Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A curve in the $xy$-plane is given parametrically by $$x(t) = e^{2t}, \quad y(t) = e^{2t} \sin(2t), \quad t \in [0, \pi/2].$$ What is the length of this curve?

Ok, actually I know what to do, but I don't know how to do it because I can't get rid of the trigonometric term.

If it were $x(t) = e^{2t}\cos(t)$, I could have done it, but it's not so I can't get rid of the trigonometric term and integrate the expression.

share|improve this question
1  
You've gotten some pretty solid answers to some of your questions. I would strongly suggest looking them over and accepting them if they meet your needs. –  user17794 Oct 23 '12 at 20:26
    
I agree with Duff. For this question, can you put the work you've done so far? It sounds like you have some idea about how to start but can't complete the argument. –  Eric Stucky Oct 23 '12 at 20:29
    
I erased everything I've done. I spent like 15 min trying to figure out what to do. –  Gladstone Asder Oct 23 '12 at 21:09

1 Answer 1

up vote 0 down vote accepted

With $x=e^{2t} \Rightarrow \dot{x} = 2 x$ and $y = e^{2t} \sin 2t = x \sin 2t \Rightarrow \dot{y} = \dot{x} \sin 2 t + 2 x \cos 2 t = 2 x \left(\sin 2 t + \cos 2 t\right)$, we want to calculate $$ \begin{eqnarray} \int_0^{\pi/2} dt \left({\dot{x}}^2 + {\dot{y}}^2\right)^{1/2} &=& \int_0^{\pi/2} dt \left[\left(2x\right)^2 + \left(2 x\right)^2 \left(\sin 2 t + \cos 2 t\right)^2\right]^{1/2} \\ &=& 2 \int_0^{\pi/2} dt \ e^{2t} \left[1 + \left(\sin 2 t + \cos 2 t\right)^2\right]^{1/2} \\ &=& \sqrt{2} \int_0^{\pi} du \ e^u \left(1 + \sin u \cos u\right)^{1/2} \end{eqnarray} $$ Wolfram cannot do that integral analytically.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.