Application of Derivatives • Topic 1 of 3

Derivative as a Rate of Change

The derivative $\dfrac{dy}{dx}$ measures how fast $y$ changes per unit change in $x$. In applications, $\dfrac{dQ}{dt}$ is the rate at which a quantity $Q$ changes with time.

Related rates

When several quantities are linked by an equation, differentiating with respect to time relates their rates. Typical steps:

  • Write the relationship between the quantities (e.g. area, volume, Pythagoras).
  • Differentiate both sides with respect to $t$.
  • Substitute the known values and solve for the unknown rate.

A positive rate means the quantity is increasing; a negative rate means it is decreasing.

Marginal quantities

In economics, the derivative of a cost function $C(x)$ is the marginal cost $\dfrac{dC}{dx}$ — the approximate cost of producing one more unit. Likewise marginal revenue is $\dfrac{dR}{dx}$.

Solved Examples
1
Worked Example
The radius of a circle increases at $3$ cm/s. How fast is the area increasing when $r=5$ cm?
Solution

$A=\pi r^2 \Rightarrow \dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$. With $r=5,\ \dfrac{dr}{dt}=3$: $\dfrac{dA}{dt}=2\pi(5)(3)=30\pi$ cm$^2$/s.

2
Worked Example
The side of a square grows at $2$ cm/s. Find the rate of change of its area when the side is $10$ cm.
Solution

$A=x^2 \Rightarrow \dfrac{dA}{dt}=2x\dfrac{dx}{dt}=2(10)(2)=40$ cm$^2$/s.

3
Worked Example
A balloon's volume $V=\tfrac{4}{3}\pi r^3$ increases. Find $\dfrac{dV}{dr}$ at $r=2$.
Solution

$\dfrac{dV}{dr}=4\pi r^2=4\pi(4)=16\pi$ cubic units per unit radius.

4
Worked Example
If $C(x)=0.005x^3-0.02x^2+30x+5000$, find the marginal cost at $x=3$.
Solution

$C'(x)=0.015x^2-0.04x+30$. At $x=3$: $C'(3)=0.015(9)-0.04(3)+30=0.135-0.12+30=30.015$. So the next unit costs about ₹30.02.

Key Points

  • $\dfrac{dy}{dx}$ is a rate of change; $\dfrac{dQ}{dt}$ is a time rate.
  • Related rates: form an equation, differentiate w.r.t. $t$, substitute, solve.
  • Positive rate $\Rightarrow$ increasing; negative $\Rightarrow$ decreasing.
  • Marginal cost $=C'(x)$, marginal revenue $=R'(x)$.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.If $A=\pi r^2$ and $\tfrac{dr}{dt}=2$, then $\tfrac{dA}{dt}$ at $r=3$ is:
Explanation: $2\pi r\,\tfrac{dr}{dt}=2\pi(3)(2)=12\pi$.
Q2.Marginal cost is:
Explanation: Marginal cost is the derivative $C'(x)$.
Q3.For $V=\tfrac43\pi r^3$, $\dfrac{dV}{dr}=$
Explanation: Differentiate: $4\pi r^2$ (the surface area).
Q4.A negative rate of change means the quantity is:
Explanation: Negative derivative $\Rightarrow$ decreasing.
Practice more MCQs → 📝 Assignment · 35 marks