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

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SUMMARY

The discussion centers on demonstrating that the tangent vector \( r'(t) \) to the curve defined by \( r(t) = t\cos(t) \mathbf{i} + t\sin(t) \mathbf{j} + \sqrt{4 - t^2} \mathbf{k} \) is always perpendicular to \( r(t) \) for \( t \in [0, 2) \). The derivative \( r'(t) \) is calculated as \( r'(t) = \left(\cos t - t\sin t, \sin t + t\cos t, \frac{-t}{\sqrt{4 - t^2}}\right) \). The dot product \( r(t) \cdot r'(t) \) simplifies to zero, confirming that \( r(t) \) is indeed perpendicular to \( r'(t) \) across the specified interval.

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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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