Question

According to the model we used to solve example 2, what happens as the top of the ladder approaches the ground? Is the model appropriate for small value of y?

Short answer

The rate of change in \(y\) with respect to \(t\) tends to \( - \infty \) as \(y\) tends to 0. This model is not appropriate for small value of\(y\).

Step-by-step solution

Draw the diagram

Let a ladder leans against a wall is sliding, which one represented by a diagram.

The objective is to find what happens if the top of the ladder approaches the ground, then check the validity of the model used to find the rate of change in \(y\), for the smaller value of \(y\).

Apply Pythagoras theorem

Here the length of the ladder is \(L\) units.

By the Pythagoras theorem,

\(\begin{aligned}{x^2} + {y^2} &= {L^2}\\{y^2} &= {L^2} - {x^2}\\y &= \sqrt {{L^2} - {x^2}} \end{aligned}\)

Put \(0\) for \(y\) in the above equation.

\(\begin{aligned}0 &= \sqrt {{L^2} - {x^2}} \\0 &= {L^2} - {x^2}\\{x^2} &= {L^2}\\x &= L\end{aligned}\)

\(\)

If the top of the ladder approaches the ground, Then the distance of the bottom of the ladder from the wall will approach to the length of the ladder.

Apply Chain rule of differentiation

Chain rule of differentiation is \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}.\frac{{du}}{{dx}}\).

Use chain rule on \({x^2} + {y^2} = {L^2}\).

\(\begin{aligned}2x\frac{{dx}}{{dt}} + 2y\frac{{dy}}{{dt}} &= 0\\2y\frac{{dy}}{{dt}} &= - 2x\frac{{dx}}{{dt}}\\y\frac{{dy}}{{dt}} &= - x\frac{{dx}}{{dt}}\\\frac{{dy}}{{dt}} &= - \frac{x}{y}\frac{{dx}}{{dt}}\end{aligned}\)

The rate of change in \(y\) with respect to \(t\) tends to \( - \infty \) as \(y\) tends to 0. This model is not appropriate for small value of\(y\).