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Find the length of the curve $x = 1 + 3t^2$, $y = 4 + 2t^3$, $0 \le t \le 1$. Length = 

          Find the length of the curve $x = 1 + 3t^2$, $y = 4 + 2t^3$, $0 \le t \le 1$.
Length =
Find the length of the curve x = 1 + 3t^2, y = 4 + 2t^3, 0 ≤ t ≤ 1. Length =

Added by Elijah H.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Find the length of the curve=1+3t2y4+2t30<t<1 Length=
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Transcript

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00:01 In this problem we have to find the length of curve given by rt equal 60i plus 8t power 3 by 2 j plus 60 square k where the t is lying between 0 to so the length of curve is length of the curve the formula of this is l equal a to b if t is lying between a to b r dash t d t so first we will find r dash t r d is derivative of r t r t is derivative of d t r t this is the component -wise differentiation here we will write it as d of d t 6 t i plus d of d t into t power 3 by 2 j plus d by d t 6 t square so finding differentiation are taking derivative we will have here 6 i and plus 8 into 3 by 2 and t power 3 by 2 minus 1 j plus 12 p k so further simplifying we will have 6 i plus 12 t power 12 t power half of j and plus 12 k 12 t k so this is r dash t now we will find here the magnitude of this r t so by putting here into this formula we will have l equal that is length of the curve and a to b is starting from 0 to 1 so 0 to 1 and r d s t so the magnitude of this r d s we will have here square root 6 square plus 12 t power half whole…
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