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International Baccalaureate · DP 2: Applications and Interpretation SL

Applications of Differentiation

⌚ ~106 min ▢ 11 blocks ✓ 45 exercises ➣ Prerequisites : Differentiation, Properties of Curves, Quadratic Functions, Modelling
I The Reciprocal Function and Average Cost
I.1 The Reciprocal Function
A delivery costs \(\dollar 60\), whatever the number \(q\) of items ordered. Shared among the items, it adds \(\dfrac{60}{q}\) dollars to each one: \(\dollar 12\) for \(5\) items, \(\dollar 3\) for \(20\), \(\dollar 0.60\) for \(100\). The more items, the smaller the share — but it never reaches zero. This is the shape of the reciprocal function \(x \mapsto \dfrac{1}{x}\), and it appears each time a fixed amount is spread over a quantity: a unit price, an average cost.
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Proposition — The Reciprocal Function at the Ends of Its Domain
The reciprocal function \(f(x) = \dfrac{1}{x}\) is defined for \(x \neq 0\).
\(x\) \(0.001\) \(0.01\) \(0.1\) \(1\) \(10\) \(100\) \(1\,000\)
\(\dfrac{1}{x}\) \(1\,000\) \(100\) \(10\) \(1\) \(0.1\) \(0.01\) \(0.001\)
  • For large values of \(x\), \(\dfrac{1}{x}\) is a small positive number, as close to \(0\) as wanted: the curve comes closer and closer to the \(x\)-axis without ever reaching it.
  • For positive values of \(x\) close to \(0\), \(\dfrac{1}{x}\) is a large positive number, as large as wanted: the curve rises along the \(y\)-axis without ever reaching it.
  • On the negative side, the same happens with negative values: \(\dfrac{1}{-1\,000} = -0.001\) and \(\dfrac{1}{-0.001} = -1\,000\).
The two axes are called the asymptotes of the curve (a hyperbola): the curve keeps getting closer to them, on each side, as far as one looks.
Method — Studying a Function \(x \mapsto ax + b + \dfrac{c}{x}\)
The steps below, written for \(x \mapsto ax + b + \dfrac{c}{x}\), apply to any sum of a polynomial function of degree at most \(3\) and of a multiple \(\dfrac{c}{x}\) of the reciprocal function (\(c \neq 0\)): only the derivative and its numerator change.
  1. Domain: \(x \neq 0\); in a context, often \((0, \infty)\) or an interval \((0, M]\).
  2. Derivative: differentiate term by term, with \(\left(\dfrac{c}{x}\right)' = -\dfrac{c}{x^2}\); for \(f(x) = ax + b + \dfrac{c}{x}\), \(f'(x) = a - \dfrac{c}{x^2}\); for \(f(x) = P(x) + \dfrac{c}{x}\) with \(P\) a polynomial, \(f'(x) = P'(x) - \dfrac{c}{x^2}\).
  3. One fraction: write the derivative over the denominator \(x^2\): \(f'(x) = \dfrac{ax^2 - c}{x^2}\), or \(\dfrac{x^2 P'(x) - c}{x^2}\). Since \(x^2 \gt 0\), the sign of \(f'(x)\) is the sign of its numerator, a polynomial whose factorised form is given, found from an obvious root, or found with technology.
  4. Table of variations: sign of the numerator, variations of \(f\), values at the stationary points.
  5. Ends of the domain: near \(0\), the term \(\dfrac{c}{x}\) takes huge values and imposes its sign (the sign of \(c\) on the right of \(0\), the opposite sign on the left); for large \(x\), \(\dfrac{c}{x}\) is negligible and \(f(x)\) is close to the polynomial part, \(ax + b\) or \(P(x)\) — the curve approaches its graph. For \(x \gt 0\), the difference \(f(x) - (ax + b) = \dfrac{c}{x}\) keeps the sign of \(c\): the curve stays on one side of the line and comes closer to it.
  6. Sketch: the curve, the graph of the polynomial part, the stationary points.
Example
Let \(f(x) = x + \dfrac{4}{x}\) for \(x \gt 0\).
  1. Compute \(f'(x)\) and check that \(f'(x) = \dfrac{(x - 2)(x + 2)}{x^2}\).
  2. Build the table of variations of \(f\) on \((0, \infty)\).
  3. Describe the behaviour of \(f\) near \(0\) and for large values of \(x\), then sketch the curve.

