Chapter 6 · Application of Derivatives

6.1 Rate of Change of Quantities

The derivative $\dfrac{dy}{dx}$ is the rate at which $y$ changes per unit change in $x$. With that reading you can find speeds, marginal costs and the rates of quantities linked by a formula — a balloon's radius, a ladder's top, the water level in a cone.

A weather balloon is being filled with helium at a steady $100\ \text{cm}^3$ every second. The volume grows at a constant rate — but the radius does not. At first the radius shoots up; later, when the balloon is big, each extra $100\ \text{cm}^3$ is spread over a huge surface and the radius creeps up only slowly. How fast is the radius growing at the moment it is $10$ cm? The two rates, $\dfrac{dV}{dt}$ and $\dfrac{dr}{dt}$, are tied together by the formula $V=\tfrac43\pi r^3$, and calculus lets us pass from one to the other.

This section starts from one sentence: a derivative is a rate of change. If $y$ depends on $x$, then $\dfrac{dy}{dx}$ tells you how many units $y$ changes per unit change of $x$, at that instant. We use it to measure how fast areas and volumes grow, how fast a particle moves, how a factory's cost rises with output, and — the most useful skill of all — how the rates of two linked quantities are related.

The Derivative as a Rate of Change

In Class 11 Mathematics, Ch. 12 — Limits and Derivatives, you defined the derivative as a limit. Let us read that limit as a statement about rates. Suppose $y=f(x)$, and $x$ changes from $x_0$ to $x_0+h$. Then $y$ changes by $f(x_0+h)-f(x_0)$, and the change in $y$ per unit change in $x$ is the quotient $$\frac{f(x_0+h)-f(x_0)}{h}.$$ This is an average rate over the interval from $x_0$ to $x_0+h$. Shrinking $h$ towards $0$ squeezes the interval to the single value $x_0$, and the average rate settles to the derivative.

📘 Definition 6.1 — Rate of change
Let $y=f(x)$. The average rate of change of $y$ with respect to $x$ over the interval from $x_0$ to $x_0+h$ $(h\ne0)$ is $\dfrac{f(x_0+h)-f(x_0)}{h}$. If $f$ is differentiable at $x_0$, the rate of change (or instantaneous rate of change) of $y$ with respect to $x$ at $x_0$ is $$\left.\frac{dy}{dx}\right|_{x=x_0}=f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}.$$ Its units are (units of $y$) per (unit of $x$). If $f'(x_0)\gt0$, $y$ is increasing at that instant; if $f'(x_0)\lt0$, it is decreasing.

The variable $x$ need not be time. "The rate of change of the area of a circle with respect to its radius" means $\dfrac{dA}{dr}$: square centimetres of area gained per centimetre of radius. When the independent variable is time $t$, the rate of change is "how fast": $\dfrac{ds}{dt}$ is the velocity of a particle whose position is $s$, and the rate of change of velocity, $\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}$, is its acceleration.

Interactive 1 — From Average Rate to Instantaneous Rate

Choose a quantity. The blue chord joins the points at $x_0$ and $x_0+h$; its slope is the average rate of change over that interval. Press Play (or drag the $h$ slider) to shrink $h$ towards $0$: the average rate settles on the instantaneous rate $f'(x_0)$, whose direction is drawn as the red line through the point. Move $x_0$ to see the rate change from point to point.

$h$—
average rate—
instantaneous rate $f'(x_0)$—
difference—
✏️ Example 1 — Areas and volumes growing with their dimensions
Find (a) the rate of change of the area of a circle with respect to its radius $r$ when $r=5$ cm; (b) the rate of change of the volume of a sphere with respect to its radius when $r=3$ cm; (c) the rate of change of the volume of a cube with respect to its edge $a$ when $a=2$ cm. Give units.
Method Write the quantity as a function of the given variable, differentiate, then substitute the value. The unit is "unit of the quantity per unit of the variable".
(a) $A=\pi r^2$, so $\dfrac{dA}{dr}=2\pi r$. At $r=5$: $\dfrac{dA}{dr}=10\pi\approx31.42\ \text{cm}^2$ per cm of radius. (Notice that $2\pi r$ is the circumference: a thin ring of width $dr$ added at the edge.)
(b) $V=\tfrac43\pi r^3$, so $\dfrac{dV}{dr}=4\pi r^2$ — the surface area. At $r=3$: $36\pi\approx113.10\ \text{cm}^3$ per cm.
(c) $V=a^3$, so $\dfrac{dV}{da}=3a^2$; at $a=2$: $12\ \text{cm}^3$ per cm.
Check (c) by an average rate: from $a=2$ to $a=2.001$ the volume changes by $2.001^3-8=0.012006$, i.e. $12.006\ \text{cm}^3$ per cm — very close to $12$ ✓.
✏️ Example 2 — Velocity and acceleration as rates
A particle moves along a straight line so that its distance from a fixed point $O$ after $t$ seconds is $s=t^3-6t^2+9t$ metres $(t\ge0)$. (a) Find its velocity and acceleration at $t=2$. (b) When is it momentarily at rest? (c) During which time interval is it moving back towards $O$?
Rates Velocity $v=\dfrac{ds}{dt}=3t^2-12t+9=3(t-1)(t-3)$ m/s; acceleration $a=\dfrac{dv}{dt}=6t-12$ m/s².
(a) $v(2)=12-24+9=-3$ m/s and $a(2)=0$ m/s². The negative velocity means $s$ is decreasing: the particle is moving towards $O$ at $3$ m/s, and at that instant its velocity is not changing.
(b) At rest when $v=0$: $3(t-1)(t-3)=0$, so $t=1$ s and $t=3$ s.
(c) It moves towards $O$ when $s$ decreases, i.e. $v\lt0$: $(t-1)(t-3)\lt0$ for $1\lt t\lt3$.
Check $s(1)=4$ m and $s(3)=0$ m: between $t=1$ and $t=3$ the particle goes from $4$ m back to $O$ itself, consistent with (c).

