Answer :
Rate of change of x with respect to [tex]\alpha[/tex] when [tex]\alpha =\frac{\pi }{3}[/tex] is 5ft/s.
Since the ladder is 10ft long resting against a vertical wall.
Let the angle between the top of the ladder and the wall is [tex]\alpha[/tex] .
And let the distance from the bottom of the ladder to the wall is x.
Thus we need to find the rate of change of x with respect to [tex]\alpha[/tex] when [tex]\alpha =\frac{\pi }{3}[/tex]
Consider L to be the length of the ladder.
Then, L = 10ft
Thus, we get the relation between the angle and distance x i.e.
x = L [tex]\sin \alpha[/tex]
On differentiating the above equation,
[tex]\frac{dx}{dt} =L \cos \alpha[/tex]
Now after putting the values [tex]\alpha =\frac{\pi }{3}[/tex] in above equation, we get,
[tex]=(10)\cos(\frac{\pi }{3} )=(10)(\frac{1}{2} )=5[/tex]
Therefore, [tex]\frac{dx}{dt} =5[/tex] ft/s
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