What is the arclength of #f(t) = (sqrt(t-2),t^2)# on #t in [2,3]#?

Question

1320 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-19T16:29:31

#L=int_2^3sqrt(1/(4(t-2))+4t^2)dtapprox5.1927#

The arc length of the parametric function #f(t)=(x(t),y(t))# on #tin[a,b]# is given through:

#L=int_a^bsqrt((x'(t))^2+(y'(t))^2)dt#

We see that:

#x(t)=sqrt(t-2)=(t-2)^(1/2)#

Then:

#x'(t)=1/2(t-2)^(-1/2)d/dt(t-2)=1/(2sqrt(t-2))#

And:

#y(t)=t^2" "=>" "y'(t)=2t#

Combining these, we see that:

#L=int_2^3sqrt((1/(2sqrt(t-2)))^2+(2t)^2)dt#

#L=int_2^3sqrt(1/(4(t-2))+4t^2)dtapprox5.1927#

This can't be integrated by hand.