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curve defined parametrically

I'm pretty sure the values of dx/dt and dy/dt are already given in the question so its just a matter of plugging into the arclenght formula for curves defined parametrically. My only concern is that this simplifies to the [t] from 1 to 0. with the final answer being simply 1. This can't seem to be right.

the last bit that was cut off by the picture is just d0`

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    $\begingroup$ Why cannot that be correct? Also, you better include the text of the problem here (use mathjax to write equations). Links can go invalid, and it will also help for people with screen readers. $\endgroup$ – mickep May 11 '17 at 6:09
  • $\begingroup$ Well it could be correct, it is just that the current professor is constantly throwing 8 mph change-ups and then with this hits us with 99 mph fastballs. It is just difficult to know if i truly have a handle on this when my answers turn out wrong due to a simple trick. I will make sure all of my future post are presented with mathjax as well thank you. $\endgroup$ – Jose Antonio Rodriguez May 11 '17 at 6:24
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You need to compute over a range $$L=\int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$$ where $$x=\int_0^t f(\theta)\, d\theta\qquad \text{and}\qquad y=\int_0^t g(\theta)\, d\theta$$ Now, using the fundamental theorem of calculus $$\frac{dx}{dt}=f(t)\qquad \text{and}\qquad \frac{dy}{dt}=g(t) $$ So, for your case, using $\sin^2(a)+\cos^2(a)=1$, you properly end with $$L=\int_{t_1}^{t_2} dt=t_2-t_1$$ and you are very correct.

You could easily show that, if $$x=\int_0^t \cos\left(\phi(\theta)\right)\, d\theta\qquad \text{and}\qquad y=\int_0^t \sin\left(\phi(\theta)\right))\, d\theta$$ you would have the same result since $$\frac{dx}{dt}=\cos\left(\phi(t)\right)\qquad \text{and}\qquad \frac{dy}{dt}=\sin\left(\phi(t)\right) $$ $$\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=1$$ for any function $\phi(\theta)$.

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This problem is almost trivial in the complex plane. Consider that

$$z=\int_0^t \cos(\pi\theta^2/2) d\theta+i\int_0^t \sin(\pi\theta^2/2) d\theta=\int_0^t e^{i\pi\theta^2/2}d\theta$$

Now, in the complex plane arc length is given by

$$s=\int |\dot z| du$$

and in the present case $\dot z=e^{i\pi\theta^2/2}$ and $|\dot z|=1$, and therefore

$$s=\int_0^1 dt=1$$

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