Marginal Cost and Marginal Revenue

Economists use rates of change all the time, under the name marginal. A factory's total cost $C(x)$ of producing $x$ units includes fixed costs (rent, machines) and costs that grow with output (material, wages, power). Its total revenue $R(x)$ is the money received from selling $x$ units. Planners want to know: at the present output, how fast is the cost rising per extra unit? How fast is the revenue rising?

📘 Definition 6.2 — Marginal cost and marginal revenue
If $C(x)$ is the total cost and $R(x)$ the total revenue when $x$ units are produced and sold, the marginal cost and marginal revenue at output $x$ are the rates of change $$\text{MC}=\frac{dC}{dx},\qquad \text{MR}=\frac{dR}{dx},$$ measured in rupees per unit. The profit is $P(x)=R(x)-C(x)$, and its rate of change is $\dfrac{dP}{dx}=\text{MR}-\text{MC}$.

Output is really counted in whole units, but for large outputs $C$ is modelled by a smooth formula, and the marginal cost at $x$ is then very close to the actual cost of producing the $(x+1)$-th unit, $C(x+1)-C(x)$. That is how businesses read it.

✏️ Example 3 — Marginal cost, revenue and profit
A small workshop making solar lanterns has total cost $C(x)=0.01x^3-0.6x^2+25x+1200$ rupees for $x$ lanterns a day, and sells each lantern for $p=200-0.5x$ rupees (the price must drop to sell more). (a) Find the marginal cost at $x=10$ and at $x=40$. (b) Find the marginal revenue at $x=50$. (c) Is the profit rising or falling with output at $x=50$?
(a) $\text{MC}=C'(x)=0.03x^2-1.2x+25$. At $x=10$: $3-12+25=16$, i.e. ₹16 per lantern. At $x=40$: $48-48+25=25$, i.e. ₹25 per lantern. The cost per extra lantern first falls (bulk buying) and later rises (overtime, crowding).
(b) Revenue $=$ (number sold) $\times$ (price): $R(x)=x(200-0.5x)=200x-0.5x^2$. So $\text{MR}=200-x$, and at $x=50$, $\text{MR}=150$, i.e. ₹150 per lantern.
(c) $\text{MC}(50)=0.03(2500)-1.2(50)+25$ $=75-60+25=40$. So $\dfrac{dP}{dx}=\text{MR}-\text{MC}=150-40=110\gt0$: profit is still rising, at about ₹110 per extra lantern.
Check $C(50)=1250-1500+1250+1200$ $=2200$ and $C(51)=1326.51-1560.6+1275+1200$ $=2240.91$. The 51st lantern really costs ₹40.91 — very close to $\text{MC}(50)=40$ rupees ✓.

Linked Rates: Two Quantities Changing Together

In the balloon problem, both $V$ and $r$ change with time, and at every instant they are tied by $V=\tfrac43\pi r^3$. We know $\dfrac{dV}{dt}$ and want $\dfrac{dr}{dt}$. The chain rule of Chapter 5 links them: differentiate the formula with respect to $t$, treating $r$ as a function of $t$.

📐 Theorem 6.1 — Linked rates
Suppose $y=f(x)$, where $f$ is differentiable, and $x=x(t)$ is a differentiable function of time $t$. Then $y$ is a differentiable function of $t$ and $$\frac{dy}{dt}=f'(x)\,\frac{dx}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}.$$ More generally, if two or more time-dependent quantities satisfy an equation at every instant, differentiating both sides of that equation with respect to $t$ gives an equation linking their rates.

