MTH1020 · Analysis of Change
Applications of Differentiation: Sketching, Related Rates & Optimisation
Week 9 of Monash MTH1020 Analysis of Change completes the differential-calculus applications. It brings together curve sketching from f′ and f″, limits at infinity and asymptotes, and the growth hierarchy (logarithms ≪ powers ≪ exponentials), then develops two modelling skills: related rates, where the chain rule links the rates of change of connected quantities, and optimisation, where a modelled quantity is maximised or minimised. Mid-semester Test 3 is sat in the Week-10 applied class, so this material is tested close to when it is taught, and the modelling questions recur in the final.
What this chapter covers
- 01Curve sketching from the calculus: domain, intercepts, f′ (increasing/decreasing), f″ (concavity), asymptotes
- 02Limits at infinity: behaviour of xⁿ, 1/xⁿ and eˣ as x → ±∞; horizontal and vertical asymptotes
- 03The growth hierarchy for a > 1, p > 0: logarithms ≪ powers ≪ exponentials, so lim(x→∞) xᵖ/aˣ = 0
- 04Related rates: linking dx/dt and dy/dt through an equation and the chain rule
- 05Related-rates strategy: diagram, name variables, write given/required rates, relate, differentiate, substitute
- 06Optimisation: identify the quantity, express it as a function of one variable with a domain, find the absolute extremum
- 07Classic setups: expanding circle/sphere, draining tank, ladder; minimal-surface cylinder (height = diameter)
- 08The Intermediate Value Theorem for locating roots on a continuous interval
Related rates: the inflating spherical balloon
- +1Write the relating equation and identify the rates. The volume of a sphere is V = (4/3)πr³. We are given dV/dt = 100 cm³/s and asked for dr/dt when r = 5 cm. Both V and r are functions of time t.
- +1Differentiate with respect to time using the chain rule. d/dt[(4/3)πr³] = (4/3)π·3r²·(dr/dt) = 4πr²·(dr/dt). So dV/dt = 4πr²·(dr/dt).
- +1Substitute the known values. 100 = 4π(5)²·(dr/dt) = 4π·25·(dr/dt) = 100π·(dr/dt).
- +1Solve for the required rate. dr/dt = 100/(100π) = 1/π ≈ 0.318 cm/s.
Key terms
- Related rates
- Problems linking the time-rates of two quantities via an equation and the chain rule. Differentiate the relation with respect to t, then substitute the instantaneous values.
- Optimisation
- Maximising or minimising a modelled quantity: express it as a function of a single variable with a stated domain, then find the absolute extremum using the derivative tests.
- Limit at infinity
- The behaviour of f(x) as x → ±∞. Finite limits give horizontal asymptotes; where a rational function's denominator → 0 there are vertical asymptotes.
- Growth hierarchy
- For a > 1 and p > 0, logarithms grow slower than powers, which grow slower than exponentials: lim(x→∞) xᵖ/aˣ = 0 and lim(x→∞) (logₐx)/xᵖ = 0.
- Curve sketching
- Assembling a graph from domain, intercepts, sign of f′ (increasing/decreasing), sign of f″ (concavity), critical and inflection points, and asymptotic behaviour.
- Intermediate Value Theorem
- If f is continuous on [a, b] and k lies between f(a) and f(b), then f(c) = k for some c in (a, b). Used to guarantee a root exists in an interval.
Applications of Differentiation: Sketching, Related Rates & Optimisation FAQ
What's the general strategy for a related-rates problem?
Draw a diagram and name the variables, write the given and required quantities as derivatives with respect to time, find an equation relating the variables (often a geometric formula), then differentiate that equation with respect to t using the chain rule. Only at the final step do you substitute the specific instantaneous values and solve for the unknown rate. The order matters: differentiate the general relation first, substitute numbers last.
Why can't I substitute the given value before differentiating?
Because the quantity is still changing. If you set r = 5 in V = (4/3)πr³ before differentiating, you replace a variable with a constant, and the derivative of a constant is zero — so you lose all the information about how V responds to r. You must differentiate the relation while r is still a variable, obtaining dV/dt = 4πr²·(dr/dt), and only then put r = 5 in to evaluate the rate at that instant.
How is an optimisation question set up?
First identify exactly what is being maximised or minimised and write it as a formula. If that formula has two variables, use a constraint from the problem to eliminate one, so you have a single-variable function — and state its domain. Then apply the derivative tests to find the absolute extremum on that domain, remembering to check endpoints where relevant. Finish with a sentence answering the actual question, in the units asked for.
Can Sia help me with related rates and optimisation?
Yes. Sia can help you set up the relating equation, keep the differentiation and the substitution in the right order, and check your units and your choice of variable to eliminate — step by step on your own practice questions. It explains the method and checks your reasoning; it does not do graded assessment for you, and Monash academic-integrity rules apply.
Exam move
Treat related rates and optimisation as fixed procedures and rehearse the classic setups (expanding sphere or circle, draining tank, ladder, minimal-surface cylinder) until the steps are automatic. For related rates, always differentiate the general relation before substituting, and carry units to catch errors. For optimisation, practise reducing to one variable with a constraint, stating the domain, and justifying the extremum — Monash marks that justification. Fold in curve sketching by combining the Week-8 shape analysis with limits at infinity and asymptotes. Because MST3 sits in the Week-10 applied class, this material is tested soon after it is taught and again in the final, so drill it now and ask Sia to generate fresh modelling problems and check each step.
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