Lengths of curves

The curve is regular, since the norm of the derivative vector equals $2$ for every $t$ and in particular never vanishes. In computing the norm one factors out $4$ under the square root and applies the identity ${\sin }^{2}t+{\cos }^{2}t=1$. The integral of a constant over the interval $[0,\pi ]$ is the product of the constant and the length of the interval, hence $L(\gamma )=2\pi$. The result is half the length $2\pi r=4\pi$ of the circle of radius $2$, in agreement with the fact that the parameter interval covers half a turn.

The curve is Cartesian, that is, of the form $(t,f(t))$ with $f(t)=t^{\frac{3}{2}}$, hence the first component of the derivative vector equals $1$ and the norm is $\sqrt{1+f^{\prime }(t{)}^2}$ with $f^{\prime }(t)=\frac{3}{2}{t}^{\frac{1}{2}}$. The square of $\frac{3}{2}{t}^{\frac{1}{2}}$ is $\frac{9}{4}t$, therefore the integrand is ${\left(1+\frac{9}{4}t\right)}^{\frac{1}{2}}$. An antiderivative is obtained with the power rule: one raises the exponent to $\frac{3}{2}$, divides by the new exponent and divides by the inner derivative $\frac{9}{4}$, which gives the factor $\frac{4}{9}\cdot \frac{2}{3}=\frac{8}{27}$. At $t=4$ the argument equals $1+9=10$ and at $t=0$ it equals $1$, hence the length is $\frac{8}{27}\left(10\sqrt{10}-1\right)$.

The derivative vector has the first two components of trigonometric type and the third one constant. In computing the norm, the contribution of the first two components is $9({\sin }^{2}t+{\cos }^{2}t)=9$ by the fundamental identity, while the third adds $16$; the sum $25$ has square root $5$. The speed is therefore constant, so the length is the product $5\cdot 2\pi =10\pi$. One turn of the helix is longer than one turn of the base circle alone, which measures $6\pi$, since the rise in height is added to the rotation.