# Related rates ladder problem; does the sign matter?

I have a 20ft long ladder leaning against a wall. The top of the ladder is sliding down the wall at a rate of 2ft/sec. I want to know the rate of change of the bottom of the ladder when the top of the ladder is 5ft from the floor. Explicitly, I'm looking for the "rate the bottom of the ladder is sliding away from the wall".

So I know a few things:

\begin{align} h &= 20ft \\ \frac{dy}{dt} &= 2ft/sec \\ \frac{dx}{dt} &= ? \end{align}

Where $$h$$ is the hypotenuse of the triangle this makes. I can find $$\frac{dx}{dt}$$ using Pythagoras' theorem and its derivative:

\begin{align} y^2 + x^2 &= 20^2 \\ 2y \frac{dy}{dt} + 2x \frac{dx}{dt} &= 0 \\ y\frac{dy}{dt} + x\frac{dx}{dt} &= 0 \\ y(2) + x\frac{dx}{dt} &= 0 \\ 2y + x\frac{dx}{dt} &= 0 \\ x\frac{dx}{dt} &= -2y \\ \frac{dx}{dt} &= \frac{-2y}{x} \end{align}

When $$x = 5$$, I know:

\begin{align} y^2 + 5^2 &= 20^2 \\ y^2 &= 20^2 - 5^2 \\ &= 400 - 25 \\ &= 375 \\ y &= \sqrt{375} \end{align}

Combining everything, I get:

\begin{align} \frac{dx}{dt} &= \frac{-2(\sqrt{375})}{5} \\ &\approx -7.746 \text{m/s} \end{align}

Is there significance to the negative sign to this answer? The software I was tested on marked $$7.746 \text{m/s}$$ as incorrect as I interpreted as it needing the magnitude (a ladder sliding down would not have the bottom going towards the wall). I believe my arithmetic is correct, so I surmise the issue is the sign.

• The negative sign is there because you wrote $\frac{dy}{dt}=2$ and not $-2$. – Andrew Chin Nov 23 '19 at 0:58

You say $$\frac{\mathrm{d}y}{\mathrm{d}t} = 2 \,\mathrm{ft}/\mathrm{sec}$$, which says the quantity $$y$$ is increasing by two feet every second. You have not stated what quantity in the problem is $$y$$. I guess that you intend $$y$$ be to the height of the point on the wall in contact with the top of the ladder. You have stated that point is rising to the ceiling.
You do not indicate what the quantity $$x$$ represents in your problem. I guess you mean that $$x$$ is the distance from the base of the wall along the floor between the wall and the point of contact of the ladder with the floor. You have concluded that, under the condition that $$y$$ is increasing at $$2$$ feet per second, $$x$$ is changing with the rate $$-2\sqrt{375}/5$$ feet per second. The sign on this result is entirely a consequence of the falsehood about the rate of change in the quantity $$y$$.