Answer:
d∅/dt = √2/5 Rad/sec
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