Calculus Two Arc Length

Question

I was using Paul's Online Notes (a really good resource - here is the link to the page I was using http://tutorial.math.lamar.edu/Classes/CalcII/ArcLength.aspx) and after deriving the arc length formula he says: In a similar fashion we can also derive a formula for $x = h\left( y \right)$ on $\left[ {c,d} \right]$. This formula is:$$L = \int_{{\,c}}^{{\,d}}{{\sqrt {1 + {{\left[ {h'\left( y \right)} \right]}^2}} \,dy}} = \int_{{\,c}}^{{\,d}}{{\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} \,dy}}$$

I don't understand what this is? I see that it is the derivative of $x$ instead of the derivative of $y$, but why is he bringing this in?

Answer 1

Notice that he's taking the integral of $x=h(y)$, not $y=f(x)$.

Apply the same logic with the right triangle hypotenuse to get that $\displaystyle \sqrt{1+\left(\frac{dx}{dy}\right)^2}$

Answer 2

Suppose you want to find the arc length of the curve $$x=y^2-y$$ where $0\le y\le1$

As you see you have $x=h(y)$ so you need to integrate with respect to $y$ and the given formulas are the arc length integrals in such a case.

Answer 3

Technically speaking, I agree with you that it's not needed since one may always transform coordinates to make our curve appear in standard position.

However, it is sometimes convenient not to do this. Then in such cases, it is simply useful to think of $x$ as a function of $y$ instead.