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Questions Answered: | 12843 |
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MBA, Ph.D in Management
Harvard university
Feb-1997 - Aug-2003
Professor
Strayer University
Jan-2007 - Present
3 x t 4
( a) , 2 t 2
2 y t 3 3 x t 4t
(b) , 2 t 2
2 y t 3t x t3 4 a 2 y t 3 x '(t ) 3t 2
y '(t ) 2t
2 (3t 2 ) 2 (2t ) 2 dt 2
2 2 0 2 0 2 2 0 2 (9t 4 ) (4t 2 ) dt 2 (9t 4 ) (4t 2 )dt 2 (9t 4 ) (4t 2 )dt (9t 4 ) (4t 2 )dt
0 9t 4 4t 2 dt
0 9t 4 dt 3 2 t 4 dt 0 (0) 3 ( 2) 3 t3 3 t 2 dt 3 3 2
3 3 3 8 8 24 3 0 3 8
3 3 3 0 2 4t 2 dt 2 0 2 t 2 dt 2 2 0 2 0 t 2 2 tdt 2 2 2 2 2 2 4 2 0 2 2 0 4 2 0 2 2 0
2 0 9t 4 4t 2 dt 8 4 12
9t 4 4t 2 dt
2 9t 4 dt 3 0 t 4 dt 2 (2) 3 (0) 3 t3 3 t 2 dt 3 3 0
3 3 3 8 8 24 3 0 3 8
3 3 3 2 0 4t 2 dt 2 2 0 t 2 dt 2 2 0 2 2 t 2 2 tdt 2 2 0 2 2 2 4 2 0 2 2 0 4 2 2 0 9t 4 4t 2 dt 8 4 12 2 2 0 (3t 2 ) 2 (2t ) 2 dt 2 2 (9t 4 ) (4t 2 ) dt (9t 4 ) (4t 2 )dt 12 12 24
0 I am to find the arc length for each curve. Compute one exactly and approximate the other numerically.
My professor said, “You did what is unacceptable by distributing the sqrt symbol over addition.
That only happens with a product NOT with a sum. When you factor out the t^2 and take to the
outside of the sqrt symbol, you will get |t|. The break the integral at 0 into one from -2 to 0; and
another from 0 to 2. Apply the definition of absolute value on each of these ranges to get rid of
the absolute value. Proceed to integrate each by substitution.”
I attempted to correct it and this is what I received back. “Unfortunately, you did not work it out.
Read the first 3 lines of my previous comments and act on them. I do not see that anywhere. It's
not as easy as you make it to be. I provided all that you need in my previous comments. Until
you follow them in the order provided do not hurry to submit what is incorrect. If you do not
understand anything I asked you to do in my previous comments, ask specifically what it is.”
I am totally mind blown and am in need of assistance please.
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