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in Derivatives by (44 points)

if x= a( øsinø + cosø) and y= b( sinø-øcosø)

find d2y/dx2 at ø=π/4

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Answer : d2y/dx at ø = π/4 = \(\frac{b8√2}{{a}^{2}\pi}\)

x=a(øsinø + cosø)      and     y = b(sinøøcosø

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/dx at π/4 = \(\frac{8√2b}{{a}^{2}π}\)

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