This is the chain rule read as a statement about rates: "($y$ per unit $x$) $\times$ ($x$ per second) $=$ $y$ per second". For the balloon, $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$. When $\dfrac{dV}{dt}=100$ and $r=10$, $\dfrac{dr}{dt}=\dfrac{100}{400\pi}=\dfrac1{4\pi}\approx0.0796$ cm/s. When $r=20$ the same pumping gives only a quarter of that — the big balloon grows slowly, just as we expected.

🧭 Problem-solving strategy — linked rates
  1. Draw and name. Sketch the situation at a general instant $t$. Give letters to every quantity that changes ($x$, $y$, $r$, $h$, $V$ …). Keep fixed lengths as numbers.
  2. List what you know and what you want, as derivatives with respect to $t$: e.g. "know $\dfrac{dx}{dt}=0.3$, want $\dfrac{dy}{dt}$ when $x=3$". A quantity that is decreasing has a negative rate.
  3. Find an equation linking the changing quantities that is true at every instant (Pythagoras, similar triangles, a volume formula). If it has an extra variable, remove it using geometry (e.g. $r=\tfrac h2$ for a cone).
  4. Differentiate both sides with respect to $t$ (chain rule on every changing quantity).
  5. Only now substitute the values at the instant asked about, and solve for the unknown rate. Give units and interpret the sign.
✏️ Example 4 — The weather balloon
Helium is pumped into a spherical balloon at $100\ \text{cm}^3/\text{s}$. Find (a) the rate at which the radius is increasing and (b) the rate at which the surface area is increasing, at the instant when the radius is $10$ cm.
Know / want $\dfrac{dV}{dt}=100\ \text{cm}^3/\text{s}$; want $\dfrac{dr}{dt}$ and $\dfrac{dS}{dt}$ when $r=10$.
(a) $V=\tfrac43\pi r^3\Rightarrow\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$. At $r=10$: $100=400\pi\dfrac{dr}{dt}$, so $\dfrac{dr}{dt}=\dfrac1{4\pi}\approx0.0796$ cm/s.
(b) $S=4\pi r^2\Rightarrow\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$ $=8\pi(10)\cdot\dfrac1{4\pi}=20\ \text{cm}^2/\text{s}$.
Check A neat shortcut: $\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}=\dfrac2r\cdot4\pi r^2\dfrac{dr}{dt}$ $=\dfrac2r\dfrac{dV}{dt}=\dfrac{2}{10}(100)=20$ ✓.
✏️ Example 5 — The sliding ladder
A $5$ m ladder leans against a vertical wall. Its foot is pulled away from the wall along the ground at $0.3$ m/s. How fast is the top sliding down the wall when the foot is $3$ m from the wall?
Name Let $x$ = distance of the foot from the wall and $y$ = height of the top, both in metres, at time $t$. The ladder's length $5$ is fixed. Know $\dfrac{dx}{dt}=0.3$; want $\dfrac{dy}{dt}$ when $x=3$.
Equation Pythagoras at every instant: $x^2+y^2=25$.
Differentiate $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$, so $\dfrac{dy}{dt}=-\dfrac xy\,\dfrac{dx}{dt}$.
Substitute When $x=3$, $y=\sqrt{25-9}=4$, so $\dfrac{dy}{dt}=-\dfrac34(0.3)=-0.225$ m/s. The top slides down at $0.225$ m/s.
Interpret The factor $\dfrac xy$ grows without bound as $y\to0$: near the ground the top moves faster and faster, even though the foot moves steadily. Watch this in Interactive 2.
✏️ Example 6 — Filling a conical tank
A water tank is an inverted cone (vertex at the bottom) of base radius $2$ m and depth $4$ m. Water flows in at $0.5\ \text{m}^3/\text{min}$. How fast is the water level rising when the water is $2$ m deep? And when it is $3$ m deep?
Name At time $t$ let the water have depth $h$ and surface radius $r$ (metres) and volume $V=\tfrac13\pi r^2h$. Know $\dfrac{dV}{dt}=0.5$; want $\dfrac{dh}{dt}$.
Remove $r$ The water forms a smaller cone similar to the tank, so $\dfrac rh=\dfrac24$, i.e. $r=\dfrac h2$. Then $V=\tfrac13\pi\cdot\dfrac{h^2}4\cdot h=\dfrac{\pi h^3}{12}$.
Differentiate $\dfrac{dV}{dt}=\dfrac{\pi h^2}{4}\dfrac{dh}{dt}$, so $\dfrac{dh}{dt}=\dfrac{4}{\pi h^2}\cdot\dfrac{dV}{dt}=\dfrac{2}{\pi h^2}$.
Substitute At $h=2$: $\dfrac{dh}{dt}=\dfrac{2}{4\pi}=\dfrac1{2\pi}\approx0.159$ m/min. At $h=3$: $\dfrac{2}{9\pi}\approx0.0707$ m/min. The level slows down as the tank widens.
Why not substitute early? Putting $h=2$ before differentiating would make $V$ a constant, and its derivative $0$ — the classic linked-rates blunder.
Interactive 2 — Linked-Rates Lab

