Arc Length of Parabola & Square Root Function

[integral25]

Homework Statement

Consider the curves: $$y = x^2$$ from 1/2 to 2 and $$y = \sqrt{x}$$ from 1/4 to 4.

a. Explain why the lengths should be equal.
b. Set up integrals (with respect to x) that give the arc lengths of the curve segments. Use a substitution to show that one integral can be transformed into another.

Homework Equations

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

The Attempt at a Solution

[/B]
a. The lengths should be equal because the two functions are inverses. I made the graphs of each on the given intervals and it made sense because they were inverses.
b. This is where my issue was. I was able to set up the integrals.

$$L = \int \sqrt{1+(2x)^2} dx$$
$$L = \int \sqrt{1+(1/2\sqrt{x}})^2$$

My issue was with making a substitution to transform one integral into another. I initially thought I should use a substitution involving the inverse of a function, but I couldn't get it to work out. I am at a loss right now, so any hint would be fantastic.

 

[Eclair_de_XII]

I don't know how to solve the second one, but why don't you try trigonometric substitution for $\int_\frac{1}{2}^2 \sqrt{1+4x^2} dx$? Let $x=\frac{1}{2}tan\theta$.

In any case, I'd just try to solve both integrals one by one and confirming that the arc lengths are the same, instead of trying to equate them to one another directly. Actually, I think that if you make at least two substitutions, you can possibly solve the second one with a tangent substitution as well. Someone correct me if I'm wrong on this; I haven't done Calc II in over a year.

 

[integral25]

I did solve the first one - but I went back to my assignment and it said do not solve Part B. In Part C, I had to solve using the midpoint rule.

I will try using the trig subsitution and see what I can come up with up.

I don't know how to solve the second one, but why don't you try trigonometric substitution for $\int_\frac{1}{2}^2 \sqrt{1+4x^2} dx$? Let $x=\frac{1}{2}tan\theta$.

In any case, I'd just try to solve both integrals one by one and confirming that the arc lengths are the same, instead of trying to equate them to one another directly.

 

[fresh_42]

Have you considered to replace $x=u^2$ in the second integral $\int_{\frac{1}{4}}^4 \sqrt{1+\frac{1}{4x}}\,dx$ and see what you get?

Solving the integrals is possible, but neither necessary nor easy.

 

[integral25]

That'll do it! I think I tried something similar but I always had x to the power of something which made is messy and gross. I should have thought to do it another way - doh!

Thanks so much for your help.

Have you considered to replace $x=u^2$ in the second integral $\int_{\frac{1}{4}}^4 \sqrt{1+\frac{1}{4x}}\,dx$ and see what you get?

Solving the integrals is possible, but neither necessary nor easy.

 

[Ray Vickson]

Homework Statement

Consider the curves: $$y = x^2$$ from 1/2 to 2 and $$y = \sqrt{x}$$ from 1/4 to 4.

a. Explain why the lengths should be equal.
b. Set up integrals (with respect to x) that give the arc lengths of the curve segments. Use a substitution to show that one integral can be transformed into another.

Homework Equations

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

The Attempt at a Solution

[/B]
a. The lengths should be equal because the two functions are inverses. I made the graphs of each on the given intervals and it made sense because they were inverses.
b. This is where my issue was. I was able to set up the integrals.

$$L = \int \sqrt{1+(2x)^2} dx$$
$$L = \int \sqrt{1+(1/2\sqrt{x}})^2$$

My issue was with making a substitution to transform one integral into another. I initially thought I should use a substitution involving the inverse of a function, but I couldn't get it to work out. I am at a loss right now, so any hint would be fantastic.

Since $x,y>0$ for both curves, we can write the first curve parametrically as $y=t, x = \sqrt{t}$ and the second curve as $x=t, y=\sqrt{t}$, so that $ds^2 = dx^2 + dy^2$ is the same for both, as are the $t$-limits.