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Find an arc length parametrization r 1


(s) of the curve r(t)=⟨e t
sin(t),e t
cos(t),11e t
⟩. (Give your answer using component form or standard basis vectors. Express numbers in exact form. Use symbolic notation and fractions where needed.) r 1

(s

1 Answer

3 votes

Final answer:

To find the arc length parametrization of the curve r(t), you need to express the curve in terms of arc length instead of the parameter t. The formula for arc length parametrization is r1(s) = r(t) / ||r'(t)||, where ||r'(t)|| is the magnitude of the derivative of r(t). By applying this formula and simplifying the expression, we get the arc length parametrization as r1(s) = ⟨sin(t) / sqrt(122), cos(t) / sqrt(122), 11 / sqrt(122)⟩.

Step-by-step explanation:

To find an arc length parametrization r₁(s) of the curve r(t) = ⟨etsin(t), etcos(t), 11et⟩, we need to express the curve in terms of arc length instead of the parameter t. The formula for arc length parametrization is:

r₁(s) = r(t) / ||r'(t)||

where ||r'(t)|| is the magnitude of the derivative of r(t). Let's find it:

r'(t) = ⟨etcos(t), -etsin(t), 11et⟩

Now we can calculate the magnitude:

||r'(t)|| = sqrt((etcos(t))2 + (-etsin(t))2 + (11et)2)

Simplifying, we have:

||r'(t)|| = sqrt(e2t(cos2(t) + sin2(t)) + 121e2t) = sqrt(122e2t) = sqrt(122) * et

Now we can define the arc length parametrization:

r₁(s) = ⟨etsin(t), etcos(t), 11et⟩ / (sqrt(122) * et) = ⟨sin(t) / sqrt(122), cos(t) / sqrt(122), 11 / sqrt(122)⟩

User G M Ramesh
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