The discussion focuses on calculating the arc length of the curve defined by the vector function r(t) = (10t², 2√10t, ln t) for the interval t = 1 to t = 8. Participants derived the components of the derivative as (20t, 2√10, 1/t) and attempted to integrate the expression ∫√(400t² + 40 + 1/t²) dt. A key insight was provided that correcting the term 400t to 400t² allows for simplification into a perfect square, facilitating the integration process.
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400t should be 400t2.Larrytsai said:Homework Statement
Find the arclength of the curve r(t)=(10t^2,2*sqrt(10)*t, ln t)
for 1<=t<=8
Homework Equations
The Attempt at a Solution
i took the derivative of each component of vector r
20t,2sqrt(10),1/t
then i square each term and square root it
int sqrt( 400t + 40 + 1/t^2)dt
If you make the change above, you'll have a perfect square that you can take the square root of.Larrytsai said:i can't integrate this to find out the length tho.