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

Question

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

Aurora Alvarado · Accepted Answer

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.

Isabella Ackerman · Answer

undefined To find the length of the curve defined by $x=1+3t^2$, $y=4+2t^3$, where $0 \leq t \leq 1$, you can use the formula for the arc length of a parametric curve:

$$ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt $$

Substitute the expressions for $x$ and $y$ in terms of $t$, then differentiate with respect to $t$, and finally integrate over the given interval $[a,b]$.

Christopher Agresta · Answer

undefined The length of the curve can be found using the formula:

$$ L = \int_{a}^{b} \sqrt{\left(\frac{{dx}}{{dt}}\right)^2 + \left(\frac{{dy}}{{dt}}\right)^2} dt $$

Substitute the given parametric equations for $ x(t) $ and $ y(t) $, and the limits of integration for $ t $, which are 0 and 1. Then compute the integral.

How do you find the length of the curve #x=1+3t^2#, #y=4+2t^3#, where #0&lt;=t&lt;=1# ?

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