How do you find the length of the curve $x=1+3t^2$ , $y=4+2t^3$ , where $0<=t<=1$ ?

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Answer 1 · 2014-10-10T14:06:22

By taking the derivative with respect to $t$ ,

${(x'(t)=6t),(y'(t)=6t^2):}$

Let us now find the length $L$ of the curve.

$L=int_0^1 sqrt{[x'(t)]^2+[y'(t)]^2}dt$

$=int_0^1 sqrt{6^2t^2+6^2t^4} dt$

by pulling $6t$ out of the square-root,

$=int_0^1 6t sqrt{1+t^2} dt$

by rewriting a bit further,

$=3int_0^1 2t(1+t^2)^{1/2}dt$

by General Power Rule,

$=3[2/3(1+t^2)^{3/2}]_0^1=2(2^{3/2}-1)$

I hope that this was helpful.

How do you find the length of the curve x=1+3t^2x=1+3t^2, y=4+2t^3y=4+2t^3, where 0<=t<=10<=t<=1 ?