Pick a situation. One rate is held constant (helium pumped in, ladder foot pulled out, water poured in); the picture on the left changes in real time, and the graph on the right plots the linked quantity against time. The slope of that graph at the moving dot is the linked rate — read it off live, compared with the formula from Theorem 6.1. Press Play or drag the time slider.

time—
quantity—
linked rate (formula)—
slope of graph (measured)—
⚠️ Common mistakes when studying alone

1. Substituting numbers too early. Differentiate the general relation first; substitute the values of the particular instant last (Example 6).

2. Losing the sign. A quantity that is decreasing has a negative rate. If a ladder's top "slides down at $0.225$ m/s", then $\dfrac{dy}{dt}=-0.225$; if a tank is drained at $0.45\ \text{m}^3/\text{min}$, $\dfrac{dV}{dt}=-0.45$. Answer in words ("falling at …") to avoid a double negative.

3. Forgetting the chain rule. $\dfrac{d}{dt}\left(r^3\right)=3r^2\dfrac{dr}{dt}$, not $3r^2$. Every changing quantity differentiated with respect to $t$ brings its own rate.

4. "With respect to" confusion. "Rate of change of $A$ with respect to $r$" is $\dfrac{dA}{dr}$ (units cm² per cm); "rate of change of $A$" alone, in a time problem, means $\dfrac{dA}{dt}$ (cm² per second).

5. Missing units. Every rate has units: (units of the quantity) per (unit of the variable).

🚀 Beyond the board syllabus (JEE Advanced)

In JEE Advanced 2026; not in the CBSE Class 12 syllabus or JEE Main. JEE Advanced uses the derivative geometrically. At a point $(x_0,y_0)$ of the curve $y=f(x)$, the line through the point with slope $m=f'(x_0)$ is the tangent to the curve there, and the line through the point perpendicular to it is the normal:

tangent: $y-y_0=f'(x_0)(x-x_0)$;   normal: $y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)$ if $f'(x_0)\ne0$.

If $f'(x_0)=0$ the tangent is horizontal and the normal is the vertical line $x=x_0$.

Worked example. Find the tangent and normal to $y=x^3-2x$ at $(1,-1)$. Solution. $y'=3x^2-2$, so the slope at $x=1$ is $1$. Tangent: $y+1=x-1$, i.e. $y=x-2$. Normal: slope $-1$, $y+1=-(x-1)$, i.e. $y=-x$. Check: both lines pass through $(1,-1)$ and their slopes multiply to $-1$ ✓.

