Ex 6.3, 5 - Find slope of normal x = a cos3, y = a sin3

Ex 6.3,5 Find the slope of the normal to the curve 𝑥=𝑎 cos﷮3﷯﷮𝜃﷯, 𝑦=𝑎 sin3 𝜃 at 𝜃= 𝜋﷮4﷯ Given 𝑥=𝑎 cos﷮3﷯﷮𝜃﷯ Differentiating w.r.t. θ 𝑑𝑥﷮𝑑𝜃﷯= 𝑑 a cos﷮3﷯﷮𝜃﷯﷯﷮𝑑𝜃﷯ 𝑑𝑥﷮𝑑𝜃﷯=𝑎 . 𝑑 cos﷮3﷯﷮𝜃﷯﷯﷮𝑑𝜃﷯ 𝑑𝑥﷮𝑑𝜃﷯=𝑎 . 3 cos﷮2﷯﷮𝜃﷯. − sin﷮𝜃﷯﷯ 𝑑𝑥﷮𝑑𝜃﷯=− 3𝑎 sin﷮𝜃 cos﷮2﷯﷮𝜃﷯﷯ We know that slope of tangent = 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝜃﷯= 𝑑𝑦﷮𝑑𝜃﷯÷ 𝑑𝑥﷮𝑑𝜃﷯ 𝑑𝑦﷮𝑑𝑥﷯= 3 𝑎 sin﷮2﷯﷮𝜃 cos﷮𝜃﷯﷯﷮− 3𝑎 sin﷮𝜃 cos﷮2﷯﷮𝜃﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯= − sin﷮𝜃﷯﷮ cos﷮𝜃﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯=− tan﷮𝜃﷯ Putting 𝜃= 𝜋﷮4﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯﷮𝜃 = 𝜋﷮4﷯﷯=−𝑡𝑎𝑛 𝜋﷮4﷯﷯ =−1 Now we know that Tangent is perpendicular to Normal Hence, Slope of tangent × Slope of Normal = −1 −1 × Slope of Normal = −1 Slope of Normal = −1﷮−1﷯ = 1 Hence Slope of Normal is 1