## rosho Group Title CalcII how to solve the arc length: a) int{a,b}sqrt(1+16x^4) b) int{a,b}sqrt(1+36cos^2(2x)) ans a)f(x)=+/-4x^3/3+C b)+/-3sin2x+C one year ago one year ago

1. rosho Group Title

@zepdrix @electrokid

2. rosho Group Title

do you want to me to draw the equation?

3. rosho Group Title

|dw:1363864864886:dw|

4. rosho Group Title

5. zepdrix Group Title

Are you sure you entered that correctly..? This is the solution Wolfram is giving, http://www.wolframalpha.com/input/?i=integral+sqrt%281%2B16x%5E4%29dx There must be a mistake somewhere in there.

6. rosho Group Title

we are trying to find the arc length. i think.

7. zepdrix Group Title

So find the arclength of this function, $$\large f(x)=\sqrt{1+16x^4}$$ From $$\large a$$ to $$\large b$$ ?

8. rosho Group Title

i'll type the text verbatim from the book. what differentiable function have an arc length on the interval [a,b] given by the following integrals?

9. zepdrix Group Title

Oh oh oh ok I see what's going on.

10. rosho Group Title

okay, that's good because i have no idea.

11. zepdrix Group Title

The formula for arc length is,$\large \int\limits ds \qquad = \qquad \int\limits \sqrt{1+\color{orangered}{\left(\frac{dy}{dx}\right)^2}}dx$ And we're given this,$\large \int\limits \sqrt{1+\color{orangered}{16x^4}}dx$

12. rosho Group Title

oh, i get it now... thanks a bunch