Exercises

I
Rates of change. Give units where the quantities have them.
  • 1 Find the rate of change of the area of a circle with respect to its radius when $r=6$ cm.
  • 2 Find the rate of change of the volume of a sphere with respect to its radius when $r=3$ cm.
  • 3 Find the rate of change of the total surface area of a cube with respect to its edge when the edge is $4$ cm.
  • 4 Find the rate of change of the area of an equilateral triangle with respect to its side when the side is $10$ cm.
  • 5 A cylinder has fixed height $7$ cm. Find the rate of change of its volume with respect to its radius when $r=2$ cm.
  • 6 For $y=x^2+3x$, find the instantaneous rate of change of $y$ with respect to $x$ at $x=2$, and the average rate of change over $[2,\,2.1]$.
  • 7 A particle moves so that $s=2t^3-15t^2+36t$ metres after $t$ seconds. Find its velocity at $t=1$, the times when it is at rest, and its acceleration at $t=4$.
  • 8 Find the rate of change of the circumference of a circle with respect to its area, when the radius is $5$ cm.
1 $\dfrac{dA}{dr}=2\pi r=12\pi\approx37.70\ \text{cm}^2$ per cm.
2 $\dfrac{dV}{dr}=4\pi r^2=36\pi\approx113.10\ \text{cm}^3$ per cm.
3 $S=6a^2$, $\dfrac{dS}{da}=12a=48\ \text{cm}^2$ per cm.
4 $A=\tfrac{\sqrt3}4a^2$, $\dfrac{dA}{da}=\tfrac{\sqrt3}2a=5\sqrt3\approx8.66\ \text{cm}^2$ per cm.
5 $V=7\pi r^2$, $\dfrac{dV}{dr}=14\pi r=28\pi\approx87.96\ \text{cm}^3$ per cm.
6 $\dfrac{dy}{dx}=2x+3=7$ at $x=2$. Average: $\dfrac{f(2.1)-f(2)}{0.1}=\dfrac{10.71-10}{0.1}=7.1$ — close to $7$, as it should be.
7 $v=6t^2-30t+36=6(t-2)(t-3)$: $v(1)=12$ m/s; at rest at $t=2$ s and $t=3$ s. $a=12t-30$, so $a(4)=18$ m/s².
8 $C=2\pi r$, $A=\pi r^2$. By the chain rule $\dfrac{dC}{dA}=\dfrac{dC/dr}{dA/dr}=\dfrac{2\pi}{2\pi r}=\dfrac1r$; at $r=5$, $\dfrac15$ cm per cm² of area.
II
Marginal cost, revenue and profit (all money in rupees, $x$ = number of units).
  • 1 $C(x)=2x^3-30x^2+200x+1000$. Find the marginal cost when $x=5$.
  • 2 For the same $C(x)$, find the marginal cost when $x=10$, and explain why it is larger than at $x=5$.
  • 3 $R(x)=60x-0.2x^2$. Find the marginal revenue at $x=50$ and at $x=100$.
  • 4 The price at which $x$ units can be sold is $p=120-2x$. Find the revenue function, the marginal revenue at $x=20$, and the output at which the marginal revenue is zero.
  • 5 $R(x)=500x$ and $C(x)=x^2+100x+2000$. Find the marginal profit at $x=150$ and interpret it.
  • 6 $C(x)=500+20x+0.1x^2$. At $x=100$ find the average cost per unit $\dfrac{C(x)}{x}$ and the marginal cost. Which is larger, and what does that suggest about increasing output?
1 $C'(x)=6x^2-60x+200$; $C'(5)=150-300+200=50$: ₹50 per unit.
2 $C'(10)=600-600+200=200$: ₹200 per unit. The $2x^3$ term dominates at higher output, so each extra unit costs more.
3 $\text{MR}=60-0.4x$: ₹40 per unit at $x=50$ and ₹20 per unit at $x=100$.
4 $R=xp=120x-2x^2$; $\text{MR}=120-4x$, which is ₹40 per unit at $x=20$, and $0$ when $x=30$ (selling more beyond $30$ units lowers revenue).
5 $P=R-C=-x^2+400x-2000$, $P'(x)=400-2x$, so $P'(150)=100$: at $150$ units, profit rises by about ₹100 per extra unit.
6 $C(100)=500+2000+1000=3500$, so average cost $=35$ rupees per unit; $C'(x)=20+0.2x=40$ rupees per unit. MC $\gt$ average cost, so each extra unit costs more than the current average and raises the average cost.
III
Linked rates.
  • 1 The radius of a circle increases at $0.4$ cm/s. How fast is its area increasing when $r=5$ cm?
  • 2 The edge of a cube increases at $3$ cm/s. How fast is its volume increasing when the edge is $10$ cm?
  • 3 The volume of a spherical balloon increases at $36\pi\ \text{cm}^3/\text{s}$. Find the rate of increase of its radius when $r=3$ cm.
  • 4 The radius of a balloon increases at $0.2$ cm/s. Find the rate of increase of its surface area when $r=15$ cm.
  • 5 The length of a rectangle decreases at $3$ cm/min and its width increases at $2$ cm/min. When the length is $10$ cm and the width $6$ cm, find the rates of change of the perimeter and of the area.
  • 6 A $10$ m ladder leans against a wall. Its foot is pulled away at $1.2$ m/s. How fast does the top slide down when the foot is $6$ m from the wall?
  • 7 A stone dropped into a pond makes circular ripples whose radius grows at $4$ cm/s. How fast is the disturbed area growing when the radius is $10$ cm?
  • 8 Water is poured at $3\ \text{cm}^3/\text{s}$ into a conical funnel (vertex down) whose radius is always half the depth of the water. How fast is the level rising when the water is $6$ cm deep?