  1. \(\begin{aligned}[t]f'(x) &= 1 - \dfrac{4}{x^2} \\ &= \dfrac{x^2 - 4}{x^2} \\ &= \dfrac{(x-2)(x+2)}{x^2}\end{aligned}\)
  2. For \(x \gt 0\), \(x + 2 \gt 0\) and \(x^2 \gt 0\): \(f'(x)\) has the sign of \(x - 2\), negative on \((0, 2)\), zero at \(2\), positive on \((2, \infty)\). The minimum is \(f(2) = 2 + 2 = 4\).
    Figure
  3. Near \(0\), \(\dfrac{4}{x}\) is huge, so \(f(x)\) is huge: the curve rises along the \(y\)-axis. For large \(x\), \(\dfrac{4}{x}\) is close to \(0\), so \(f(x)\) is close to \(x\): the curve approaches the line \(y = x\).
    Figure

Skills to practice
  • Differentiating a Combination with the Reciprocal Function
I.2 Average Cost
Definition — Average Cost
When producing \(q\) units costs \(C(q)\) in total (\(q \gt 0\)), the average cost is the cost per unit $$A(q) = \dfrac{C(q)}{q}.$$ If the total cost is \(C(q) = aq^2 + bq + c\), with a fixed cost \(c\) paid even for \(q\) close to \(0\), then $$A(q) = aq + b + \dfrac{c}{q}:$$ the fixed cost spread over the \(q\) units is the reciprocal term. In the same way, buying \(q\) items at \(p\) dollars each with a fixed fee \(F\) gives a unit price \(p + \dfrac{F}{q}\).
Example
A workshop's daily cost for \(q\) units, in dollars, is \(C(q) = q^2 + 20q + 900\), for \(0 \lt q \leq 100\).
  1. Express the average cost \(A(q)\).
  2. Compute \(A'(q)\) and check that \(A'(q) = \dfrac{(q - 30)(q + 30)}{q^2}\).
  3. Build the table of variations of \(A\) and give the production that makes the average cost minimal, with this minimal cost.
  4. Interpret: why is the average cost high for a small production, and high again for a large one?

  1. \(A(q) = \dfrac{q^2 + 20q + 900}{q} = q + 20 + \dfrac{900}{q}\).
  2. \(\begin{aligned}[t]A'(q) &= 1 - \dfrac{900}{q^2} \\ &= \dfrac{q^2 - 900}{q^2} \\ &= \dfrac{(q-30)(q+30)}{q^2}\end{aligned}\)
  3. For \(q \gt 0\) the sign of \(A'(q)\) is that of \(q - 30\): \(A\) decreases on \((0, 30]\) and increases on \([30, 100]\). The minimal average cost is reached for \(30\) units: \(A(30) = 30 + 20 + 30 = 80\), that is \(\dollar 80\) per unit.
    Figure
  4. For a small production, the fixed cost \(900\) is shared among few units: the term \(\dfrac{900}{q}\) is large (\(A(5) = 205\)). For a large production, the term \(q\), which grows with the quantity (overtime, wear), takes over (\(A(100) = 129\)). Between the two, at \(30\) units, the two effects balance.

Skills to practice
  • Interpreting the Average Cost
  • Solving Problems with a Reciprocal Term
II Optimisation
II.1 Optimisation Problems
Optimisation is one of the most important applications of differential calculus. It involves finding the maximum or minimum value that a function can take, subject to given conditions. In many problems, this corresponds to finding the global maximum or minimum of that function on a specified domain.
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Definition — Optimisation
Optimisation is the process of finding a maximum or minimum value of a function, known as the objective function, subject to a set of constraints (or over a specified domain). The solution that achieves this is called the optimal solution.
Example
In machine learning, a key goal is to minimise a loss (error) function by adjusting model parameters. This optimisation process underlies how modern neural networks are trained.
Method — Solving Optimisation Problems
  1. Model: Identify the quantity to be optimised and express it as a function of a single variable (for example \(f(x)\)). Use any given constraints to eliminate other variables. Determine the feasible domain for \(x\).
  2. Differentiate: Check that \(f\) is differentiable on the domain, then find the first derivative of the function, \(f'(x)\).
  3. Find Stationary Points: Solve \(f'(x)=0\) for \(x\) in the domain.
  4. Test and Justify: Use the sign of the derivative (sign table of \(f'\), table of variations) to decide whether each stationary point is a local maximum, a local minimum, or neither.
  5. Check Endpoints: If the domain is a closed interval \([a, b]\), evaluate the function at the stationary points and also at the endpoints, \(f(a)\) and \(f(b)\), to find the global extremum on that interval.
  6. Conclusion: State the final answer clearly in the context of the problem, including units.
Example
A rectangle has a perimeter of 12 cm. Find its dimensions to maximise the area.

