M1數學 (easy) 設x=(1/t^3) - 3e^(2t) 而y=(4/t^3) + e^(-3t) 如果dy/dx=4 而t不=0 求t?

x = t^(-3) - 3e^(2t)
dx/dt = -3t^(-4) - 6e^(2t)

y = 4t^(-3) + e^(-3t)
dy/dt = -12t^(-4) - 3e^(-3t)

4 = dy/dx = (dy/dt)/(dx/dt) = [ - 12t^(-4) - 3e^(-3t) ] / [ - 3t^(-4) - 6e^(2t) ]
4[ - 3t^(-4) - 6e^(2t) ] = - 12t^(-4) - 3e^(-3t)
- 12t^(-4) - 24e^(2t) = - 12t^(-4) - 3e^(-3t)
e^(-3t) = 8e^(2t)
e^(-3t) / e^(2t) = 8
e^(-5t) = 8
- 5t = ln 8
t = - ( ln 8 ) / 5 ..... Ans