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Calculus 2 - length of arc and volume questions


Radam

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2 hours ago, studiot said:

Are you not aware of this formula?

 


L=ba(1+(dydx)2)dx

 

Yes I am familiar with it. I'm not sure how I can reduce the square roots, enabling me to integrate. 

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For the first, you are given the graph of y= x4+ (1/32)x-2.  The derivative is y'= 4x3- (1/16)x-3.  Notice the "3" and "-3" powers.  When you square that, something nice happens: y'2= (4x3- (1/16)x-3)2= 16x6- 2(4x3)(1/16)(x-3)+ (1/256)x^-6.  I wrote that middle term out In detail to make clear that the "x" terms cancel leaving 16x6- (1/2)+ (1/256)x-6.  When we add 1 that "-1/2" becomes "+1/2" so that is simply (4x3+ (1/16)x-3)2 , a "perfect square".   sqrt{1+ (dy/dx)2}= 4x3+ (1/16)x-3.  Integrate that from 1 to 2.

Edited by Country Boy
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