Question

Find the length of the curve: y = integral of x to 1, sqrt (sqrt(t) -1)dt where 1≤x≤16

How do I get this in terms that I can understand?

Answer 1

use the draw button to try to write that out better, hard to read the information

Answer 2

or if you can take a screenshot or pic of the problem that would be good as well

Answer 3

\[\int\limits_{1}^{x}\sqrt{\sqrt{t}-1}dt\] where 1≤x≤16
is the original equation

Answer 4

interesting ....

Answer 5

\[ds=\sqrt{(x')^2+(y')^2}\]
the length of a curve is just:\[\int ds\]
you say:\[y=\int\limits_{1}^{x}\sqrt{\sqrt{t}-1}dt\]
\[y'=\sqrt{\sqrt{x}-1}(x')-\sqrt{\sqrt{1}-1}(1')\]

Answer 6

y'^2 = sqrt(x)-1

x'^2 = 1

ds = sqrt(1+sqrt(x)-1), from 1 to 16

Answer 7

\[L=\int_{1}^{16}\sqrt{x^{1/2}}~dx\]
\[L=\int_{1}^{16}x^{1/4}~dx\]

Answer 8

Oh, I got it now! So the derivative of a derivative is plugging in the points of the integral and subtracting both ends?

Answer 9

in general:\[F(x)=\int_{\alpha}^{\beta}f(t)~dt\]
\[F(x)=F(\beta)-F(\alpha)\]
taking the derivative invokes the chain rule
\[\frac{d}{dx}F(x)=f(\beta)~\beta'-~f(\alpha)~\alpha'\]

Answer 10

in this case: 1' = 0, and x' = 1 leaving us the original inset expression in terms of x

Answer 11

Thank you so much!

Answer 12

youre welcome