Why do I get two different answers when solving for arclength? 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?
 A: 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.
A: 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 $$\int _0^1 \sqrt {1+\tan^2(t)}(8t\cos(t))dt = \int _0^1 8tdt=4$$
A: 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 $x$)
$$
\int_0^18t\,dt = 4. 
$$