  • 9 A man $1.8$ m tall walks away from a $6$ m lamp post at $1.4$ m/s. How fast is his shadow lengthening, and how fast is the tip of his shadow moving?
  • 10 A point moves along $y=x^2$ so that its $x$-coordinate increases at $2$ units/s. At which point is its $y$-coordinate increasing at $16$ units/s?
1 $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}=2\pi(5)(0.4)$ $=4\pi\approx12.57\ \text{cm}^2/\text{s}$.
2 $\dfrac{dV}{dt}=3a^2\dfrac{da}{dt}=3(100)(3)=900\ \text{cm}^3/\text{s}$.
3 $36\pi=4\pi(9)\dfrac{dr}{dt}$, so $\dfrac{dr}{dt}=1$ cm/s.
4 $\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}=8\pi(15)(0.2)$ $=24\pi\approx75.40\ \text{cm}^2/\text{s}$.
5 $P=2(l+w)$: $\dfrac{dP}{dt}=2(-3+2)=-2$ cm/min (perimeter decreasing). $A=lw$: $\dfrac{dA}{dt}=l\dfrac{dw}{dt}+w\dfrac{dl}{dt}$ $=10(2)+6(-3)=2\ \text{cm}^2/\text{min}$ (area increasing).
6 $x^2+y^2=100$; at $x=6$, $y=8$. $\dfrac{dy}{dt}=-\dfrac xy\dfrac{dx}{dt}=-\dfrac68(1.2)=-0.9$: the top slides down at $0.9$ m/s.
7 $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}=2\pi(10)(4)$ $=80\pi\approx251.3\ \text{cm}^2/\text{s}$.
8 $r=\tfrac h2$ gives $V=\dfrac{\pi h^3}{12}$ and $\dfrac{dV}{dt}=\dfrac{\pi h^2}4\dfrac{dh}{dt}$. At $h=6$: $3=9\pi\dfrac{dh}{dt}$, so $\dfrac{dh}{dt}=\dfrac1{3\pi}\approx0.106$ cm/s.
9 Let $x$ = man's distance from the post and $s$ = shadow length. Similar triangles: $\dfrac{s}{1.8}=\dfrac{x+s}{6}$, so $4.2s=1.8x$ and $s=\tfrac37x$. Then $\dfrac{ds}{dt}=\tfrac37(1.4)=0.6$ m/s, and the tip (at distance $x+s$) moves at $1.4+0.6=2$ m/s.
10 $\dfrac{dy}{dt}=2x\dfrac{dx}{dt}=4x$. Setting $4x=16$ gives $x=4$: the point $(4,16)$.
IV
Applications and modelling.
  • 1 Epidemic. In a town of $1000$ people, the number who have caught a flu after $t$ days is modelled by $N(t)=\dfrac{1000}{1+99e^{-0.5t}}$. Find the rate of spread at $t=0$, show that $N'(t)=0.5N\left(1-\dfrac N{1000}\right)$, and find the rate of spread when $500$ people are infected.
  • 2 Medicine. The concentration of a drug in the blood is $C(t)=20e^{-0.2t}$ mg/L, $t$ hours after an injection. Find the rate of change of concentration at $t=5$ h. (Recall that $\log x=\log_e x$ and $\dfrac{d}{dt}e^{kt}=ke^{kt}$.)
  • 3 Rocket tracking. A rocket rises vertically from a launch pad. A tracking camera on the ground is $3$ km from the pad. At the instant when the rocket is $4$ km high and rising at $0.3$ km/s, find how fast (a) the distance from the camera to the rocket and (b) the camera's angle of elevation $\theta$ are changing.
  • 4 Water supply. A cylindrical tank of radius $1.5$ m is drained at $0.45\ \text{m}^3/\text{min}$. How fast is the water level falling?
  • 5 Braking. After the brakes are applied, a car travels $s=20t-t^2$ metres in $t$ seconds. How long does it take to stop, and how far does it travel while braking?
  • 6 Kite. A kite flies at a constant height of $60$ m and the wind carries it horizontally away from the girl flying it at $2$ m/s. How fast must she let out string when $100$ m of string is out?
1 Write $N=1000(1+99e^{-0.5t})^{-1}$. Chain rule: $N'=1000\cdot\dfrac{49.5e^{-0.5t}}{(1+99e^{-0.5t})^2}$. At $t=0$: $\dfrac{49500}{10000}=4.95$ people/day. Since $\dfrac N{1000}=\dfrac1{1+99e^{-0.5t}}$ and $1-\dfrac N{1000}=\dfrac{99e^{-0.5t}}{1+99e^{-0.5t}}$, the product $0.5N\left(1-\tfrac N{1000}\right)$ equals the expression for $N'$. When $N=500$: $N'=0.5(500)(0.5)=125$ people/day — the fastest spread.
2 $C'(t)=-4e^{-0.2t}$; $C'(5)=-4e^{-1}\approx-1.47$ mg/L per hour: the concentration is falling at about $1.47$ mg/L per hour.
3 Let $h$ = height, $D$ = distance (km). (a) $D^2=9+h^2\Rightarrow D\dfrac{dD}{dt}=h\dfrac{dh}{dt}$; at $h=4$, $D=5$: $\dfrac{dD}{dt}=\dfrac45(0.3)=0.24$ km/s. (b) $\tan\theta=\dfrac h3\Rightarrow\sec^2\theta\,\dfrac{d\theta}{dt}=\dfrac13\dfrac{dh}{dt}$; $\sec^2\theta=1+\tfrac{16}9=\tfrac{25}9$, so $\dfrac{d\theta}{dt}=\dfrac9{25}\cdot0.1=0.036$ rad/s.
4 $V=\pi(1.5)^2h=2.25\pi h$, so $\dfrac{dV}{dt}=2.25\pi\dfrac{dh}{dt}$. With $\dfrac{dV}{dt}=-0.45$: $\dfrac{dh}{dt}=-\dfrac{0.2}{\pi}\approx-0.0637$ m/min — falling at about $6.4$ cm per minute.
5 $v=20-2t=0$ at $t=10$ s; $s(10)=200-100=100$ m.
6 $L^2=x^2+60^2$; when $L=100$, $x=80$. $\dfrac{dL}{dt}=\dfrac xL\dfrac{dx}{dt}=\dfrac{80}{100}(2)=1.6$ m/s.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask a 1-mark MCQ on a rate, a 2- or 3-mark linked-rates problem (balloon, cube, cone, ladder, two moving objects) and often a case study built on a growing shape; marks go for the relation, the differentiation and the units. JEE combines a linked rate with a curve or with similar triangles.

