# Thread: Finding equation for a curve, given length integral

1. ## Finding equation for a curve, given length integral

Hey,

So I'm a bit confused on how an answer was reached. The question states:

Find a curve through the point (1,1) whose length integral is given by:

$L = \int_1^4 \sqrt{1+\frac{1}{4x}}dx$. From here I try to use point-slope formula, where m = $\frac{1}{2\sqrt{x}}$ (which comes from taking the square root of $\frac{1}{4x}$. Now, I apply the point-slope formula to try and get the equation of the curve, using:

y - y1 = m(x-x1) ==> y - (1) = $\frac{1}{2\sqrt{x}} (x - (1))$

After adding 1 to both sides and distributing the $\frac{1}{2\sqrt{x}}$ to the (x - 1), I end up with something like:

$y = \frac{x+2\sqrt{x}-1}{2\sqrt{x}}$... which isn't anything like the y = $\sqrt{x}$ answer it SHOULD be.

Can someone point out where I went wrong?

2. ## Re: Finding equation for a curve, given length integral

Originally Posted by Calcme
Hey,

So I'm a bit confused on how an answer was reached. The question states:

Find a curve through the point (1,1) whose length integral is given by:

$L = \int_1^4 \sqrt{1+\frac{1}{4x}}dx$. From here I try to use point-slope formula, where m = $\frac{1}{2\sqrt{x}}$ (which comes from taking the square root of $\frac{1}{4x}$. Now, I apply the point-slope formula to try and get the equation of the curve, using:

y - y1 = m(x-x1) ==> y - (1) = $\frac{1}{2\sqrt{x}} (x - (1))$

After adding 1 to both sides and distributing the $\frac{1}{2\sqrt{x}}$ to the (x - 1), I end up with something like:

$y = \frac{x+2\sqrt{x}-1}{2\sqrt{x}}$... which isn't anything like the y = $\sqrt{x}$ answer it SHOULD be.

Can someone point out where I went wrong?
1. The length of a curve with the equation y = f(x) is calculated by:

$L=\int(\sqrt{1+(y')^2})dx$

2. Therefore $(y')^2=\frac1{4x}~\implies~y' = \sqrt{\frac1{4x}} = \frac1{2\sqrt{x}}$

Thus $y = \int\left(\frac1{2\sqrt{x}} \right)dx = \sqrt{x} + c$

3. ## Re: Finding equation for a curve, given length integral

Ahh, I see... we're just working all the way backwards until we get our original curve/function.

Thanks a lot.