Calculus (3rd Edition)

$\approx 6.26$
The first derivative of the given vector is given by: $r'(t) =i+2j+2t \ k \implies ||r'(t)||=\sqrt {}(1)^2+(2)^2+(2t)^2=\sqrt {5+4t^2}$ We calculate the length by integration: $L=\int_{p}^{q}\|r'(t)\|dt=\int_{0}^{2}\sqrt{4t^2+5} dt$ Let us apply the substitution method such that: $a=2t \implies dt=\dfrac{da}{2}$ and $b=\sqrt 5$ Now, $L=\dfrac{1}{2} \times \int_0^4 \sqrt {a^2+b^2} \ da =\dfrac{1}{4} [a \sqrt {a^2+5}+5 \ln (a+\sqrt {a^2+5})]_0^4\\=\dfrac{1}{4} [4 \sqrt {21}+5 \ln (4+\sqrt {21})-5 \ln \sqrt 5] \\ \approx 6.26$