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Question: Answered & Verified by Expert
A ladders $5 \mathrm{~m}$ long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall, at the rate of $2 \mathrm{~m} / \mathrm{sec}$. The speed at which its height on the wall decreases when the foot of the ladder is $4 \mathrm{~m}$ away from the wall is
MathematicsApplication of DerivativesCOMEDKCOMEDK 2015
Options:
  • A $\frac{3}{8} \mathrm{~m} / \mathrm{sec}$
  • B $\frac{8}{3} \mathrm{~m} / \mathrm{sec}$
  • C $\frac{5}{3} \mathrm{~m} / \mathrm{sec}$
  • D $\frac{2}{3} \mathrm{~m} / \mathrm{sec}$
Solution:
2972 Upvotes Verified Answer
The correct answer is: $\frac{8}{3} \mathrm{~m} / \mathrm{sec}$
Let $A B$ be the ladder of length $5 \mathrm{~m}$. We are given, $\frac{d x}{d t}=2 \mathrm{~m} / \mathrm{sec}$



In $\triangle A B C$
$$
\begin{aligned}
& A B^{2}=A C^{2}+B C^{2} \\
\Rightarrow \quad &(5)^{2}=x^{2}+h^{2} \quad \text{...(i)}
\end{aligned}
$$
Differentiating Eq. (i) w.r.t. $t$, we get
$$
\begin{aligned}
0 &=2 x \frac{d x}{d t}+2 h \frac{d h}{d t} \\
\Rightarrow \frac{d h}{d t} &=\frac{-x}{h} \frac{d x}{d t} \quad \text{...(ii)}
\end{aligned}
$$
Now, from Eq. (i), when $x=4$
$\quad h^{2}=25-16$ $\Rightarrow h^{2}=9 \Rightarrow h=3 \mathrm{~m}$
From Eq. (ii), we get $\left[\frac{d h}{d t}\right]=\frac{-4}{3} \times 2=\frac{-8}{3}$
$\because$ The negative sign shows the height decreases and decreasing rate is $\frac{8}{3} \mathrm{~m} / \mathrm{sec}$.

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