Question

To show the tangent of the angle \((\psi )\) between the tangent line and radial line is the ratio of the radial distance \((r)\) and its angular derivative \(\left( {\frac{{dr}}{{d\theta }}} \right)\).

Short answer

It is proved that \(\tan \psi = \frac{r}{{\frac{d}{{do}}}}\).

Step-by-step solution

Given data

The angle between the tangent line of the curve \(r = f(\theta )\) and the radial line \(OP\) at the point \(P\) is \(\psi \) and \(\psi = f - \theta \).

Concept of Tangent line equation

The equation of the tangent line can be found by the use of the formula\({\bf{y}}{\rm{ }}--{\rm{ }}{{\bf{y}}_{\bf{1}}}\; = {\bf{m}}{\rm{ }}\left( {{\bf{x}}{\rm{ }}--{\rm{ }}{{\bf{x}}_{\bf{1}}}} \right)\)where m is the slope and\(({x_1}\;,{\rm{ }}{y_1})\)is the coordinate points of the line.

Calculation for the function \(\tan (f - \theta )\)

Substitute \(\psi = f - \theta \) in \(\tan \psi \) as \(\tan \psi = \tan (f - \theta )\).

Apply formula \(\left( {\tan (A - B) = \frac{{\tan A - \tan B}}{{1 + \tan A\tan B}}} \right)\) in the above equation.

\(\tan (f - \theta ) = \frac{{\tan f - \tan \theta }}{{1 + \tan f\tan \theta }}\) …… (1)

Take slope of the tangent line for the curve \(r = f(\theta )\) as \(\tan f = \frac{{dy}}{{dx}}\).

The chain rule for \(\frac{{dy}}{{dx}}\) is, \(\frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dy}}} \right)}}{{\left( {\frac{{dx}}{{dy}}} \right)}}\).

Substitute \(\tan f = \frac{{dy}}{{dx}}\) in the above equation.

\(\tan f = \frac{{\left( {\frac{{dy}}{{d\underline {xy} }}} \right)}}{{\left( {\frac{{dx}}{{dy}}} \right)}}\)

Substitute \(\tan f\) in equation (1).

\(\tan (f - \theta ) = \frac{{\left( {\frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} - \tan \theta } \right)}}{{\left( {1 + \frac{{\left( {\frac{{dy}}{{dv}}} \right)}}{{\left( {\frac{{dx}}{{dy}}} \right)}}\tan \theta } \right)}}\)

Calculation for the polar equations

Multiply \(\left( {\frac{{dx}}{{d\theta }}} \right)\) in above equation at numerator and denominator.

\(\tan (f - \theta ) = \frac{{\frac{{dx}}{{dy}}\left( {\frac{{\left( {\frac{{dy}}{{di}}} \right)}}{{\left( {\frac{{dx}}{{di}} - \tan \theta } \right)}}} \right.}}{{\frac{{dx}}{{dy}}\left( {1 + \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}}\tan \theta } \right)}}\)

\(\tan (f - \theta ) = \frac{{\left( {\frac{{dy}}{{dy}}} \right) - \left( {\frac{{dx}}{{dy}}} \right)\tan \theta }}{{\left( {\frac{{dx}}{{dy}}} \right) + \left( {\frac{{dy}}{{dy}}} \right)\tan \theta }}\) …… (2)

The polar equation for the coordinate \(x\) is as, \(x = r\cos \theta \). …… (3)

The polar equation for the coordinate \(y\) is as, \(y = r\sin \theta \). …… (4)

Differentiate the equation (3) with respect to \(\theta \).

\(\begin{aligned}{l}\frac{{dx}}{{d\theta }} = - r(\sin \theta ) + \left( {\frac{{dr}}{{d\theta }}} \right)\cos \theta \\\frac{{dx}}{{d\theta }} = \left( {\frac{{dr}}{{dA}}} \right)\cos \theta - r\sin \theta \end{aligned}\)

Differentiate the equation (4) with respect to \(\theta \).

