This answer is restricted. Please login to view the answer of this question.

Login NowSolution:

Given,

y^{2} = x^{3}

y = x^{3/2}

\(\frac{dy}{dx} = \frac{3}{2} x^{\frac{1}{2}}\)

The arc length formula gives

L = \(\int_1^4 \sqrt{1 + \left ( \frac{dy}{dx} \right )^2 dx}\)

= \(\int_1^4 \sqrt{1 + \frac{9}{4}x dx}\)

If we substitute

u = \(1 + \frac{9}{4}x\), then \(du = \frac{9}{4} dx\)

when x = 4 then u = 10

when x = 1 then u = 13/4

Therefore

L = \(\frac{4}{9} \int_{\frac{13}{4}}^{10} \sqrt{u} \enspace du\)

= \(\left [ \frac{4}{9} . \frac{2}{3} u^{\frac{3}{2}} \right ]_{\frac{13}{3}}^{10}\)

= \(\frac{8}{27} \left [ 10^{\frac{3}{2}} – \left ( \frac{13}{4} \right )^{\frac{3}{2}} \right ]\)

= \(\frac{1}{27} \left ( 80 \sqrt{10} – 13\sqrt{13} \right )\)

If you found any type of error on the answer then please mention on the comment or report an answer or submit your new answer.

Click here to submit your answer.

HAMROCSIT.COM

## Discussion