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  • A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?

10. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

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Let h be the height of the ladder and x be the distance of the foot of the ladder from the wall
It is given  that \frac{dx}{dt} = 2 \ cm/s
We need to find  the rate at  which the height of the ladder decreases (\frac{dh}{dt})
length of ladder(L) = 5m and x = 4m    (given)
By Pythagoras theorem, we can say that 
h^{2}+x^{2} = L^{2}
h^{2} = L^{2} - x^{2}
  h  = \sqrt{L^{2} - x^{2}}
Differentiate on both sides w.r.t.  t
\frac{dh}{dt} = \frac{d(\sqrt{L^{2} -x^{2}})}{dx}.\frac{dx}{dt} = \frac{1}{2}\frac{-2x}{\sqrt{5^{2}-x^{2}}}.\frac{dx}{dt} = \frac{-x}{\sqrt{25-x^{2}}}\frac{dx}{dt}
at x = 4

\frac{dh}{dt} = \frac{-4}{\sqrt{25-16}}\times 2 = \frac{-4}{3} \times 2 =\frac{-8}{3} \ cm/s
Hence, the rate at which the height of ladder decreases is \frac{8}{3} \ cm/s

Posted by

Gautam harsolia

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