  • Model: Let the length be \(x\) and the width be \(y\).
    Figure
    The perimeter is \(2x+2y=12\), which simplifies to \(x+y=6\) and hence \(y=6-x\). The area is \(A=xy\). We write \(A\) as a function of \(x\): $$ A(x) = x(6-x) = 6x-x^2. $$ Since lengths must be non-negative, \(x \ge 0\) and \(y=6-x \ge 0 \implies x \le 6\). The domain is \(x \in [0,6]\).
  • Derivative: $$A'(x) = 6-2x.$$
  • Stationary point: $$ A'(x) = 0 \implies 6-2x=0 \implies 2x=6 \implies x=3. $$
  • Justification: sign of \(A'\). \(A'(x) = 6 - 2x\) is positive for \(x < 3\) and negative for \(x > 3\), so \(A\) is increasing on \([0, 3]\) and decreasing on \([3, 6]\): \(x = 3\) gives the maximum of \(A\).
  • Check endpoints: The domain is the closed interval \([0,6]\). We check the area at the stationary point and at the endpoints, where \(x = 0\) and \(x = 6\) give flattened rectangles of area \(0\):
    • \(A(0) = 6(0)-0^2 = 0\)
    • \(A(3) = 6(3)-3^2 = 18-9 = 9\)
    • \(A(6) = 6(6)-6^2 = 0\)
    The area is maximised at \(x=3\).
  • Conclusion: The area is maximised when the length is \(x=3\) cm. The corresponding width is \(y=6-3=3\) cm. The rectangle with maximum area is therefore a square of side 3 cm, with an area of \(9\text{ cm}^2\).

Example
A closed storage container with a square base must contain a volume of \(8\) m\(^3\). Find the dimensions that minimise the total surface area of the container.

  • Model: Let \(x\) be the side of the square base and \(h\) the height. The volume constraint gives $$x^2 h = 8 \implies h = \frac{8}{x^2}.$$ The surface consists of the base, the top and four lateral faces: $$S = 2x^2 + 4xh = 2x^2 + 4x \cdot \frac{8}{x^2} = 2x^2 + \frac{32}{x}.$$ The domain is \(x \in (0, \infty)\) — an open interval: there are no endpoints to check, so the justification must come from the variations.
  • Derivative: $$S'(x) = 4x - \frac{32}{x^2} = \frac{4x^3 - 32}{x^2}.$$
  • Stationary point: \(S'(x) = 0 \iff 4x^3 = 32 \iff x^3 = 8 \iff x = 2\).
  • Test and justify: the sign of \(S'(x)\) is the sign of \(4x^3 - 32\): negative for \(0 < x < 2\), positive for \(x > 2\). So \(S\) decreases on \((0, 2]\) and increases on \([2, \infty)\): the variation table shows that \(x = 2\) gives the global minimum on the whole domain — no endpoint check is possible, and none is needed.
  • Conclusion: \(x = 2\) m and \(h = \frac{8}{4} = 2\) m: the optimal container is a cube of side \(2\) m, with total surface \(S(2) = 2(4) + \frac{32}{2} = 24\) m\(^2\).

Skills to practice
  • Setting Up an Optimisation Problem
  • Solving Optimisation Problems
  • Solving Optimisation Problems in Context
II.2 An Optimum at an End Point
Example
A wire \(20\) cm long is cut into two pieces. A piece of length \(x\) cm is bent into a square, and the other piece into a circle. The values \(x = 0\) and \(x = 20\) mean that the whole wire makes a single shape, so \(0 \le x \le 20\).
  1. Show that the total area enclosed is \(A(x) = \dfrac{x^2}{16} + \dfrac{(20 - x)^2}{4\pi}\) cm\(^2\).
  2. Find the value of \(x\) for which \(A'(x) = 0\), and show that it gives the least total area.
  3. Where should the wire be cut to enclose the greatest total area?

