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Answered

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A ladder 10 m long is learning against a vertical wall. If the bottom of the ladder is pushed horizontally towards the wall at 8/5 m/sec, how fast is the top of the ladder sliding up the wall when the bottom of the ladder is 6 m from the wall?​

Sagot :

Jiminswifieuidn

Answer:

Let AB be the ladder, where AB=10 metres,. Let at time t seconds, the end A of the ladder be x metres from the wall and the end B be y metres from the ground.

Since, OAB is a right angled triangle, by Pythagoras theorem.

x

2

+y

2

=10

2

i.e.. y

2

=100−x

2

Differentiating w.r.t t we get

2y

dt

dy

=0−2x

dt

dx

∴

dt

dy

=

y

x

.

dt

dx

...(1)

Now,

dt

dx

=

sec

12 metres

is the rate at wh the bottom of the ladder s pulled horizontally and

dt

dy

is the rate which the top of ladder B is sliding. which the top of ladder B is sliding.

When x=6,y

2

=100−36=64

∴y=8

∴(1) gives,

dt

dy

=−

8

6

(1.2)

=

8

6

×

10

12

−

10

9

=−0.9

Hence, the top of the ladder is sliding down the wall, at the rate of

sec

0.9 metre

Kimnyupagodnaidn

Answer:

Solution

[verified]

Let AB be the ladder, where AB=10 metres,. Let at time t seconds, the end A of the ladder be x metres from the wall and the end B be y metres from the ground.

Since, OAB is a right angled triangle, by Pythagoras theorem.

x

2

+y

2

=10

2

i.e.. y

2

=100−x

2

Differentiating w.r.t t we get

2y

dt

dy

=0−2x

dt

dx

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x

2

+y

2

=10

2

i.e.. y

2

=100−x

2

Differentiating w.r.t t we get

2y

dt

dy

=0−2x

dt

dx

∴

dt

dy

=

y

x

.

dt

dx

...(1)

Now,

dt

dx

=

sec

12 metres

is the rate which the top of ladder B is sliding. which the top of ladder B is sliding.

When x=6,y

2

=100−36=64

∴y=8

∴(1) gives,

dt

dy

=−

8

6

(1.2)

=

8

6

×

10

12

−

10

9

=−0.9

Hence, the top of the ladder is sliding down the wall, at the rate of

sec

0.9 metre

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Step-by-step explanation:

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