Math Help - Arc Length Problem

evaluated at upper bound 1 and lower bound 0





















evaluated at upper bound 1 and lower bound 0

evaluated at upper bound 1 and lower bound 0

evaluated at upper bound 1 and lower bound 0

- Is this right?

Why have you written your derivative in such an odd format?

$\displaystyle \begin{align*} y &= \sqrt{ 2 - x^2} \\ \frac{ \mathrm{d}y }{\mathrm{d}x} &= -\frac{2x}{2 \sqrt{2 - x^2} } \\ &= -\frac{x}{\sqrt{2 - x^2}} \end{align*}$

so your arclength is

$\displaystyle \begin{align*} L &= \int_a^b{ \sqrt{1 + \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 } \, \mathrm{d}x } \\ &= \int_0^1{ \sqrt{ 1 + \left( -\frac{x}{\sqrt{2 - x^2} } \right) ^2 } \, \mathrm{d}x } \end{align*}$

Go from here...

Sorry for delay in responding.















Using trig substitution you get to this point:









- What next it seems like all the thetas cancel out, so how can we evaluate it?

Originally Posted by Jason76

Sorry for delay in responding.

Fixed that step

Using trig substitution you get to this point:

NO! Here your $\displaystyle \begin{align*} a = \sqrt{2} \end{align*}$ so the substitution you must make is $\displaystyle \begin{align*} x = \sqrt{2}\sin{ \left( \theta \right) } \implies \mathrm{d}x = \sqrt{2}\cos{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$

- What next it seems like all the thetas cancel out, so how can we evaluate it?

Even though you are out by a significant factor from the above mistake, you come to the right conclusion that the "functional" part cancels out. So how would you evaluate $\displaystyle \begin{align*} \int{ \mathrm{d}x } \end{align*}$? Hint: $\displaystyle \begin{align*} \int{ \mathrm{d}x} = \int{ 1 \, \mathrm{d}x } \end{align*}$...

Would the answer be evaluated at 1 upper bound and 0 lower bound?

Well, if you mean evaluated where x is between 0 and 1, then yes.

But you will be better off changes your bounds to $\displaystyle \begin{align*} \theta \end{align*}$ bounds, so that you don't have to convert back to a function of x.

How could I convert back to a function of x without all the trig functions? If I had those, then all I would have to do is make a right triangle.

You NEED the trig functions to convert back to a function of x.

Convert your bounds, it's a much easier process.

Originally Posted by Prove It

You NEED the trig functions to convert back to a function of x.

Convert your bounds, it's a much easier process.

how to convert bounds? hint?

You know that $\displaystyle \begin{align*} x = \sqrt{2}\sin{ \left( \theta \right) } \end{align*}$. If x = 0 then what is $\displaystyle \begin{align*} \theta \end{align*}$? If x = 1 then what is $\displaystyle \begin{align*} \theta \end{align*}$?

Originally Posted by Prove It

You know that $\displaystyle \begin{align*} x = \sqrt{2}\sin{ \left( \theta \right) } \end{align*}$. If x = 0 then what is $\displaystyle \begin{align*} \theta \end{align*}$? If x = 1 then what is $\displaystyle \begin{align*} \theta \end{align*}$?

right I see. The sin of 0 is 0 and the sin of 1 is ?? Is it

Answer would be ??

NO! You are trying to figure out which ANGLES give those sine values! What angle gives the sine value as 0? (You have correctly identified it as 0). What angle gives $\displaystyle \begin{align*} \sqrt{2}\sin{ \left( \theta \right) } = 1 \implies \sin{ \left( \theta \right) } = \frac{1}{\sqrt{2}} \end{align*}$. This is another one that you should have committed to memory.

Wouldn't it be?

Don't listen to me then...

is not the answer on homework.