[tex](a)\\\\\\\text{Given that,}~~\\\\ f(t) =\dfrac 6{\sqrt{t-8}}\\\\\\\\\implies f'(t) = 6\left[\dfrac{\left(\sqrt{t-8}\right) \cdot 0 - \dfrac 1{2\sqrt{t-8 }}}{\left(\sqrt{t-8}\right)^2}\right]\\\\\\\\\implies f'(t) = -6\left(\dfrac {\dfrac 1{2\sqrt{t-8}}}{t-8}\right)\\\\\\\implies f'(t) = -6\left(\dfrac{1}{2\sqrt{t-8} (t-8)}\right)\\\\\\\implies f'(t) =-\dfrac 3{(t-8)^{\tfrac 32}}[/tex]
(b)
[tex]\text{Given that,}\\\\y=f(t)\\\\\text{Slope of y,} ~~ f'(t) =-\dfrac 3{(t-8)^{\tfrac 32}}\\\\\text{At point (33, 6/5)}\\\\\\f'(t) = -\dfrac 3{(33-8)^{\tfrac 32}} = - \dfrac 3{ (25)^{ \tfrac 32}} = - \dfrac 3{5^3} = -\dfrac 3{125}\\\\\\\text{Equation with given points,}\\\\y - \dfrac 65 = -\dfrac 3{125} ( t - 33)\\\\\\\implies y =-\dfrac 3{125} t +\dfrac{99}{125} + \dfrac 65\\\\\\\implies y = -\dfrac 3{125} t +\dfrac{249}{125}[/tex]
