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Novice
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Date Posted: 5/12/2008 12:54:27 AM  Status: Live
calculus 1
Course Textbook Chapter Problem
Calculus N/A N/A N/A
Question Details:
the length of a rectangle is decreasing at the rate of 8ft/s while the width is increasing at the rate of 4ft/s. At a particular time, the length is 16ft and the diagonal is 400ft. At this particular time, tind the rate of change of the area.

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Sage
Karma Points: 3,725
Date Posted: 5/12/2008 1:10:52 AM  Status: Live
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Response:
Area = Length x width  ---------> A = L.w
 
Let's differentiate this expression in terms of t (time)
 
 
we know that :   and 
 
=======>   dA/dt = -8w + 4L
 
Knowing  L = 16 ft and w , we can just plug them into the expression and find (dA/dt)  (in ft2/s)
 
at a particular time:
 
we know that : D2 = L2 + w2 --------> w = √(D2 - L2) = √ (400)2 - (16)2 ˜ 400 ft
 
can you double check the value of the diagonal it seems the value is to high
 
hope this helps
 



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