Question

find d^y/dx^2 if x = 2acos^3θ ; y = 3asin^2θ at θ = π/6

Answer 1

Answer:

Your answer attached in the photo

Answer 2

Step-by-step explanation:

$x = 2a {cos}^{3} \theta, \: y = 3asin^2θ \\ \ \ \\\\ \ \ x = 2a {cos}^{3} \theta \\ \frac{dx}{d \theta} = \frac{d}{d \theta} (2a {cos}^{3} \theta) \\ = 2a\frac{d}{d \theta}( {cos}^{3} \theta ) \\ = - 6a {cos}^{2} \theta sin \theta \\ \\ y = 3asin^2θ \\ \frac{dy}{d \theta} = \frac{d}{d \theta} (3asin^2θ) \\ = 3a\frac{d}{d \theta}( {sin}^{2} \theta ) \\ = 6asin \theta cos \theta \\ \\ \boxed{ \frac{dy}{d x} = \frac{ \frac{dy}{d \theta} }{ \frac{dx}{d \theta} } }\\\\\frac{dy}{d x} = \frac{ \frac{dy}{d \theta} }{ \frac{dx}{d \theta} }\\ \:\: \:\: \:\: \:\: \:\: \: \:\: \:\: = \frac{ \cancel{6a} \cancel{sin \theta} \cancel{cos \theta }}{- \cancel{6a }{cos}^{ \cancel{2}} \theta \cancel{ sin \theta} } \\ = - \frac{ 1}{cos \theta} \\ = - sec \theta \\\\\\ \frac{dy}{dx} = - sec \theta \\ \frac{ {d}^{2}y }{d {x}^{2} } = \frac{d}{dx} ( - sec \theta) \\ = - \frac{d}{dx}(sec \theta) \\ = - sec \theta tan \theta \frac{d \theta}{dx} \\ = \frac{ sec \theta tan \theta}{ 6a{cos}^{2} \theta sin \theta } \\ \\\\ \\ \left. \dfrac{ {d}^{2} y }{d {x}^{2} } \right]_{ \theta = \frac{\pi}{6} } =\frac{ sec \frac{\pi}{6} tan \frac{\pi}{6} }{ 6a{cos}^{2} \frac{\pi}{6} sin \frac{\pi}{6} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ \frac{2}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ 3 } }{6a \times \frac{ 3 }{4} \times \frac{1}{2} } \\ = \frac{ \frac{2}{3} }{ \frac{9a}{4} } \\ = \frac{2}{3} \times \frac{4}{9a} \\ = \frac{8}{27a}$