What is the arclength of function r(t)=<2t^(3/2),2t-1,t> 0<t<1 I know you have t

Question

What is the arclength of function r(t)=<2t^(3/2),2t-1,t> 0<t<1

I know you have to find the integral of the derivative of the function r. So I end up with integral from 0 to 1of sqrt(9t+5)dt.

Now this is the part where I'm confused. When you find the integral, you use substitute for u so you end up with u=9t+5 and thus the integral of sqrt(u)du, with du being 9. The next part is sqrt(u)(1/9du). Where does the 1/9 come from? I know it has something to do with du being 9, but why is it the reciprocal? Can someone please explain how to do integrals again?

Explanation / Answer

Ok man its very easy

when you put : u=9t +5 then u have to eliminate the dt also in terms of du

therefore you do du/dt = 9

=>du = 9 dt

=>dt = du/9

then u substitute dt as (1/9)du

Hope u understood

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.