Suppose $k$ is the signed curvature of a plane curve $C$ (parametrized by $\gamma$, say) expressed in terms of its arc length. Show that, if $C_a$ is the image of $C$ under the dilation mapping $\vec{v}\to a\vec{v}$ where $a$ is a non-zero constant, the signed curvature of $C_a$ in terms of its arc-length $s$ is $\frac{1}{a}k(\frac{s}{a})$.
This is problem $2.2.4$ in Andrew Pressley's Elementary Differential Geometry. My question about this problem is simply whether the formula should be $\frac{1}{|a|}k(\frac{s}{|a|})$ where the $a$ is replaced by its absolute value. My calculation showing this is as follows:
We may assume $\gamma(t)=(f(t),g(t))$ where $t(u)=\int_{u_0}^u|\dot{\gamma}(w)|dw$ is the arc-length function of $\gamma$ corresponding to the base point $\gamma(u_0),$ since the signed curvature $k$ of $\gamma$ is expressed in terms of the arc-length of $\gamma$. Hence $\gamma$ is unit-speed and its signed curvature $k(t)$ satisfies $\ddot{\gamma}=k(t)\cdot n_s$ where $n_s$ is the vector $\dot{\gamma}$ rotated counterclockwise by $90$ degrees i.e. $n_s=(-g'(t),f'(t)).$
Now the arc-length $s$ of the curve $\beta=a\gamma(t)$ is given by $$s=\int_0^t|\dot{\beta}(w)|\space dw=\int_0^t|a\dot\gamma(w)|\space dw=|a|\int_0^t|\dot\gamma(w)|\space dw=|a|t$$ since $\gamma$ is unit-speed and since $t(u_0)=0.$ Then $t=\frac{s}{|a|}$ and the signed curvature $k_a$ of $C_a$ satisfies $\ddot{\beta}=k_a\cdot n_{s,a}$ where $\ddot{\beta}=\frac{a}{|a|^2}(f''(\frac{s}{|a|}),g''(\frac{s}{|a|}))$ and $n_{s,a}=\frac{a}{|a|}(-g'(\frac{s}{|a|}),f'(\frac{s}{|a|}))$. In other words $$\ddot{\beta}=\frac{a}{|a|^2}\ddot{\gamma}(t)\space\space\text{ and }\space\space n_{s,a}=\frac{a}{|a|}n_s(t).$$
Thus it follows $\ddot{\beta}=k_a\cdot n_{s,a}\iff\frac{a}{|a|^2}\ddot{\gamma}(t)=k_a\cdot\frac{a}{|a|}n_s(t)\iff \ddot{\gamma}=(|a|k_a)\cdot n_s(t)$ therefore $$|a|k_a=k(t)\iff k_a(s)=\frac{1}{|a|}k(t)=\frac{1}{|a|}k\left(\frac{s}{|a|}\right).$$
Is there a way to remove the absolute values?