The sides of an equilateral triangle are increasing at the rate of $2$ cm/sec. The rate at which the area increases, when the side is $10$ cm, is: (A) $10$ cm$^2$/s (B) $\sqrt{3}$ cm$^2$/s (C) $10\sqrt{3}$ cm$^2$/s (D) $\dfrac{10}{3}$ cm$^2$/s

> [!TLDR] > The area of an equilateral triangle is $A = \frac{\sqrt{3}}{4} s^2$. Differentiating with respect to time gives $\frac{dA}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt}$. Substituting $s = 10$ cm and $\frac{ds}{dt} = 2$ cm/s yields $\frac{dA}{dt} = 10\sqrt{3}$ cm$^2$/s. The correct option is **(C)**. This is a classic **Related Rates** problem. The core idea: when two quantities are linked by a formula (here, area and side length of a triangle), their rates of change with respect to time are also linked. If you know how fast one is changing, you can find how fast the other changes — by differentiating the relationship with respect to time. The key step is always the same: write the relationship, differentiate both sides with respect to $t$ (using the chain rule where needed), then plug in the known values. Let’s walk through it. 1. **Write the formula for the area of an equilateral triangle.** For a triangle with side length $s$, the area is $$A = \frac{\sqrt{3}}{4} s^2.$$ This comes from the standard formula $A = \frac{1}{2} \times \text{base} \times \text{height}$, where the height of an equilateral triangle is $\frac{\sqrt{3}}{2}s$. 2. **Differentiate both sides with respect to time $t$.** Since $s$ changes with time, $A$ also changes with time. Using the chain rule: $$\frac{dA}{dt} = \frac{d}{dt} \left( \frac{\sqrt{3}}{4} s^2 \right) = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt}.$$ This is the general formula linking the rate of change of area to the rate of change of side length. 3. **Plug in the given values.** We are told: - $\frac{ds}{dt} = 2$ cm/s (the side is increasing at this rate), - $s = 10$ cm (the side length at the moment we care about). Substituting: $$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 10 \times 2 = \frac{\sqrt{3}}{2} \times 20 = 10\sqrt{3}.$$ 4. **Interpret the result.** …

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increa

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Concept understanding — Related Rates

Related Rates

The idea: quantities that change together

Many real situations involve two or more quantities that all vary with time, linked by a fixed relationship. Inflate a balloon and its radius and volume both grow; slide a ladder down a wall and the top's height and the foot's distance both change. A related-rates problem gives you the rate at which one quantity is changing and asks for the rate of another, at some instant.

The key insight: if the quantities are tied together by an equation, then their rates are tied together too. We uncover that link by differentiating the equation with respect to time $t$ .

The core mechanism: differentiate with respect to time

Every variable is a function of $t$ , so differentiating brings in the chain rule — each variable's derivative picks up a factor of its own rate. For example, if the volume of a sphere is $V=\tfrac{4}{3}\pi r^{3}$ , then differentiating both sides with respect to $t$ gives

\frac{dV}{dt} = 4\pi r^{2}\,\frac{dr}{dt}.

This single equation connects the rate the volume grows, $\dfrac{dV}{dt}$ , to the rate the radius grows, $\dfrac{dr}{dt}$ . Knowing one (and the current $r$ ) gives the other.

The standard procedure

t

Worked example

Air is pumped into a spherical balloon at $\dfrac{dV}{dt}=100\ \text{cm}^3/\text{s}$ . How fast is the radius increasing when $r=5\ \text{cm}$ ?

From $\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$ , substitute $\dfrac{dV}{dt}=100$ and $r=5$ :

$100 = 4\pi(5)^2\,\frac{dr}{dt} = 100\pi\,\frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{\pi}\ \text{cm/s}.$ …

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The area of an equilateral triangle is $A = \frac{\sqrt{3}}{4} s^2$ . Differentiating with respect to time gives $\frac{dA}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt}$ . Substituting $s = 10$ cm and $\frac{ds}{dt} = 2$ cm/s yields $\frac{dA}{dt} = 10\sqrt{3}$ cm $^2$ /s. The correct option is (C).

This is a classic Related Rates problem. The core idea: when two quantities are linked by a formula (here, area and side length of a triangle), their rates of change with respect to time are also linked. If you know how fast one is changing, you can find how fast the other changes — by differentiating the relationship with respect to time.

The key step is always the same: write the relationship, differentiate both sides with respect to $t$ (using the chain rule where needed), then plug in the known values.

Let’s walk through it.

Write the formula for the area of an equilateral triangle. For a triangle with side length $s$ , the area is

A = \frac{\sqrt{3}}{4} s^2.

This comes from the standard formula $A = \frac{1}{2} \times \text{base} \times \text{height}$ , where the height of an equilateral triangle is $\frac{\sqrt{3}}{2}s$ .

Differentiate both sides with respect to time $t$ . Since $s$ changes with time, $A$ also changes with time. Using the chain rule:

\frac{dA}{dt} = \frac{d}{dt} \left( \frac{\sqrt{3}}{4} s^2 \right) = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt}.

This is the general formula linking the rate of change of area to the rate of change of side length.

Plug in the given values.

We are told:
- $\frac{ds}{dt} = 2$ cm/s (the side is increasing at this rate),
- $s = 10$ cm (the side length at the moment we care about).
Substituting:

\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 10 \times 2 = \frac{\sqrt{3}}{2} \times 20 = 10\sqrt{3}.

Interpret the result. …

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Q.The sides of an equilateral triangle are increasing at the rate of $2$ cm/sec. The rate at which the area increases, when the side is $10$ cm, is:
(A) $10$ cm $^2$ /s
(B) $\sqrt{3}$ cm $^2$ /s
(C) $10\sqrt{3}$ cm $^2$ /s
(D) $\dfrac{10}{3}$ cm $^2$ /s

Concept understanding — Related Rates

Related Rates

The idea: quantities that change together

The core mechanism: differentiate with respect to time

The standard procedure

Worked example

Method: Related Rates for a Quantity Defined by a Single Changing Length

Steps

Common Mistakes

Q.The sides of an equilateral triangle are increasing at the rate of 222 cm/sec. The rate at which the area increases, when the side is 101010 cm, is: (A) 101010 cm2^22/s (B) 3\sqrt{3}3​ cm2^22/s (C) 10310\sqrt{3}103​ cm2^22/s (D) 103\dfrac{10}{3}310​ cm2^22/s

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Q.The sides of an equilateral triangle are increasing at the rate of $2$ cm/sec. The rate at which the area increases, when the side is $10$ cm, is:
(A) $10$ cm $^2$ /s
(B) $\sqrt{3}$ cm $^2$ /s
(C) $10\sqrt{3}$ cm $^2$ /s
(D) $\dfrac{10}{3}$ cm $^2$ /s

Unlock everything free for 14 days