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.
Best Answer
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.
This is false.
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 $y$.
While you may be capable of intuiting the correct signs of your results in early problems such as this one, as the situations become more complex, relying on your intuition is hazardous. Others will be right to question why they should believe your intuition when calculating with the correct signs is straightforward. It appears you would benefit from drawing a diagram where you unambiguously assign consistent signs to the quantities in your equations, which will force consistent signs on the derivatives. Drawing this diagram also avoids the difficulties spelled out above for a reader to interpret how your symbols are related to concrete quantities.