  1. The square has perimeter \(x\), so its side is \(\dfrac{x}{4}\) and its area \(\dfrac{x^2}{16}\). The circle has circumference \(2\pi\rho = 20 - x\), so its radius is \(\rho = \dfrac{20 - x}{2\pi}\) and its area is $$\begin{aligned}[t] \pi\rho^2 &= \pi \times \frac{(20 - x)^2}{4\pi^2} && \text{(area of a disc, } \rho = \tfrac{20 - x}{2\pi}\text{)} \\ &= \frac{(20 - x)^2}{4\pi}. \end{aligned}$$ Adding the two areas gives \(A(x) = \dfrac{x^2}{16} + \dfrac{(20 - x)^2}{4\pi}\).
  2. Since \((20 - x)^2 = 400 - 40x + x^2\) has derivative \(-40 + 2x = -2(20 - x)\), $$\begin{aligned}[t] A'(x) &= \frac{2x}{16} - \frac{2(20 - x)}{4\pi} && \text{(power rule, term by term)} \\ &= \frac{x}{8} - \frac{20 - x}{2\pi} \\ &= \frac{\pi x - 4(20 - x)}{8\pi} && \text{(common denominator } 8\pi\text{)} \\ &= \frac{(\pi + 4)x - 80}{8\pi}. \end{aligned}$$ So \(A'(x) = 0\) for \(x_0 = \dfrac{80}{\pi + 4} \approx 11.2\). Since \(8\pi > 0\) and \(\pi + 4 > 0\), \(A'(x)\) has the sign of \((\pi + 4)x - 80\): negative for \(x < x_0\), positive for \(x > x_0\).
    Figure
    \(A\) decreases, then increases: \(x_0\) gives the least total area. With \(\dfrac{x_0}{4} = \dfrac{20}{\pi + 4}\) and \(20 - x_0 = \dfrac{20\pi}{\pi + 4}\), $$\begin{aligned}[t] A(x_0) &= \frac{400}{(\pi + 4)^2} + \frac{400\pi^2}{4\pi(\pi + 4)^2} \\ &= \frac{400 + 100\pi}{(\pi + 4)^2} \\ &= \frac{100}{\pi + 4} \approx 14.0 \text{ cm}^2. \end{aligned}$$
  3. The only stationary point is a minimum: the greatest value of \(A\) on \([0, 20]\) is at an end point. Compare the values at the end points: $$A(0) = \frac{400}{4\pi} = \frac{100}{\pi} \approx 31.8 \qquad \text{and} \qquad A(20) = \frac{400}{16} = 25.$$ Since \(\pi < 4\), \(\dfrac{100}{\pi} > \dfrac{100}{4} = 25\). The greatest total area, \(\dfrac{100}{\pi} \approx 31.8\) cm\(^2\), is reached for \(x = 0\): the wire is not cut, and all of it is bent into a circle.
Figure
Setting \(A'(x) = 0\) found the worst cut, not the best one: on a closed interval, the values at the end points are always compared with the values at the stationary points (step 5 of the Method).

Skills to practice
  • Finding an Optimum at an End Point
II.3 Optimisation with Technology
Example
An oil platform \(P\) stands \(4\) km off a straight coast; \(A\) is the point of the coast nearest to \(P\), and a refinery \(R\) stands on the coast, \(10\) km from \(A\). A pipeline runs straight under the sea from \(P\) to a point \(M\) of the coast between \(A\) and \(R\), with \(AM = x\) km, then along the coast from \(M\) to \(R\). Laying the pipeline costs \(5\) million dollars per km under the sea and \(3\) million dollars per km on land.
Figure
  1. Show that the total cost, in millions of dollars, is \(C(x) = 5\sqrt{16 + x^2} + 3(10 - x)\), for \(0 \le x \le 10\).
  2. Use technology to find where the pipeline should reach the coast, and the least cost.

