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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?

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1 Answer

  • 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
answered Aug 8, 2013 by thanvigandhi_1

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