\(\begin{aligned}{l}\frac{{dy}}{{d\theta }} &= r\cos \theta + \frac{{dr}}{{d\theta }}\sin \theta \\\frac{{dy}}{{d\theta }} &= \frac{{dr}}{{d\theta }}\sin \theta + r\cos \theta \end{aligned}\)

Substitute \(\left( {\frac{{dr}}{{d\theta }}\cos \theta - r\sin \theta } \right)\) for \(\left( {\frac{{dx}}{{d\theta }}} \right)\) and \(\left( {\frac{{dr}}{{d\theta }}\sin \theta + r\cos \theta } \right)\) for \(\left( {\frac{{dy}}{{d\theta }}} \right)\) in equation (2).

\(\begin{aligned}{l}\tan (f - \theta ) &= \frac{{\left( {\frac{{d\theta }}{{d\theta }}\sin \theta + r\cos \theta } \right) - \left( {\frac{{dr}}{{d\theta }}\cos \theta - r\sin \theta } \right)\tan \theta }}{{\left( {\frac{{dr}}{{d\theta }}\cos \theta - r\sin \theta } \right) + \left( {\frac{{dr}}{{d\theta }}\sin \theta + r\cos \theta } \right)\tan \theta }}\\\tan (f - \theta ) &= \frac{{\frac{d}{d}\sin \theta + r\cos \theta - \frac{{dr}}{{d\theta }}\cos \theta \tan \theta + r\sin \theta \tan \theta }}{{\frac{d}{{dx}}\cos \theta - r\sin \theta + \frac{{dr}}{{d\theta }}\sin \theta \tan \theta + r\cos \theta \tan \theta }}\end{aligned}\)

Calculation for the function \(\tan \psi \)

Substitute \(\left( {\frac{{\sin \theta }}{{\cos \theta }}} \right)\) for \((\tan \theta )\) in above equation.

\(\begin{aligned}{l}\tan (f - \theta ) &= \frac{{\frac{{dr}}{{di}}\sin \theta + r\cos \theta - \frac{d}{d}\cos \theta \frac{{\sin \theta }}{{\cos \theta }} + r\sin \theta \frac{{\sin \theta }}{{\cos \theta }}}}{{\frac{d}{{d\theta }}\cos \theta - r\sin \theta + \frac{d}{d}\sin \theta \frac{{\sin \theta }}{{\cos \theta }} + r\cos \theta \frac{{\sin \theta }}{{\cos \theta }}}}\\\tan (f - \theta ) &= \frac{{\frac{d}{{di}}\sin \theta + r\cos \theta - \frac{{dr}}{{d\theta }}\sin \theta + r\frac{{{{\sin }^2}\theta }}{{\cos \theta }}}}{{\frac{d}{{di}}\cos \theta - r\sin \theta + \frac{{dr}}{{dy}}\frac{{\sin {r^2}\theta }}{{{{\cos }^2}\theta }} + r\sin \theta }}\\\tan (f - \theta ) &= \frac{{r\cos \theta + r\frac{{{{\sin }^2}\theta }}{{{{{\mathop{\rm cis}\nolimits} }^\theta }\theta }}}}{{\frac{{{\mathop{\rm dr}\nolimits} }}{{{\mathop{\rm din}\nolimits} }}\cos \theta + \frac{{\sin 2\theta }}{{\cos \theta }}}}\\\tan (f - \theta ) &= \frac{{r{{\cos }^2}\theta + r{{\sin }^2}\theta }}{{\frac{{dr}}{{di}}{{\cos }^2}\theta + \frac{{d{s^2}{{\sin }^2}\theta }}{{di}}}}\end{aligned}\)

Simplify further as shown below.

\(\begin{aligned}{l}\tan (f - \theta ) &= \frac{{r\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)}}{{\frac{{dr}}{{d\theta }}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)}}\\\tan (f - \theta ) &= \frac{r}{{\frac{{dr}}{{d\theta }}}}\end{aligned}\)

Substitute \(\psi \) for \(f - \theta \) in above equation.

\(\tan \psi = \frac{r}{{\frac{{dr}}{{d\theta }}}}\)

Hence, it is proved that \(\tan \psi = \frac{r}{{\frac{{dr}}{{dv}}}}\).