go 1-3 d 2 Cos (so t- 30e) 4 -d LS Cos 50t dt

If x=a (cos t + t sin t) and y = a (sin t - t cos t ) , find (d^2x)/(dt^2),(d^2y)/(dt^2) and (d^2y)/(dx^2) . Also mention the domain of validity.

If x = sin t and y = tan t , then (dy)/(dx) is equal to a) cos ^(2) t b) (1)/( cos ^(3) t ) c) (1)/( cos ^(2) t ) d) sin ^(2) t

int sqrt(2+ sin 3t) cos 3t dt

If x = a ( cos t + t sint) and y = a (sin t - t cost), 0 < t < pi/2 , find (d^2x)/(dt^2),(d^2y)/(dt^2)

If x = a sin 2t(1 + cos 2t) and y = b cos 2 t(1 – cos 2 t), show that ((d y)/(d x))_(at t =pi/4)=b/a.

Suppose we consider friction between string and the pulley while still considering the string to be massless. So, in such a theoretical case, the tension in the string in contant with the pulley will not be constant and its variation is calculated as follows. If the string is just on the verge of slipping on the pulley, then friction acting is limiting and if T_(2) gt T_(1) then friction will act towards T_(1) in tangential direction as shown. d N = (T + df) sin (d theta//2) + (T + dT) sin (d theta//2) dN = T sin (d theta//2) + df sin (d del//2) + T sin (d theta//2) + d T sin (d theta//2) dN ~~ 2T sin (d theta//2) d N ~~ 2T (d theta)/(2) d N = T d theta ........(1) Alsso, (T + df) cos ((d theta)/(2)) = (T + d T) cos ((d theta)/(2)) T cos ((d theta)/(2)) + df cos ((d theta)/(2)) = T cos ((d theta)/(2)) 1 dT cos ((d theta)/(2)) df = dT = mu d N .........(2) from (1) & (2) dT = mu T d theta int_(T_(1))^(T_(2)) (dT)/(T) = mu int_(0)^(theta) d theta ln ((T_(2))/(T_(1))) = mu theta T_(2) = T_(1) e^(mu theta) , where theta is the angle of contact between the string and the pulley Based on above information, answer the following questions. Two masses m_(1) kg and m_(2) kg passes over an atwood machine. Find the ratio of masses m_(1) and m_(2) so that string passing over the pulley will just start slipping over its surface. The friction coefficient between the string and pulley surface is 0.2.

cos t dx/dt + sin t = 1