We are given the second derivative of #f(t)#
#color(red)(f''(t) = t^(-3/2))#
That is, #color(red)((d^2y)/(dt^2) = t^(-3/2))#
We must find #color(blue)(f(t)# using the given details.
Solution Process being used:
We will integrate #color(red)(f''(t) = t^(-3/2))# to obtain the first derivative of #f(t)# and get #color(blue)(f'(t)#
Then we will integrate #color(blue)(f'(t)# to obtain #color(blue)(f(t)#
#color(green)(Step.1)#
To find the first derivative, we integrate #color(red)(f''(t)#
Hence, #color(red)(f'(t) = int " " f''(t)*dt#
#f'(t) = int " "t ^ (-3/2)*dt#
#rArr t ^ (-3/2+1)/(-3/2+1)+C#
#rArr t^(-1/2)/(-1/2)+C#
#rArr -t^(-1/2)/(1/2)+C#
#rArr -2t^(-1/2)+C#
#rArr -2/sqrt(t)+C#
#color(blue)(f'(t) = int " "t ^ (-3/2)*dt = -2/sqrt(t)+C)#
That is, #color(red)((dy)/(dt) = -2/sqrt(t)+C)#
#color(green)(Step.2)#
To obtain the f(t), we integrate #color(red)(f'(t)#
Hence, #color(red)(f(t) = int " " f'(t)*dt#
We will now work on
#color(red)(f(t) = int " " f'(t)*dt =int " " -2/sqrt(t)+C *dt#
We will pull the constant term out to get
#-2int " " 1/sqrt(t)+C*dt#
#rArr -2 int t ^ (-1/2)+C* dt#
#rArr -2.[t^(-1/2+1]/(-1/2+1]]+(C/1)*t^1#
#rArr -2.[t^(1/2]/(1/2]]+Ct#
#rArr -2.(2/1)*[t^(1/2)]+Ct#
#rArr -4.sqrt(t)+Ct#
We must also add a constant term to our answer.
Hence,
#color(blue)(f(t) = -4sqrt(t)+Ct+D)#
is our required function
