A three meters steel pipe is leaning against a vertical wall and the other end is on the horizontal floor. If the lower end slides away from the wall at the rate of 2 cm/s, how fast the upper end is sliding down when the lower end is 2 m away from the wall?
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Step-by-step explanation:
Let ym be the height of the wall at which the ladder touches. Also, let the foot of the ladder be xm away from the wall.
Then, by Pythagoras theorem, we have:
x
2
+y
2
=25 [Length of the ladder =5m]
⇒y=
25−x
2
Then, the rate of change of height (y) with respect to time (t) is given by,
dt
dy
=
25−x
2
−x
⋅
dt
dx
It is given that
dt
dx
=2cm/s
∴
dt
dy
=
25−x
2
−2x
Now, when x=4m, we have:
dt
dy
=
25−4
2
−2×4
=−
3
8
Hence, the height of the ladder on the wall is decreasing at the rate of
3
8
cm/s.
just change the no.
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