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

Question

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

anonymous · Answer

@zepdrix @electrokid

anonymous · Answer

do you want to me to draw the equation?

anonymous · Answer

|dw:1363864864886:dw|

anonymous · Answer

start with that problem, i'll draw the second when someone has explained the answer.

zepdrix · Answer

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.

anonymous · Answer

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

zepdrix · Answer

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

anonymous · Answer

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?

zepdrix · Answer

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

anonymous · Answer

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

zepdrix · Answer

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,$$\large \int\limits \sqrt{1+\color{orangered}{16x^4}}dx$$

anonymous · Answer

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