Question

Find dy/dx and d²y/dx² of;

x = theta + cos theta, y = 1 + sin theta

; theta = pi/6

Answer 1

i will use parameter t instead of theta

for parametric function

first derivative $y'_x$ can be found as ${\frac {y'_t} {x'_t}}$

second derivative $y''_{xx}$ can be found as ${\frac {(y'_x)'_t} {x'_t}}$

$y'_t=cos(t)$

$x'_t=1-sin(t)$

so,

$y'_x={\frac {cos(t)} {1-sin(t)}}\implies y'_x({\frac {\pi} 6})={\frac {cos({\frac {\pi} 6})} {1-sin({\frac {\pi} 6})}}=\sqrt3$

$(y'_x)'_t={\frac {-sin(t)+sin^2(t)+cos^2(t)} {(1-sin(t))^2}}={\frac 1 {(1-sin(t))}}$

so,

$y''_{xx}={\frac 1 {(1-sin(t))^2}}\implies y''_{xx}({\frac {\pi} 6})={\frac1 {(1-sin({\frac {\pi} 6}))^2}}=4$