Asked by Erica
1.) Find the arc length y=2/3x ^3/2 + 2 over (14,16)
2.) Find the arc length y=8x^3/2+7 over (0,6)
2.) Find the arc length y=8x^3/2+7 over (0,6)
Answers
Answered by
Reiny
The two questions are very similar in setup, I will do the first one, you do the second
length = integral √( 1 + (y')^2) dx from a to b
yours:
y = (2/3)x^(3/2) + 2
y' = x^(1/2)
(y')^2= x
length = integral √( 1 + x) dx from 14 to 16
= [ (2/3)(1+x)^(3/2) ] from 14 to 16
= (2/3)(17^(3/2) ) - (1/2)(15^(3/2) )
= appr 7.999
length = integral √( 1 + (y')^2) dx from a to b
yours:
y = (2/3)x^(3/2) + 2
y' = x^(1/2)
(y')^2= x
length = integral √( 1 + x) dx from 14 to 16
= [ (2/3)(1+x)^(3/2) ] from 14 to 16
= (2/3)(17^(3/2) ) - (1/2)(15^(3/2) )
= appr 7.999
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.