Question

How to find the arc length of y=1+6x^(3/2), 0 14 years ago

Answer 1

measure the hypotenuse of a very tiny triangle from 0 to 1

Answer 2

\[\int_{D}^{}ds=\int_{0}^{1}\sqrt{1+[f'(x)]^2 }dx\]

Answer 3

y=1+6x^(3/2)

y' =9x^(1/2) i think

Answer 4

\[\int_{0}^{1}\sqrt{1+(9x^{1/2})^2 }dx\]
\[\int_{0}^{1}\sqrt{1+81x}\ dx\]

Answer 5

Thank you, I appreciate your help!

Answer 6

youre welocme, with any luck, that made sense :)

Answer 7

and was typed up correctly lol