The derivative as a rate of change

What you will be able to do

• explain the derivative as an instantaneous rate of change, with its units;
• find the rate of change of an area or a volume with respect to a length;
• find velocity and acceleration from a distance formula;
• read a marginal cost as the extra cost of one more item.

From average speed to speed at an instant

A bus covers $60\ \text{km}$ in $1.5\ \text{h}$. Its average speed is $\dfrac{60}{1.5} = 40\ \text{km/h}$. But the speedometer does not show an average. It shows the speed at this very moment.

To find it, we squeeze the time interval. Let $s(t)$ be the distance travelled after time $t$. Over a short interval $h$, the average speed is

As $h$ gets smaller, this settles to a number. That number is the derivative $s'(t) = \dfrac{ds}{dt}$, the speed at the instant $t$.

On the graph of $s$ against $t$, the average speed is the gradient of the line joining two points. The speed at an instant is the gradient of the tangent, the line that just touches the curve. The word gradient is used in A-Level and IB. NCERT, AP and most books in the US say slope. They mean the same number, and we use both words.

The derivative with its units

If $y$ depends on $x$, then $\dfrac{dy}{dx}$ is the rate of change of $y$ with respect to $x$. Its unit is the unit of $y$ divided by the unit of $x$.

• Distance $s$ in metres and time $t$ in seconds: $\dfrac{ds}{dt}$ is in $\text{m/s}$.
• Volume $V$ in $\text{cm}^3$ and radius $r$ in cm: $\dfrac{dV}{dr}$ is in $\text{cm}^3/\text{cm}$, which is $\text{cm}^2$.

A positive derivative means $y$ grows as $x$ grows. A negative derivative means $y$ shrinks.

Rates with respect to a length

The area of a circle is $A = \pi r^2$. Then

At $r = 5\ \text{cm}$ this is $10\pi \approx 31.42\ \text{cm}^2/\text{cm}$. If the radius grows by a tiny amount, say $0.01\ \text{cm}$, the area grows by about $31.42 \times 0.01 = 0.3142\ \text{cm}^2$.

Notice that $2\pi r$ is the circumference. For a ball, $V = \tfrac{4}{3}\pi r^3$ gives $\dfrac{dV}{dr} = 4\pi r^2$, the surface area. This is no accident. A thin new layer on a ball has the area of the ball times its thickness.

Velocity and acceleration

For a particle moving along a line, with position $s(t)$:

• velocity $v = \dfrac{ds}{dt}$, which is negative when the particle moves backwards;
• acceleration $a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$.

The particle is at rest when $v = 0$.

Marginal cost

A workshop's cost of making $x$ chairs is $C(x)$ rupees. The marginal cost is $C'(x)$. It is the extra cost of one more chair, to a good approximation, because one chair is a small change in $x$. In the same way, the marginal revenue is $R'(x)$.

A worked example

A pond ripple is a circle whose radius grows at $4\ \text{cm/s}$. How fast does the area grow when $r = 15\ \text{cm}$?

1. The link. $A = \pi r^2$, so $\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}$. This is the chain rule: the area depends on $r$, and $r$ depends on $t$.
2. Put in the numbers. $\dfrac{dA}{dt} = 2\pi \times 15 \times 4 = 120\pi$.
3. Answer. $120\pi \approx 376.99\ \text{cm}^2/\text{s}$.

Practise with new numbers

Try it in Python

The program squeezes the time interval and prints the average speed each time. It settles on the derivative.

Common mistakes

• Leaving out the units. A rate needs units, such as $\text{cm}^2/\text{s}$ or rupees per chair.
• Confusing the average rate with the rate at an instant. The average uses two points. The derivative uses one.
• Mixing up the two variables. $\dfrac{dA}{dr}$ and $\dfrac{dA}{dt}$ are different rates.
• Forgetting the chain rule. When $r$ changes with $t$, then $\dfrac{dA}{dt} = \dfrac{dA}{dr}\,\dfrac{dr}{dt}$.
• Reading a negative velocity as slow. It means the particle moves in the negative direction.

Remember these

• $\dfrac{dy}{dx}$ is the rate of change of $y$ with respect to $x$. Its unit is the unit of $y$ per unit of $x$.
• Average rate $= \dfrac{\Delta y}{\Delta x}$. The derivative is its limit.
• $v = \dfrac{ds}{dt}$ and $a = \dfrac{dv}{dt}$.
• Marginal cost $= C'(x)$, the cost of one more item.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A stone has fallen $s = t^3\ \text{m}$ after $t$ seconds in a test rig. Find the average speed between $t = 1\ \text{s}$ and $t = 3\ \text{s}$, and the speed at $t = 2\ \text{s}$.

2. Find the rate of change of the volume of a sphere with respect to its radius at $r = 3\ \text{cm}$.

3. The cost of making $x$ bags is $C(x) = 0.01x^2 + 5x + 300$ dollars. Find the marginal cost at $x = 100$.

4. $V$ is in $\text{m}^3$ and $t$ is in minutes. What are the units of $\dfrac{dV}{dt}$?

5. A particle moves with $s = 2t^3 - 12t^2 + 18t$ metres. When is it at rest?

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 6 (Application of Derivatives), the section on the rate of change of quantities.
• OpenStax, Calculus Volume 1, Section 3.4 (Derivatives as Rates of Change).