What is the magnitude of r'(t) if r'(t) is <1,2t,3t2>?

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SUMMARY

The magnitude of the vector function r'(t) = <1, 2t, 3t²> is calculated using the formula |r'(t)| = √(1² + (2t)² + (3t²)²). This simplifies to |r'(t)| = √(1 + 4t² + 9t⁴). The discussion focuses on the challenge of using this magnitude as a denominator for r(t) = , emphasizing that the expression does not simplify further.

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Homework Statement


whats |r'(t)| if r'(t) is <1,2t,3t2>


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The Attempt at a Solution



im finding normal and binormal vectors and i need to divide r(t)/r'(t)

|r'(t)| = sqrt (12+(2t)2+(3t2)2)

how does that simplify down to something that can be an appropriate denominator for r(t)=<t,t2,t3>
 
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It is
[tex]\sqrt{1+ 4t^2+ 9t^4}[/tex]
but it really does not reduce any more than that.
 
yeah. just looks like its going to be a longwinded ridiculous question
 

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