I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 12.10 : Curvature
1. Find the curvature of \(\vec r\left( t \right) = \left\langle {\cos \left( {2t} \right), - \sin \left( {2t} \right),4t} \right\rangle \).
Show All Steps Hide All Steps
Start SolutionWe have two formulas we can use here to compute the curvature. One requires us to take the derivative of the unit tangent vector and the other requires a cross product.
Either will give the same result. The real question is which will be easier to use. Cross products can be a pain to compute but then some of the unit tangent vectors can be quite messy to take the derivative of. So, basically, the one we use will be the one that will probably be the easiest to use.
In this case it looks like the unit tangent vector won’t be that bad to work with so let’s go with that formula. Here’s the unit tangent vector work.
\[\begin{array}{c}\vec r'\left( t \right) = \left\langle { - 2\sin \left( {2t} \right), - 2\cos \left( {2t} \right),4} \right\rangle \hspace{0.5in}\left\| {\vec r'\left( t \right)} \right\| = \sqrt {4{{\sin }^2}\left( {2t} \right) + 4{{\cos }^2}\left( {2t} \right) + 16} = \sqrt {20} = 2\sqrt 5 \\ \vec T\left( t \right) = \frac{1}{{2\sqrt 5 }}\left\langle { - 2\sin \left( {2t} \right), - 2\cos \left( {2t} \right),4} \right\rangle = \left\langle { - \frac{{\sin \left( {2t} \right)}}{{\sqrt 5 }}, - \frac{{\cos \left( {2t} \right)}}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right\rangle \end{array}\] Show Step 2Now, what we really need is the magnitude of the derivative of the unit tangent vector so here is that work,
\[\vec T'\left( t \right) = \left\langle { - \frac{2}{{\sqrt 5 }}\cos \left( {2t} \right),\frac{2}{{\sqrt 5 }}\sin \left( {2t} \right),0} \right\rangle \hspace{0.25in}\hspace{0.25in}\left\| {\vec T'\left( t \right)} \right\| = \sqrt {\frac{4}{5}{{\cos }^2}\left( {2t} \right) + \frac{4}{5}{{\sin }^2}\left( {2t} \right)} = \frac{2}{{\sqrt 5 }}\] Show Step 3The curvature is then,
\[\kappa = \frac{{\left\| {\vec T'\left( t \right)} \right\|}}{{\left\| {\vec r'\left( t \right)} \right\|}} = \frac{{{}^{2}/{}_{{\sqrt 5 }}}}{{2\sqrt 5 }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{5}}}\]So, in this case, the curvature will be independent of \(t\). That won’t always be the case so don’t expect this to happen every time.