2 – 50\)
\(\frac{dP}{dx} = 20 – 0.6x\)
Set to zero: \(20 – 0.6x = 0 \Rightarrow 0.6x = 20 \Rightarrow x = \frac{20}{0.6} = \frac{100}{3} \approx 33.33\)
Check: \(\frac{d^2P}{dx^2} = -0.6 < 0\), so maximum
Question 5: A 13 ft ladder leans against a wall. The bottom is pulled away at 2 ft/s. How fast is the top sliding down when the bottom is 5 ft from the wall?
Answer:
Let \(x\) = distance from wall, \(y\) = height
\(x^2 + y^2 = 13^2 = 169\)
Given: \(\frac{dx}{dt} = 2\), find \(\frac{dy}{dt}\) when \(x=5\)
When \(x=5\): \(25 + y^2 = 169 \Rightarrow y^2 = 144 \Rightarrow y = 12\)
Differentiate: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\)
\(2(5)(2) + 2(12)\frac{dy}{dt} = 0\)
\(20 + 24\frac{dy}{dt} = 0\)
\(\frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6} \approx -0.833\) ft/s (sliding down)

🎯 Key Concepts Summary

  • Motion Problems:
    • Displacement \(s(t)\): position from origin
    • Velocity \(v(t) = \frac{ds}{dt}\): rate of change of position
    • Acceleration \(a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}\): rate of change of velocity
    • At rest: \(v = 0\)
    • Maximum height/minimum: \(v = 0\), check sign of \(a\)
  • Related Rates Strategy:
    • Identify given rate and rate to find
    • Find equation relating variables
    • Differentiate with respect to time
    • Substitute known values and solve
  • Optimization Strategy:
    • Identify quantity to maximize/minimize
    • Write as function of one variable (use constraints)
    • Find derivative, set to zero
    • Verify maximum/minimum (second derivative test)
    • Answer in context
  • Economics Applications:
    • Cost \(C(x)\), Revenue \(R(x) = p(x) \cdot x\), Profit \(P(x) = R(x) – C(x)\)
    • Marginal Cost \(C'(x)\), Marginal Revenue \(R'(x)\), Marginal Profit \(P'(x)\)
    • Maximum profit when \(P'(x) = 0\) or \(R'(x) = C'(x)\)
  • Common Geometric Formulas:
    • Circle: Area \(A = \pi r^2\), Circumference \(C = 2\pi r\)
    • Sphere: Volume \(V = \frac{4}{3}\pi r^3\), Surface Area \(S = 4\pi r^2\)
    • Cylinder: Volume \(V = \pi r^2 h\), Surface Area \(S = 2\pi r^2 + 2\pi rh\)
    • Rectangle: Area \(A = lw\), Perimeter \(P = 2l + 2w\)

CSEC Exam Strategy: When solving rates of change problems: (1) Read carefully to understand the real-world context, (2) Draw diagrams for geometry problems, (3) Clearly define variables and write what’s given/what’s needed, (4) Show all steps including differentiation, (5) Include units in final answers, (6) Check if your answer makes sense in context. Remember that derivatives represent instantaneous rates of change, so your answers should be in units like “m/s”, “$ per item”, etc.

Common Mistakes to Avoid: 1. Forgetting to differentiate with respect to time in related rates
2. Not using the Chain Rule correctly in related rates
3. Forgetting to include all surfaces in surface area problems
4. Not checking if critical points are maxima/minima
5. Confusing displacement, velocity, and acceleration
6. Forgetting units or giving incorrect units
7. Not answering the question asked (e.g., finding x when asked for area)
8. Not considering practical constraints (e.g., dimensions must be positive)

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