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.