Sample 2022-23 Class 12 Maths Q22 Solution: A man walks away from a street light: rates of his shadow

The question

A man 1.6 m tall walks at the rate of 0.3 m/sec away from a street light that is 4 m above the ground. At what rate is the tip of his shadow moving? At what rate is his shadow lengthening?

Exam answer

Let AB be the street light of height 4 m and ED the man of height 1.6 m. At any time $t$ seconds, let the man be at a distance $x$ m from AB and let his shadow EC be $y$ m long.

The street light AB, the man ED and his shadow EC.

Triangles ABC and EDC are similar, so

\frac{AB}{ED} = \frac{AC}{EC} \Rightarrow \frac{4}{1.6} = \frac{x + y}{y}

Why?

The light pole AB and the man ED both stand vertically, so AB is parallel to ED. The two triangles also share the angle at C. Two equal angles make the triangles similar, so their corresponding sides are in the same ratio.

The tip C of the shadow is beyond the man, so

AC = AE + EC = x + y

\begin{aligned} 4y &= 1.6x + 1.6y \\ \Rightarrow 2.4y &= 1.6x \\ \Rightarrow 3y &= 2x \end{aligned}

Why?

Cross-multiply: $4y = 1.6(x + y)$ ⁠. Then take $1.6y$ to the left: $4y - 1.6y = 2.4y$ ⁠.

Dividing both sides by 0.8 clears the decimals, since $2.4 = 3 \times 0.8$ and $1.6 = 2 \times 0.8$ ⁠.

\begin{aligned} \frac{4}{1.6} &= \frac{x + y}{y} \\ 4 \times y &= 1.6 \times (x + y) \\ 4y &= 1.6x + 1.6y \\ 4y - 1.6y &= 1.6x \\ 2.4y &= 1.6x \\ \frac{2.4y}{0.8} &= \frac{1.6x}{0.8} \\ 3y &= 2x \end{aligned}

Differentiating both sides w.r.t. $t$ ⁠, with $\frac{dx}{dt} = 0.3$ m/s,

\begin{aligned} 3\frac{dy}{dt} &= 2\frac{dx}{dt} \\ \Rightarrow \frac{dy}{dt} &= \frac23 \times 0.3 = 0.2 \end{aligned}

Why?

Both $x$ and $y$ change with time, so we differentiate the relation $3y = 2x$ with respect to $t$ ⁠. The man walks away from the light at 0.3 m/s, so $x$ increases at that rate: $\frac{dx}{dt} = 0.3$ ⁠.

\begin{aligned} \frac{d}{dt}(3y) &= \frac{d}{dt}(2x) \\ 3\frac{dy}{dt} &= 2\frac{dx}{dt} \\ 3\frac{dy}{dt} &= 2(0.3) = 0.6 \\ \frac{dy}{dt} &= \frac{0.6}{3} = 0.2 \end{aligned}

Watch out: find the relation between $x$ and $y$ first, and only then differentiate. Never put numbers in before differentiating.

The tip of the shadow is at a distance $(x + y)$ m from AB. So the tip moves at the rate

\begin{aligned} \frac{d}{dt}(x + y) &= \frac{dx}{dt} + \frac{dy}{dt} \\ &= 0.3 + 0.2 = 0.5 \end{aligned}

Why?

The tip C moves away from the fixed pole AB, and its distance from AB is $AC = x + y$ ⁠. The rate of change of a sum is the sum of the rates.

The length of the shadow is $y$ ⁠, so the shadow lengthens at $\frac{dy}{dt}$ ⁠. It is positive, so the shadow is getting longer.

The shadow lengthens at the rate $\frac{dy}{dt} = 0.2$ m/s.

$\therefore$ The tip of the shadow moves at 0.5 m/s, and the shadow lengthens at 0.2 m/s.

Where this comes from

Chapter 6, Application of Derivatives: Rate of change: a man's shadow from a street light.

Question type: Rates in time

How to spot it: click to see

Type 2. Rates in time: how fast is it changing now?

Spot it: words like "per second", "how fast", "at the rate of", and one moment, such as "when the radius is 10 cm".

Here: He walks at 0.3 m/s, so $\frac{dx}{dt} = 0.3$ ⁠. Similar triangles link his distance and his shadow, $3y = 2x$ ⁠: differentiate it with respect to $t$ for $\frac{dy}{dt}$ ⁠, and the tip moves at $\frac{dx}{dt} + \frac{dy}{dt}$ ⁠.

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Q22: A man walks away from a street light: rates of his shadow