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, $$r(t)\cdot r'(t)=t\cos^2t-t^2\sin t\cos t+t\sin^2t+t^2\sin t\cos t-t=\\t(\cos^2t+\sin^2t)-t=t-t=0\Rightarrow r(t)\mbox{ is perpendicular to }r'(t)$$
 
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