MHB Joko's question at Yahoo Answers (r(t) perpendicular to r'(t))

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The discussion centers on demonstrating that the tangent vector r'(t) is always perpendicular to the curve r(t) defined by the equation r(t) = tcos(t) i + tsin(t) j + sqrt(4 − t^2) k for 0 < t < 2. The calculation shows that the dot product of r(t) and r'(t) equals zero, confirming their perpendicularity. The derivative r'(t) is computed as r'(t) = (cos t - t sin t, sin t + t cos t, -t/sqrt(4 - t^2)). The conclusion is that for all t in the specified interval, the tangent vector is indeed perpendicular to the curve. This mathematical relationship is essential for understanding the geometric properties of the curve.
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Here is the question:

r (t) = tcos(t) i + tsin(t) j + sqrt(4 − t^2) k, 0<t<2 (should be less then equal to signs in the constraint)

Show, by calculation, that the tangent vector to this curve is always perpendicular
to r (t).

I have struggled to much with this question, not really sure where to start?

Here is a link to the question:

Please help, How to show if a tangent vector is perpendicular to r(t)? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Joko103,

For all $t\in [0,2)$ we have $$r(t)=(t\cos t,t\sin t,\sqrt{4 − t^2})\\r'(t)=\left(\cos t-t\sin t\cos t,\sin t+t\cos t,\dfrac{-t}{\sqrt{4 − t^2}}\right)$$ Then, $$r(t)\cdot r'(t)=t\cos^2t-t^2\sin t\cos t+t\sin^2t+t^2\sin t\cos t-t=\\t(\cos^2t+\sin^2t)-t=t-t=0\Rightarrow r(t)\mbox{ is perpendicular to }r'(t)$$
 
Good morning I have been refreshing my memory about Leibniz differentiation of integrals and found some useful videos from digital-university.org on YouTube. Although the audio quality is poor and the speaker proceeds a bit slowly, the explanations and processes are clear. However, it seems that one video in the Leibniz rule series is missing. While the videos are still present on YouTube, the referring website no longer exists but is preserved on the internet archive...

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