Question

find the first and the second derivative (dy/dx and d(^2)y/d(^2)x) for the parametric question. x=1-sin(t) and y=t-cos(t)

Answer 1

\[dx/dt=-\cos(t);
dy/dt=1+\sin(t);
dy/dx=dy/dt \div dx/dt=(1+\sin(t))/(-\cos(t))\]
\[d^2y/dx^2=[(-\cos(t))(\cos(t))-(1+\sin(t))(\sin(t))]/(-\cos(t))^2\]
\[d^2y/dx^2=(-\cos^2(t)-\sin(t)-\sin^2(t))/\cos^2(t)\]
\[d^2y/dx^2=(-1-\sin(t))/\cos^2(t)\]
\[d^2y/dx^2=(-1/\cos^2(t))+(-\sin(t)/\cos^2(t))\]
\[d^2y/dx^2=-\sec^2(t)+(-\tan(t)/\cos(t))\]
\[d^2y/dx^2=-\sec^2(t)-\sec(t)\tan(t)\]
\[d^2y/dx^2=-\sec(t)[\sec(t)+\tan(t)]\]