Question

Show that the curves \(r = a\sin \theta \)_and\(r = a\cos \theta \)_{intersect at right angles.}

Short answer

Both the curves intersect each other at right angle.

Step-by-step solution

Step 1:

We have equations of curves \(r = a\sin \theta \) -----(1)

And \(r = a\cos \theta \) -----(2)

Step 2:

Let \({m_1}\) be the slope of the tangent to curve \(r = a\sin \theta \)

Then\({m_1} = \frac{{dy}}{{dx}}\)

We get \(\frac{{dr}}{{d\theta }} = a\cos \theta \)

Step 3:

Then

Step 4:

Let \({m_2}\) be the slope of the tangent to curve \(r = a\cos \theta \)

Then\({m_2} = \frac{{dy}}{{dx}}\)

We get \(\frac{{dr}}{{d\theta }} = - a\sin \theta \)

Step 5:

Then

\(\begin{aligned}{l}\frac{{dy}}{{dx}} &= \frac{{\frac{{dr}}{{d\theta }}\sin \theta + r\cos \theta }}{{\frac{{dr}}{{d\theta }}\cos \theta - r\sin \theta }}\\ &= \frac{{ - a\sin \theta \sin \theta + a\cos \theta \cos \theta }}{{ - a\sin \theta \cos \theta - a\cos \theta \sin \theta }}\\ &= \frac{{a\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)}}{{ - a\left( {2\sin \theta \cos \theta } \right)}} &= - \frac{{\cos 2\theta }}{{\sin 2\theta }}\\\frac{{dy}}{{dx}} &= - \cot 2\theta \\ \Rightarrow {m_2} &= - \cot 2\theta \end{aligned}\)

Step 6:

Both curves will intersect each other at right angle when \({m_1} \cdot {m_2} = - 1\)

We have\({m_1} = \tan 2\theta \)and\({m_2} = - \cot 2\theta \)

Then\({m_1} \cdot {m_2} = - \tan \left( {2\theta } \right)\cot \left( {2\theta } \right) = - 1\)

Thus both the curves intersect each other at right angle.