MHB Therefore, the curvature at any point on the curve is $\boxed{\pi|t|}$.

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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

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[Chris L T521]

This week's problem was correctly answered by Ackbach, BAdhi, MarkFL and Pranav. You can find Mark's solution below.

[sp]We are given:

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

To compute the unit tangent, we need:

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

Differentiating with respect to $t$, we obtain:

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

Hence, the curvature is given by:

$$\kappa=\frac{\left|\textbf{T}'(t) \right|}{\left|\textbf{r}'(t) \right|}=\pi|t|$$[/sp]