Show that the curves rho_1=f(phi) and rho_2=a+f(phi) have

Show that the curves $\rho_1=f(\varphi)$ and $\rho_2=a+f(\varphi)$ have equal polar subnormals at points lying on the same radius vector (i.e.\ at points with the same value of $\varphi$).

Step 1: The polar subnormal formula

For $\rho=g(\varphi)$, the length of the polar subnormal is $S_N=\left|d\rho/d\varphi\right|$, which depends only on the curve's rate of change, not on $\rho$ itself.

Step 2: Differentiate both curves

\[ \frac{d\rho_1}{d\varphi} = f'(\varphi), \qquad \frac{d\rho_2}{d\varphi} = \frac{d}{d\varphi}\bigl[a+f(\varphi)\bigr] = 0+f'(\varphi) = f'(\varphi). \tag{1} \]

The additive constant $a$ contributes nothing to the derivative.

Step 3: Verification

\[f(\varphi)=\sin\varphi+\varphi^2,\ a=3.7,\ \varphi=1.1\text{: } S_{N_1}=S_{N_2}=|f'(1.1)|\approx2.65453\text{ exactly}\ \checkmark\]