Oil moves through a pipeline such that the distance s, it moves and the time t?

s^3-t^2=7t





Find the velocity of the oil for s=4.01 m and t=5.25 s.Oil moves through a pipeline such that the distance s, it moves and the time t?
Greetings





take derivatives of both sides with respect to t





3s^2s' - 2t = 7


3(4.01)^2*s' - 2(5.25) = 7


48.24s' - 10.5 = 7


48.24s' = 17.5


s' = 0.363 m/s





velocity of the oil is about 0.363 m/s





RegardsOil moves through a pipeline such that the distance s, it moves and the time t?
s^3-t^2=7t


3s^2ds-2tdt=7dt


3s^2ds=dt(2t+7)


ds/dt=(2t+7)/3s


For s=4.01 m and t=5.25s.,


ds/dt=(2*5.25+7)/3*4.01


ds/dt=(10.5+7)/12.03


ds/st=17.5/12.03


ds/dt= 1.45 m/sec.