How do you find the exact arc length of a parametric equation ? #x(t)=2 + 9t^2# #y(t)=9 + 6t^3#

Question

3201 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-05T00:59:34

The length
#L=int_0^1sqrt[(18t)^2+(18t^2)^2]*dt=[6(t^2+1)^(3/2)]_0^1=10.97056#

if a curve is given by a parametric equations

#x(t)=2 + 9t^2#

#y(t)=9 + 6t^3#

where #0 ≤ t ≤ 1#

the length of the curve is given by

#L=int_a^bsqrt[((dx)/dt)^2+((dy)/dt)^2]*dt#

#(dx)/dt=18t#

#(dy)/dt=18t^2#

#L=int_0^1sqrt[(18t)^2+(18t^2)^2]*dt=int_0^1sqrt[324t^2+324t^4]*dt#

#L=int_0^1sqrt[324t^2+324t^4]*dt=int_0^1sqrt[(324t^2)*(1+t^2)]*dt=int_0^1(18*t)*sqrt(1+t^2)*dt#

#L=int_0^1(18*t)*sqrt(1+t^2)*dt#

#=[6*(t^2+1)^(3/2)]_0^1=18*(2^(3/2)/3-1/3)=3*2^(5/2)-6=10.97056#