Why do I get two different answers when solving for arclength?

Question

I am given that $\frac{dx}{dt}=8t\cos(t)$ and $\frac{dy}{dt}=8t\sin(t)$. I tried solving for the arclength from $t=0$ to $t=1.$

Method 1: $$\text{Arclength} = \int_{0}^{1} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dx = 4.$$

Method 2:

$$\text{Arclength} = \int_{0}^{1} \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx.$$ However, when I solve using method 2, I get $1.22619,$ when the answer should be $4.$ What is causing this difference?

Answer 1

Your second formula applies when you see $y$ as a function of $x$; you don't say how you found $dy/dx$.

Playing a bit loose with differentials, we have $$ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{8t\sin t}{8t\cos t}=\tan t. $$ Then $$ \sqrt{1+\left(\frac{dy}{dx} \right)^2}\,dx=\sqrt{1+\tan ^2 t}\ \,dx=\frac1{\cos t}\,dx =\frac1{\cos t}\,8t\,cos t\,dt=8t\,dt. $$ So your second integral is (now the limits are on $t$, note that we don't easily know the limits on $x$) $$ \int_0^18t\,dt = 4. $$

Answer 2

Your first method requires a change. (It is $dt$ not $dx$)

$$I = \int^1_0 \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt = \int^1_0 \sqrt{(8t)^2(cos^2t + sin^2t)}dt =\int^1_0 8tdt = 4[t^2]^1_0 = 4$$

Now, for the 2nd method.

It is actually an equivalence of the first one. It can be deduced like this.

$$\int \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt = \int \frac{dx}{dt}\sqrt{1 + \frac{(\frac{dy}{dt})^2}{(\frac{dx}{dt})^2}}dt = \int\sqrt{1+(\frac{dy}{dx})^2}dx$$

So, the second method also yields 4.

Answer 3

The second method should give you the correct answer as well.

Note that $$ \sqrt{1+(\frac{dy}{dx})^2} dx =\sqrt {1+\tan^2(t)}(8t\cos(t))dt$$

so the arc length is $$\int _0^1 \sqrt {1+\tan^2(t)}(8t\cos(t))dt = \int _0^1 8tdt=4$$