Answer
a) $4 \sqrt 5$ b) $\dfrac{17 \sqrt {17}-1}{12}$
Work Step by Step
a ) Here, we have
$ ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
and $ds= \dfrac{\sqrt 5}{2} dt$
This implies that
$L=\int_{0}^{4} (t) ( \dfrac{\sqrt 5}{2} dt)$
or, $ \dfrac{\sqrt 5}{2} [\dfrac{t^2}{2}]_{0}^{4}=4 \sqrt 5$
b) Here, we have
$ ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
and $ds=\sqrt {1+4t^2} dt$
This implies that
$L=\int_{0}^{2} (t) ( \sqrt {1+4t^2}) dt$
Plug $1+4t^2 =p \implies t dt =\dfrac{dp}{8}$
or, $\int_{1}^{17} (\dfrac{1}{8}) p^{1/2} dp= \dfrac{1}{8} [\dfrac{2}{3}u^{3/2}]_{1}^{17}=\dfrac{17 \sqrt {17}-1}{12}$
Hence, a) $4 \sqrt 5$ b) $\dfrac{17 \sqrt {17}-1}{12}$