  1. The triangle \(PAM\) is right-angled at \(A\), so by Pythagoras' theorem \(PM = \sqrt{AP^2 + AM^2} = \sqrt{16 + x^2}\) km, and \(MR = 10 - x\) km. The cost is \(5\) times the length under the sea plus \(3\) times the length on land: $$C(x) = 5\sqrt{16 + x^2} + 3(10 - x), \qquad 0 \le x \le 10.$$
  2. Enter \(Y_1 = 5\sqrt{16 + X^2} + 3(10 - X)\) with the window \(0 \le x \le 10\), \(40 \le y \le 56\). The curve falls, then rises: the minimum tool, with bounds on either side of the lowest point, displays \(X \approx 3\) and \(Y \approx 46\) (a value such as \(X = 2.9999998\) is the tool's approximation: the tool finds the minimum to its precision, not exactly).
    Figure
    The values at the end points are larger: \(C(0) = 5 \times 4 + 30 = 50\) and \(C(10) = 5\sqrt{116} \approx 53.9\). The pipeline should reach the coast about \(3\) km from \(A\), that is about \(7\) km before the refinery, for a least cost of about \(46\) million dollars. A table of values agrees with the reading: \(C(3) = 5\sqrt{25} + 21 = 46\), while \(C(2.9) \approx 46.003\) and \(C(3.1) \approx 46.003\) are slightly larger.
The curve is very flat near its lowest point: the table of values gives \(C(2.5) \approx 46.08\) and \(C(3.5) \approx 46.08\), so landing the pipeline half a kilometre on either side of the best point costs less than \(0.1\) million dollars more. Zooming in on the graph, or reading a table, shows how sensitive the cost is to the choice of \(x\) — information that the single optimal value does not give.

Skills to practice
  • Solving Optimisation Problems with Technology
III Rates of Change
III.1 Rates of Change with Respect to Time
When a quantity \(Q\) depends on the time \(t\), the derivative \(Q'(t) = \dfrac{\mathrm{d}Q}{\mathrm{d}t}(t)\) is the instantaneous rate of change of \(Q\): how fast \(Q\) changes at the instant \(t\), the value approached by the average rate \(\dfrac{\Delta Q}{\Delta t}\) over shorter and shorter durations. Its unit is the unit of \(Q\) divided by the unit of time. Speed is the best-known example; physics and chemistry give two more.
  • Power: if \(E(t)\) is the energy received by a system from the instant \(0\) to the instant \(t\), the power at the instant \(t\) is \(P(t) = \dfrac{\mathrm{d}E}{\mathrm{d}t}(t)\), in watts (\(\mathrm{W}\)) when \(E\) is in joules (\(\mathrm{J}\)) and \(t\) in seconds. When the power is constant, the energy received during a duration \(\Delta t\) is \(\Delta E = P \times \Delta t\).
  • Reaction rates: during a chemical reaction, the concentration \([A](t)\) of a reactant \(A\) decreases, and its rate of disappearance is \(-\dfrac{\mathrm{d}[A]}{\mathrm{d}t}(t)\), which is non-negative; the concentration \([B](t)\) of a product \(B\) increases, and its rate of appearance is \(\dfrac{\mathrm{d}[B]}{\mathrm{d}t}(t)\). With concentrations in \(\mathrm{mol \cdot L^{-1}}\) and \(t\) in minutes, these rates are in \(\mathrm{mol \cdot L^{-1} \cdot min^{-1}}\).
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Example
During a test, the energy received by a water heater from the instant \(0\) to the instant \(t\) is \(E(t) = 2t^2 + 10t\), in joules, for \(0 \leq t \leq 10\) (\(t\) in seconds).
  1. Compute the power \(P(t)\), then \(P(0)\) and \(P(10)\).
  2. Compute the average power over the test, \(\dfrac{E(10) - E(0)}{10 - 0}\), and compare it with \(P(0)\) and \(P(10)\).

  1. $$\begin{aligned}[t] P(t) &= E'(t) &&\text{(the power is the derivative of the energy)}\\ &= 2 \times 2t + 10 &&\text{(} (t^2)' = 2t\text{, } (t)' = 1\text{)}\\ &= 4t + 10.\end{aligned}$$ The expression of \(E\) is a polynomial, differentiable at every real \(t\), including \(t = 0\) and \(t = 10\). So \(P(0) = 10~\mathrm{W}\) and \(P(10) = 50~\mathrm{W}\): the power grows during the test.
  2. \(E(10) = 200 + 100 = 300~\mathrm{J}\) and \(E(0) = 0\), so the average power is \(\dfrac{300}{10} = 30~\mathrm{W}\), between \(P(0) = 10~\mathrm{W}\) and \(P(10) = 50~\mathrm{W}\). A constant power of \(30~\mathrm{W}\) would give the same energy: \(30 \times 10 = 300~\mathrm{J}\).

Skills to practice
  • Interpreting a Rate of Change
  • Computing a Rate of Change in Context
IV Going further
Skills to practice
  • Solving Exam-Style Problems on Optimisation and Rates of Change
  • Quiz