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Differential Calculus

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.

Shreya Gupta

7 Years agoGrade

3 Answers

Susmita

7 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

7 Years ago

Sorry.it would be

(25/20)*(1/5)*10=5/2

Forgot to do l²..................................................

Susmita

Approved Tutor Answer7 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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