WBJEE · Maths · Application of Derivatives
A ladder \(20 \mathrm{ft}\) long leans against a vertical wall. The top end slides downwards at the rate of \(2 \mathrm{ft}\) per second. The rate at which the lower end moves on a horizontal floor when it is \(12 \mathrm{ft}\) from the wall is
- A \(\frac{-8}{3}\)
- B \(\frac{6}{5}\)
- C \(\frac{3}{2}\)
- D \(\frac{17}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{-8}{3}\)
Step-by-step Solution
Detailed explanation
\[ x^{2}+y^{2}=(20)^{2}=400 \] We have. \(\frac{d y}{d t}=2 \mathrm{ft} / \mathrm{sec}\) When \[ x=12 \] then \(\left(13^{2}+y^{2}=400\right.\) \(\Rightarrow \quad 144+y^{2}=400\) \(\Rightarrow \quad y^{2}=400-144=256\) \(\Rightarrow \quad y=16\) Now,…
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