Miscellaneous · Q2: Isosceles triangle on a fixed base $b$⁠: the area falls at $\sqrt{3}\,b$ cm²/s

Basics you need

Solution

1. Name the variables, units and the given rate

   Let $t$ be the time in seconds, $x$ cm the length of each equal side at time $t$⁠, and $A$ cm² the area. The base $b$ cm is a constant. We are told

   $$\frac{dx}{dt} = -3 \text{ cm/s},$$

   and we want $\frac{dA}{dt}$ at the instant when $x = b$⁠.

   The rate is negative because "decreasing" means $x$ gets smaller as $t$ grows. Writing the sign in from the start lets the answer's sign tell us whether the area grows or shrinks.
   Rate of change as a derivative (Basic 1). Lesson: rate of change of quantities

2. Height from Pythagoras

   In an isosceles triangle the altitude from the apex meets the base at its midpoint (the two right triangles it creates are congruent, by RHS: same hypotenuse $x$⁠, common side $h$⁠). So each half of the base is $\frac b2$⁠, and in either right triangle

   $$\begin{aligned} h^2 + \left(\tfrac b2\right)^2 &= x^2 \\ h &= \sqrt{x^2 - \tfrac{b^2}{4}} = \tfrac12\sqrt{4x^2 - b^2}. \end{aligned}$$

   This needs $4x^2 > b^2$⁠, that is $x > \frac b2$⁠: two sides of a triangle must together be longer than the third. At the instant we care about, $x = b > \frac b2$⁠, so all is well.

   The area needs a height, and the height is not given directly. The half-triangle is a right triangle whose sides we know in terms of $x$ and $b$⁠, so Pythagoras supplies $h$⁠. Writing $\sqrt{x^2 - \frac{b^2}{4}}$ as $\frac12\sqrt{4x^2 - b^2}$ clears the fraction inside the root (multiply inside by 4, so take $\frac12$ outside).
   Pythagoras' theorem (Basic 3). Basics: area and volume formulas

3. The area as a function of $x$

   $$A = \frac12 \cdot b \cdot h = \frac{b}{4}\sqrt{4x^2 - b^2}.$$
   With $b$ fixed, $A$ depends on $x$ alone. That is exactly what we need: one changing quantity whose rate we know.

4. Differentiate with respect to time (chain rule)

   The inner function is $u = 4x^2 - b^2$⁠, with $\frac{du}{dt} = 8x\,\frac{dx}{dt}$ (the constant $b^2$ has derivative 0). The outer function is $\frac b4\sqrt{u}$⁠, with derivative $\frac{b}{4}\cdot\frac{1}{2\sqrt u}$⁠. So

   $$\begin{aligned} \frac{dA}{dt} &= \frac{b}{4}\cdot\frac{1}{2\sqrt{4x^2 - b^2}}\cdot 8x\,\frac{dx}{dt} \\ &= \frac{bx}{\sqrt{4x^2 - b^2}}\,\frac{dx}{dt}. \end{aligned}$$
   $x$ is itself a function of $t$⁠, so differentiating $x^2$ with respect to $t$ gives $2x\,\frac{dx}{dt}$⁠, not just $2x$⁠. Then $\frac b4 \cdot \frac12 \cdot 8 = b$⁠, which tidies the constants.
   Chain rule, Theorem 4, Section 5.3.1 (Basic 2); derivative of $\sqrt u$ (Basic 4). Basics: related rates

5. Substitute the instant $x = b$

   At $x = b$⁠: $\sqrt{4b^2 - b^2} = \sqrt{3b^2} = \sqrt3\,b$ (as $b > 0$⁠). With $\frac{dx}{dt} = -3$⁠:

   $$\begin{aligned} \frac{dA}{dt} &= \frac{b \cdot b}{\sqrt3\,b}\cdot(-3) \\ &= -\frac{3b}{\sqrt3} = -\sqrt3\,b. \end{aligned}$$
   The instant enters only now, after differentiating. $\frac{3}{\sqrt3} = \sqrt3$ because $3 = \sqrt3\cdot\sqrt3$⁠. At this instant the triangle is equilateral (all sides $b$⁠), which is a good mental picture.

6. Interpret the sign and state the units

   $\frac{dA}{dt} = -\sqrt3\,b$ is negative, so the area is decreasing. Its units are cm² per second (cm² of area per second of time). So the area is decreasing at $\sqrt3\,b$ cm²/s.

   The question asks "how fast is the area decreasing", so the answer is the size of the rate, $\sqrt3\,b$⁠, with the word "decreasing" carrying the minus sign.
   A negative rate means the quantity is decreasing (Note after Example 3, Section 6.2). Basics: rate of change as a derivative

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