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?

  • 1
    $\begingroup$ About your second method: Can you tell us under what conditions it is guaranteed to work? [You can click "edit" below your question to add your answer to this question.] Also: exactly how did you compute $dy/dx$? $\endgroup$ May 13, 2019 at 1:39
  • $\begingroup$ Your expressions are equally. The problem is likely to be with your steps. Edit your question to show how you solve each of them $\endgroup$
    – Tojra
    May 13, 2019 at 1:39
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    $\begingroup$ The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$. $\endgroup$ May 13, 2019 at 1:40
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    $\begingroup$ A previous edit corrected the $dx$ in the first formula to $dt$ (among many other changes, mostly unnecessary). This error is a significant part of the question, so I restored it. The appropriate way to deal with an error like this is to inform the OP about it in an answer (as someone did) rather than stealthily "fixing" the problem. $\endgroup$
    – David K
    May 13, 2019 at 10:47
  • $\begingroup$ Why is it a significant part of the question? It is not, if OP had integrated the first formulare w.r.t. to $x$, then his result would depend on $t$. $\endgroup$ May 13, 2019 at 14:40

3 Answers 3


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 $x$) $$ \int_0^18t\,dt = 4. $$


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.


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


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