Find the locus of the feet of the perpendiculars dropped

Find the locus of the feet of the perpendiculars dropped from the origin onto the tangents of the curve $x=a(\cos t+t\sin t),\ y=a(\sin t-t\cos t)$ (the involute of a circle of radius $a$).

Step 1: Velocities

\[ \dot x = a(-\sin t+\sin t+t\cos t) = at\cos t, \qquad \dot y = a(\cos t-\cos t+t\sin t) = at\sin t. \tag{1} \]

Step 2: The tangent line

The slope is $\dot y/\dot x=\tan t$. Writing the tangent through $T=\bigl(a(\cos t+t\sin t),\,a(\sin t-t\cos t)\bigr)$ with this slope and multiplying by $\cos t$, the $-a\sin t\cos t$ terms cancel and $y_0\cos t-x_0\sin t=-at$ (since $\cos^2t+\sin^2t=1$), leaving

\[ x\sin t - y\cos t = at. \tag{2} \]

Step 3: The perpendicular from the origin

Slope $-\cot t$, through the origin: \[x\cos t + y\sin t = 0. \tag{3}\]

Step 4: The foot $M$

Equations (2)–(3) form an orthogonal (rotation-type) linear system in $x,y$. Multiplying (2) by $\sin t$ and (3) by $\cos t$ and adding eliminates $y$; multiplying (2) by $\cos t$ and (3) by $\sin t$ and subtracting eliminates $x$:

\[ x_M = at\sin t, \qquad y_M = -at\cos t. \tag{4} \]

Step 5: Polar form of the locus

\[ x_M^2+y_M^2 = a^2t^2 \;\Longrightarrow\; \rho = a|t|. \tag{5} \]

From (4), $\cos\varphi=x_M/\rho=\pm\sin t$ and $\sin\varphi=y_M/\rho=\mp\cos t$ (sign matching that of $t$); for $t>0$ this gives $\cos\varphi=\sin t,\ \sin\varphi=-\cos t$, i.e.\ $\varphi=t-\dfrac\pi2$, so $t=\varphi+\dfrac\pi2$. Substituting into $\rho=at$:

\[ \rho = a\!\left(\varphi+\frac\pi2\right). \tag{6} \]

Step 6: Verification

\[a=2,\ t=0.3,1.2,2.5\text{: }\rho=at\text{ matches }\sqrt{x_M^2+y_M^2}\text{ exactly, and }\varphi=t-\pi/2\text{ to machine precision}\ \checkmark\]

\[t=-0.7\text{: }\rho=a|t|=1.4\text{ (not }at=-1.4\text{), confirming }\rho=a|t|\text{ for negative }t\text{ too}\ \checkmark\]