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A ladder,5cm long , standing on a horizontal floor ,leans against a vertical wall. If the top if the ladder slides down at a rate of 10cm/s,find rate at which the angle between the floor and the ladder is decreasing when the lower end of the ladder is 2m from the wall.

A ladder,5cm long , standing on a horizontal floor ,leans against a vertical wall. If the top if the ladder slides down  at a rate of 10cm/s,find rate at which the angle between the floor and the ladder is decreasing when the lower end of the ladder is 2m from the wall.

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3 Answers

Susmita
425 Points
5 years ago
It is a right angle triangle with side adjucent to right angle x,hypotaneus l=5cm,and base length say y.
given dx/dt=10cm/s.
sinA=x/l
Or,A=sin-1(x/l)
We are asked to find out dA/dt when y=2m=200cm.
\frac{dA}{dt}=\frac{1}{\sqrt{1-\frac{x^2}{l^2}}} \frac{d(\frac{x}{l})}{dt}
        \frac{dA}{dt}=\frac{l^2}{\sqrt{l^2-x^2}} [\frac{1}{l}\frac{dx}{dt}-\frac{x}{l^2}\frac{dl}{dt}]
As l is fixed so dl/dt=0.
and y=200cm
So l2=x2+y2
Or,l2-x2=400
\frac{dA}{dt}=\frac{5}{20}*\frac{1}{5}*10
=1/2
if it helps please approve.thank u.
 
Susmita
425 Points
5 years ago
Sorry.it would be
(25/20)*(1/5)*10=5/2
Forgot to do l2..................................................
Susmita
425 Points
5 years ago
Oh 
y=200cm
So l2-x2=y2=(200)2
root(l2-x2)=200 and
That wont be l2.
\frac{dA}{dt}=\frac{l}{\sqrt{(l^2-x^2)}}\frac{1}{l}\frac{dx}{dt}
       =(5/200)*(1/5)*10=1/20
 
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