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a 10 foot ladder rests against a vertical wall if the bottom of the ladder slides away from the wall at a speed of 2 ft/s how fast is the angle betwween the top of the ladder and the wall changing when that angle is

Sagot :

Answer:

d∅/dt = √2/5 Rad/sec

Step-by-step explanation:

According to the Question,

  • Given That, a 10-foot ladder rests against a vertical wall if the bottom of the ladder slides away from the wall at a speed of 2 ft/s how fast is the angle between the top of the ladder and the wall changing when that angle is π/4.

Solution,

Let x be the Distance between the base of the wall and the bottom of the ladder.

and let ∅ be the angle between the top of the ladder and the wall.

Then, Sin∅ =x/10  so, x=sin∅ *10

Differentiating with respect to time t we get,

dx/dt = 10 * cos∅ * d∅ /dt

  • We have given that dx/dt = 2 ft/s and ∅ =π/4

Now, Put these value we get

2 = 10 *(cos(π/4))* d∅/dt

2 = 10/√2 * d∅/dt

d∅/dt = √2/5 Rad/sec

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