Question

Find the arclength of the curve r(t)=⟨62√t,e6t,e−6t⟩r(t)=⟨62t,e6t,e−6t⟩, 0≤t≤10≤t≤1

Answer 1

The answer is e3−e−3.

Recall that the arclength for parametric curves is:

L=∫ba√(dxdt)2+(dydt)2dt

So,

dxdt=et−e−t
dydt=−2

Now substituting:

L=∫30√(et−e−t)2+(−2)2dt
=∫30√e2t−2+e−2t+4dt expand
=∫30√e2t+2+e−2tdt simplify
=∫30√(et+e−t)2dt factor
=∫30(et+e−t)dt simplify
=et−e−t∣∣30 integrate
=e3−e−3−(e0−e0) evaluate
=e3−e−3