# Problem of the Week #94 - January 13th, 2014

Status
Not open for further replies.

#### Chris L T521

##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!

-----

Problem: Find the curvature of the curve with the parametric equations
$x(t) = \int_0^t \sin\left(\tfrac{1}{2}\pi \theta^2\right)\,d\theta,\qquad y(t) = \int_0^t\cos\left(\tfrac{1}{2}\pi\theta^2\right)\,d\theta$

-----

#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by Ackbach, BAdhi, MarkFL and Pranav. You can find Mark's solution below.

We are given:

$$\displaystyle \textbf{r}(t)=\left\langle\int_0^t\sin\left(\frac{\pi}{2}\theta^2 \right)\,d\theta, \int_0^t\cos\left(\frac{\pi}{2}\theta^2 \right)\,d\theta \right\rangle$$

To compute the unit tangent, we need:

$$\displaystyle \textbf{T}(t)=\frac{\textbf{r}'(t)}{\left|\textbf{r}'(t) \right|}=\frac{\left\langle\sin\left(\frac{\pi}{2}t^2 \right), \cos\left(\frac{\pi}{2}t^2 \right) \right\rangle}{1}= \left\langle\sin\left(\frac{\pi}{2}t^2 \right), \cos\left(\frac{\pi}{2}t^2 \right) \right\rangle$$

Differentiating with respect to $t$, we obtain:

$$\displaystyle \textbf{T}'(t)=\left\langle \pi t\cos\left(\frac{\pi}{2}t^2 \right), -\pi t\sin\left(\frac{\pi}{2}t^2 \right) \right\rangle$$

Hence, the curvature is given by:

$$\displaystyle \kappa=\frac{\left|\textbf{T}'(t) \right|}{\left|\textbf{r}'(t) \right|}=\pi|t|$$

Status
Not open for further replies.