Exercise 6.1 · Q10: A 5 m ladder slides: how fast does the top fall?

Basics you need

Solution

1. Name the variables and fix the units

   At time $t$ seconds, let

   • $x$ = distance of the foot of the ladder from the wall, in cm;
   • $y$ = height of the top of the ladder above the ground, in cm.

   Both change with time. The ladder's length does not: it is $5$ m $= 500$ cm. The foot moves away from the wall, so $x$ increases:

   $$\frac{dx}{dt} = 2 \text{ cm/s}.$$

   The instant we care about is when $x = 4$ m $= 400$ cm.

   The rate is given in cm/s, so working in cm means the answer comes out directly in cm/s. Mixing metres and centimetres in one formula is the most common way to get this question wrong by a factor of 100. We also assume, as the question intends, that the wall is vertical, the ground horizontal, and the top of the ladder stays in contact with the wall while the foot slides.
   $\frac{dx}{dt} > 0$ because $x$ is growing. A decreasing quantity would get a negative rate, as with the length in Example 4 ($\frac{dx}{dt} = -3$ cm/min). Basics: rate of change as a derivative

2. The relation between $x$ and $y$

   The wall and the ground meet at a right angle, and the ladder is the hypotenuse of the right triangle they form (see the figure). By Pythagoras,

   $$x^2 + y^2 = 500^2 = 250000 \quad\text{for every } t.$$
   To connect the two rates we first need an equation connecting the two quantities themselves, and it must hold at every moment, not just at the instant. The ladder's fixed length gives exactly such an equation. Only an equation true for all $t$ may be differentiated with respect to $t$⁠.
   Pythagoras theorem: (leg)$^2$ + (leg)$^2$ = (hypotenuse)$^2$⁠.

3. The height at the instant

   Put $x = 400$ into the relation:

   $$\begin{aligned} y^2 &= 250000 - 400^2 \\ &= 250000 - 160000 = 90000, \\ y &= \sqrt{90000} = 300 \text{ cm}. \end{aligned}$$

   We take the positive square root because $y$ is a height. So at the instant the top is 3 m up the wall: the triangle is the familiar 3-4-5 right triangle.

   The formula for $\frac{dy}{dt}$ in Step 4 will involve both $x$ and $y$⁠, and the question gives only $x$⁠. The relation from Step 2 supplies $y$ at the same moment.

4. Differentiate the relation with respect to time

   Differentiate both sides of $x^2 + y^2 = 250000$ with respect to $t$⁠. Each square is a function of a quantity that depends on $t$⁠, so use the chain rule on each:

   $$\begin{aligned} \frac{d}{dt}\left(x^2\right) &= 2x\,\frac{dx}{dt}, \\ \frac{d}{dt}\left(y^2\right) &= 2y\,\frac{dy}{dt}, \\ \frac{d}{dt}(250000) &= 0. \end{aligned}$$

   So

   $$2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0,$$

   and, dividing by $2y$ (allowed because $y \ne 0$ while the top is above the ground),

   $$\frac{dy}{dt} = -\frac{x}{y}\,\frac{dx}{dt}.$$
   This is the heart of a related-rates problem: one equation between the quantities becomes one equation between their rates. We differentiate before substituting the instant's numbers. If we put $x = 400$ first, $x$ would look like a constant and its rate would wrongly become 0. The right-hand side 250000 is the square of the ladder's length, which never changes, so its rate is 0.
   Chain rule (Theorem 4, Section 5.3.1) with the power rule $\frac{d}{dx}\left(x^2\right) = 2x$ (Table 5.3). Basics: related rates (chain rule)

5. Substitute the values at the instant

   Now use $x = 400$⁠, $y = 300$ and $\frac{dx}{dt} = 2$⁠:

   $$\begin{aligned} \frac{dy}{dt} &= -\frac{400}{300}\times 2 \\ &= -\frac{4}{3}\times 2 = -\frac{8}{3} \text{ cm/s}. \end{aligned}$$

   As a decimal, $-\frac{8}{3} \approx -2.67$ cm/s.

   Every symbol now has a value at the same instant, so the formula from Step 4 gives a number. The units work out: $\frac{x}{y}$ is cm divided by cm, a pure number, and $\frac{dx}{dt}$ is in cm/s, so $\frac{dy}{dt}$ is in cm/s.

6. Read the sign and answer the question asked

   $\frac{dy}{dt}$ is negative, so the height is decreasing, exactly as the question expects. The question asks how fast it is decreasing, which is the size of the rate:

   $$\begin{aligned} \text{rate of decrease} &= \frac{8}{3} \text{ cm/s} \\ &\approx 2.67 \text{ cm/s}. \end{aligned}$$

   Notice the top falls faster than the foot moves out ($\frac{8}{3} > 2$⁠), because at this moment $x > y$⁠, so the factor $\frac{x}{y} = \frac{4}{3}$ is bigger than 1.

   "How fast is it decreasing" already contains the direction, so the answer is stated as a positive speed of decrease. Saying "decreasing at $-\frac{8}{3}$ cm/s" would be a double negative. The ratio $\frac{x}{y}$ also shows what happens later: as the foot moves out, $y$ shrinks and $\frac{x}{y}$ grows without bound, so near the ground the formula predicts an ever faster fall. In reality the top leaves the wall before that, so the model is only trusted while the top is in contact with the wall.
   Negative rate means decreasing quantity (the Note after Example 3, Section 6.2). Lesson: rate of change of quantities

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