A ladder 5 m long is leaning against a wall . the bottom of the ladder is pulled along the grand away from the wall at the rate of 2cm/sec.how fast is its height on the wall decreasing when the foot of the ladder isv4m away frm wall
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Step-by-step explanation:
ANSWER
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.
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