Comment
Share
Q)

# A ladder, 5 meter long,standing on a horizontal floor,leans against a vertical wall.If the top of the ladder slides downwards at the rate of 10cm/sec,then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 meters from the wall measure in "radian/seconds" is?

$(A)\;\frac{1}{10}radian/sec\quad(B)\;\frac{1}{20}radian/sec\quad(C)\;20\;radian/sec$$(D)\;10\;radian/sec$

Comment
A)
Toolbox:
• To obtain the absolute maxima or minima for the function $f(x)$
• (i) Find $f'(x)$ and put $f'(x)=0$
• (ii) Obtain the points from $f'(x)=0$
• (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
Length of the ladder = $500$ cm
Let $\theta$ be the angle between the floor and ladder.
$\sin \theta = \large\frac{y}{500}$
On differentiating w.r.t $t$ we get,
$\cos \theta \large\frac{d\theta}{dt}=\large\frac{1}{500} \times \large\frac{dy}{dt}$
It is given $\large\frac{dy}{dt}=10\: cm/sec$
$\cos \theta = \large\frac{x}{500}$ when $x = 200$
$\cos \theta = \large\frac{200}{500} = \large\frac{2}{5}$
Step 2
Substituting the respective values we get,
$\large\frac{2}{5} \large\frac{d\theta}{dt} = \large\frac{1}{500} \times 10$
$\Rightarrow \large\frac{ d\theta}{dt} = \large\frac{1}{50} \times \large\frac{5}{20}$
$= \large\frac{1}{20}$ rad / sec
Hence the correct option is B