IB Analysis and Approaches HL · Differential calculus · Section 9 of 14

Rate of change problems

3 worked examples · By Álvaro López

Key idea

In a rate of change problem several quantities depend on time and are linked by a formula. If you know how one of them changes, the derivative tells you how another changes at a particular instant. At IB it is often asked in Paper 1, almost always with circles, spheres, cones or right-angled triangles.

Differentiating with respect to time

If a quantity $y$ changes with time $t$ , its rate of change is $\frac{d y}{d t}$ . If it is positive, $y$ is increasing; if it is negative, $y$ is decreasing. Its units are those of $y$ divided by those of $t$ (for example cm $^{3}$ /s).

When two quantities are related, for example the area $A$ of a circle and its radius $r$ with $A = π r^{2}$ , both depend on $t$ . The chain rule gives the relationship between their rates of change.

The reason is this: if the radius changes at a certain rate, the area changes in proportion to how much each unit of radius contributes to the area, which is $\frac{d A}{d r}$ .

Ladder 5 metres long leaning against a wall: the base is x metres from the wall and the top is y metres above the ground

Worked examples

Example 1 Basic

An oil spill

The radius of a circular oil spill grows at a constant rate of $2$ cm/s. Find the rate of change of the area of the spill at the instant when the radius is $10$ cm.

Solution

We have $A = π r^{2}$ and $\frac{d r}{d t} = 2$ .

Differentiating with respect to $t$ : $\frac{d A}{d t} = 2 π r \frac{d r}{d t}$ .

With $r = 10$ : $\frac{d A}{d t} = 2 π \cdot 10 \cdot 2 = 40 π$ cm $^{2}$ /s $\approx 125.7$ cm $^{2}$ /s.

The area is increasing at $40 π$ cm $^{2}$ /s at that instant.

Example 2 Medium

A balloon being inflated

A spherical balloon is inflated so that its volume increases at a rate of $100$ cm $^{3}$ /s. At the instant when the radius is $5$ cm:

a) Find the rate of change of the radius.

b) Find the rate of change of the surface area of the balloon, $S = 4 π r^{2}$ .

Solution

a) $V = \frac{4}{3} π r^{3}$ , so $\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ . With $\frac{d V}{d t} = 100$ and $r = 5$ : $100 = 4 π \cdot 25 \cdot \frac{d r}{d t} = 100 π \frac{d r}{d t}$ , and therefore $\frac{d r}{d t} = \frac{1}{π}$ cm/s $\approx 0.32$ cm/s.

b) $\frac{d S}{d t} = 8 π r \frac{d r}{d t} = 8 π \cdot 5 \cdot \frac{1}{π} = 40$ cm $^{2}$ /s.

Example 3 Hard

The ladder that slides

A 5 m ladder leans against a vertical wall. Its base moves away from the wall at $0.6$ m/s. Let $x$ be the distance of the base from the wall and $y$ the height of the top, both in metres.

a) Find the speed at which the top moves down when the base is $3$ m from the wall.

b) Find that speed when the top is $3$ m above the ground.

c) Find the rate of change of the area of the triangle formed by the ladder, the wall and the ground when $x = 3$ .

Solution

The relation is $x^{2} + y^{2} = 25$ , with $\frac{d x}{d t} = 0.6 = \frac{3}{5}$ .

Differentiating: $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$ , so $\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t}$ .

a) With $x = 3$ : $y = 25 - 9 = 4$ . Then $\frac{d y}{d t} = - \frac{3}{4} \cdot \frac{3}{5} = - \frac{9}{20} = - 0.45$ m/s. The negative sign means it is going down.

b) With $y = 3$ : $x = 4$ . Then $\frac{d y}{d t} = - \frac{4}{3} \cdot \frac{3}{5} = - \frac{4}{5} = - 0.8$ m/s. The lower it gets, the faster it falls.

c) The area is $A = \frac{x y}{2}$ . Differentiating with the product rule: $\frac{d A}{d t} = \frac{1}{2} (\frac{d x}{d t} y + x \frac{d y}{d t})$ . With $x = 3$ , $y = 4$ , $\frac{d x}{d t} = \frac{3}{5}$ and $\frac{d y}{d t} = - \frac{9}{20}$ : $\frac{d A}{d t} = \frac{1}{2} (\frac{3}{5} \cdot 4 - 3 \cdot \frac{9}{20}) = \frac{1}{2} (\frac{12}{5} - \frac{27}{20}) = \frac{1}{2} \cdot \frac{21}{20} = \frac{21}{40} = 0.525 m^{2} / s .$

In the exam: you must write the equation linking the variables, differentiate with the chain rule and only then substitute. If there is a negative sign, say what it means (going down, decreasing). Marks are lost for omitted units.

Common mistakes

Summary

The rate of change of $y$ is $\frac{d y}{d t}$ : positive if increasing, negative if decreasing.
Find an equation linking the quantities at any instant and differentiate it with respect to $t$ using the chain rule.
Substitute the values for the instant asked after differentiating.
Common formulas: circle, sphere, cylinder, cone and Pythagoras.
Answer with the sign (what it means) and the units.

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