Answer : d2y/dx2 at ø = π/4 = \(\frac{b8√2}{{a}^{2}\pi}\)
x=a(øsinø + cosø) and y = b(sinø - øcosø)
dx/dø = a(sinø + øcosø - sinø) = aøcosø
dy/dø = b(cosø - cosø + øsinø) = bøsinø
\(\frac{dy}{dx} = \frac{\frac{dy}{d\phi}}{\frac{dx}{d\phi}}\)
\(\frac{b\phi\sin\phi}{a\phi\cos\phi}\)
\(= \frac{btan\phi}{a}\)
\(\therefore \frac{dy}{dx} = \frac{btan\phi}{a} \)
\(\frac{{d}^{2}y}{d{x}^{2}} = \frac{d(\frac{btan\phi}{a})}{d\phi} × \frac{d\phi}{dx}
\)
\(= \frac{b{sec}^{2}\phi}{a} × \frac{1}{a\phi cos\phi}\)
\(= \frac{b}{{a}^{2}\phi{cos}^{3}\phi}\)
\(\therefore \frac{{d}^{2}y}{d{x}^{2}}=\frac{b}{{a}^{2}\phi{cos}^{3}\phi}\)
Putting ø = π/4 in above we get
d2y/dx2 at π/4 = \(\frac{8√2b}{{a}^{2}π}\)