E1. Board · 1 mark The radius of a circle is increasing at $3$ cm/s. The rate at which its area is increasing when the radius is $10$ cm is  (A) $30\pi\ \text{cm}^2/\text{s}$  (B) $60\pi\ \text{cm}^2/\text{s}$  (C) $20\pi\ \text{cm}^2/\text{s}$  (D) $6\pi\ \text{cm}^2/\text{s}$

Idea $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}=2\pi(10)(3)=60\pi$.
Answer (B). Trap (C): $20\pi$ is $\dfrac{dA}{dr}$ — it forgets the factor $\dfrac{dr}{dt}=3$. (A) halves the correct value (using $\pi r\dfrac{dr}{dt}$); (D) is $2\pi\dfrac{dr}{dt}$, the rate of the circumference.

E2. Board · 2 marks The volume of a cube is increasing at $9\ \text{cm}^3/\text{s}$. How fast is its surface area increasing when the edge is $10$ cm?

Edge rate $V=a^3\Rightarrow\dfrac{dV}{dt}=3a^2\dfrac{da}{dt}$: $9=300\dfrac{da}{dt}$, so $\dfrac{da}{dt}=0.03$ cm/s. (1 mark)
Answer $S=6a^2\Rightarrow\dfrac{dS}{dt}=12a\dfrac{da}{dt}$ $=120(0.03)=3.6\ \text{cm}^2/\text{s}$. (1 mark)

E3. Board · 3 marks Car A is $3$ km east of a road junction and driving towards it at $40$ km/h along the east road. At the same moment car B is $4$ km north of the junction and driving away from it at $50$ km/h along the north road. Is the distance between the cars increasing or decreasing at that moment, and at what rate?

Set-up Let $x$, $y$ be the distances of A and B from the junction and $D$ the distance between them: $D^2=x^2+y^2$. Here $\dfrac{dx}{dt}=-40$ (A approaches), $\dfrac{dy}{dt}=+50$. (1 mark)
Differentiate $D\dfrac{dD}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$. At the moment, $x=3$, $y=4$, $D=5$. (1 mark)
Answer $\dfrac{dD}{dt}=\dfrac{3(-40)+4(50)}5=\dfrac{80}5=16$ km/h: the distance is increasing at $16$ km/h. (1 mark) The most common error is to take $\dfrac{dx}{dt}=+40$, which gives $\dfrac{320}{5}=64$ km/h.

E4. Board · Case study A spherical weather balloon is inflated at a steady $1600\pi\ \text{cm}^3/\text{s}$. (i) How fast is the radius increasing when $r=20$ cm? [2 marks] (ii) How fast is the surface area increasing at that instant? [1 mark] (iii) Find the rate of change of the volume with respect to the radius when $r=20$ cm. [1 mark]

(i) $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$: $1600\pi=4\pi(400)\dfrac{dr}{dt}$, so $\dfrac{dr}{dt}=1$ cm/s.
(ii) $\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}=160\pi\approx502.7\ \text{cm}^2/\text{s}$.
(iii) $\dfrac{dV}{dr}=4\pi r^2=1600\pi\ \text{cm}^3$ per cm. (Note the different units from (i): this is a rate with respect to $r$, not $t$.)

E5. JEE Main A particle moves along the curve $y=\tfrac23x^3+1$. The points at which its $y$-coordinate is changing twice as fast as its $x$-coordinate are  (A) $\left(1,\tfrac53\right)$ only  (B) $\left(1,\tfrac53\right)$ and $\left(-1,\tfrac13\right)$  (C) $\left(\sqrt2,\tfrac{4\sqrt2}3+1\right)$ only  (D) $\left(-1,\tfrac13\right)$ only

Method $\dfrac{dy}{dt}=2x^2\dfrac{dx}{dt}$. We need $\dfrac{dy}{dt}=2\dfrac{dx}{dt}$ (with $\dfrac{dx}{dt}\ne0$), so $2x^2=2$, $x=\pm1$. Then $y=\tfrac23+1=\tfrac53$ or $y=-\tfrac23+1=\tfrac13$.
Answer (B). Trap (A) forgets the root $x=-1$ (since $x^2$ appears, both signs work); (C) solves $x^2=2$ by setting $2x^2=4$. ⏱ Fast route: the condition is $\dfrac{dy}{dx}=2$, i.e. $2x^2=2$.

