Q.The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ?

Concept: Related Rates — we connect the rate of change of the area to the given rate of change of the equal sides using the geometry of the triangle.

Let the equal sides be $a$ and the base $b$ (fixed). The height $h = \sqrt{a^2 - (b/2)^2}$. Area $A = \frac{1}{2} b h = \frac{b}{2} \sqrt{a^2 - \frac{b^2}{4}}$.

Differentiate with respect to time $t$:

Given $\frac{da}{dt} = -3$ cm/s (decreasing). When $a = b$, the height becomes $h = \sqrt{b^2 - b^2/4} = \frac{\sqrt{3}}{2}b$. Substitute:

This is a related-rates problem: with the base $b$ fixed, the area depends on the equal side $x$ through the height.

1. Express the area.

Let each equal side have length $x$. The altitude to the base is $h = \sqrt{x^2 - \dfrac{b^2}{4}}$, so

2. Differentiate with respect to time.

3. Substitute the given data.

The sides decrease at $3$ cm/s, so $\dfrac{dx}{dt} = -3$. When $x = b$,

Therefore

The negative sign shows the area is shrinking.

Method: Related Rates via a Geometric Area Relation

This method applies whenever a quantity built from other changing quantities (here, the area of a shape whose side lengths change with time) needs its rate of change found at a specific instant.

Steps

Step 1: Express the target quantity as a function of the changing variable(s)

Identify which lengths are fixed and which vary with time, then write the quantity you want the rate of (here, area $A$) purely in terms of the one varying length, using geometry (Pythagoras for the height of an isosceles triangle, or a standard area/volume formula).

Step 2: Differentiate both sides with respect to time $t$

Every length that changes with time picks up a $\dfrac{d(\cdot)}{dt}$ factor via the chain rule — never substitute a specific numeric value for the variable before this step, or its rate will vanish from the equation.

Step 3: Substitute the given rate and the instant's values

Plug in the known $\dfrac{dx}{dt}$ (with the correct sign — decreasing means negative) and the value of $x$ at the instant described in the question, then simplify.

Step 4: Interpret the sign

A negative result means the quantity is decreasing at that instant; state the answer with the correct sign and units, matching what the question asks (e.g. "how fast is the area decreasing" wants the magnitude, with the negative sign explaining why it is decreasing).

Common Mistakes

Mistake 1: Substituting the given numeric condition before differentiating

Why it's wrong: setting $x=b$ into the area formula first turns $x$ into a constant, so its derivative $\dfrac{dx}{dt}$ disappears from the equation entirely and the chain-rule link between the rates is lost. Correct approach: differentiate the general relation $A(x)$ with respect to $t$ first, and only substitute the specific value of $x$ afterward.

Mistake 2: Dropping or misreading the sign of the given rate

Why it's wrong: "decreasing at 3 cm/s" means $\dfrac{dx}{dt} = -3$, not $+3$; using the wrong sign flips the final answer from decreasing to increasing. Correct approach: always translate "increasing/decreasing at rate $r$" into a signed $\dfrac{d(\cdot)}{dt} = \pm r$ before substituting.

Mistake 3: Mixing up which side is the "equal side" versus the "base" in the height formula

Why it's wrong: writing $h=\sqrt{b^2-(x/2)^2}$ instead of $h=\sqrt{x^2-(b/2)^2}$ swaps which length is halved, giving an entirely wrong area function. Correct approach: draw the isosceles triangle, drop the altitude to the fixed base $b$, and confirm the half-base $b/2$ is one leg of the right triangle with the equal side $x$ as hypotenuse.