E6. JEE Advanced (Numerical answer.) Water is poured at $3\pi\ \text{cm}^3/\text{s}$ into an inverted conical vessel whose semi-vertical angle is $30^\circ$. When the water is $3$ cm deep, the area of the water surface is increasing at $k\pi\ \text{cm}^2/\text{s}$. Find $k$.

Geometry Surface radius $r=h\tan30^\circ=\dfrac h{\sqrt3}$, so $V=\tfrac13\pi r^2h=\dfrac{\pi h^3}9$ and the surface area is $A=\pi r^2=\dfrac{\pi h^2}3$.
Level rate $\dfrac{dV}{dt}=\dfrac{\pi h^2}3\dfrac{dh}{dt}$: at $h=3$, $3\pi=3\pi\dfrac{dh}{dt}$, so $\dfrac{dh}{dt}=1$ cm/s.
Answer $\dfrac{dA}{dt}=\dfrac{2\pi h}3\dfrac{dh}{dt}=2\pi$, so $k=\mathbf 2$. ⏱ Fast route: $\dfrac{dV}{dt}=A\dfrac{dh}{dt}$ always (the new water is a thin layer of area $A$), and $\dfrac{dA}{dt}=\dfrac{2A}{h}\dfrac{dh}{dt}=\dfrac2h\dfrac{dV}{dt}=2\pi$.
🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. When a bottle or a funnel is filled from a tap at a steady rate, how fast does the water level rise — and does it follow the linked-rates model $\dfrac{dh}{dt}=\dfrac{1}{A(h)}\dfrac{dV}{dt}$, where $A(h)$ is the area of the water surface at depth $h$?

Why it matters. Water utilities, irrigation engineers and chemical plants read the level in a tank to know how much is flowing in or out. Rain gauges, fuel gauges in vehicles and drip-irrigation tanks all convert a level rate into a volume rate using the shape of the container — exactly Theorem 6.1.

What to do.

  1. Choose two containers: one straight-sided (a jar or a cylindrical bottle) and one that widens upwards (a funnel with its spout closed, a conical cup, or a flared glass). Measure their dimensions.
  2. Find the tap's flow rate $Q=\dfrac{dV}{dt}$ by timing how long it takes to fill a $1$ L measuring jug (repeat three times and average).
  3. Stick a paper scale on each container and record a phone video while it fills at the same tap setting. From the video, read the depth $h$ every two seconds.
  4. Estimate $\dfrac{dh}{dt}$ at several depths from your table (change in $h$ over $4$ s, centred on each reading) and compare with the model $\dfrac{Q}{A(h)}$, using $A=\pi r^2$ with $r$ measured or computed from the container's shape.

Tools: Measuring jug, ruler, phone camera, spreadsheet (columns $t$, $h$, measured rate, model rate).

Hand in: Your table and a graph of measured and model $\dfrac{dh}{dt}$ against $h$ for both containers, with two or three sentences explaining the differences you see (tap fluctuations, splashing, reading errors, the container not being a perfect cone).

Going further. Design a container (by its radius as a function of height) whose level would rise at a constant rate when filled at a constant rate — and one whose level would rise faster and faster.

Instructor note: for a cylinder $\dfrac{dh}{dt}=\dfrac{Q}{\pi R^2}$ is constant, so its $h$–$t$ graph should be a straight line; for a cone with $r=kh$, $h\propto t^{1/3}$ and the rate falls like $\dfrac1{h^2}$. A typical kitchen tap gives $Q\approx0.1$–$0.2$ L/s. The going-further answers: any cylinder (constant area); any container narrowing upwards, such as an upright cone.

🧪 Self-check quiz

0/5 correct

Q1. The rate of change of the volume of a sphere with respect to its radius, when $r=2$ cm, is

Q2. A cost $C(x)$ is in rupees and the output $x$ in kilograms. The units of the marginal cost are

Q3. A particle has position $s=t^3-3t$ metres at time $t\ge0$ seconds. It is momentarily at rest when

Q4. A $5$ m ladder slides with $x^2+y^2=25$ ($x$ = foot, $y$ = top). If $\dfrac{dx}{dt}=3$ m/s when $x=4$, then $\dfrac{dy}{dt}$ is

Q5. In a linked-rates problem, the values at the particular instant (such as $r=10$) should be substituted

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