Question

Please Help - Suppose f(t)=6t−8−−−−√.

(a) Find the derivative of f.

f′(t) =

-3/(sqrt(t-8)(t-8))



(b) Find an equation for the tangent line to the graph of y=f(t) at the point (t,y)=(33,6/5).

Tangent line: y =

Answer 1

`(a)\\\\\\\text{Given that,}~~\\\\ f(t) =\frac 6{√(t-8)}\\\\\\\\\implies f'(t) = 6\left[\frac{\left(√(t-8)\right) \cdot 0 - \frac 1{2√(t-8 )}}{\left(√(t-8)\right)^2}\right]\\\\\\\\\implies f'(t) = -6\left(\frac {\frac 1{2√(t-8)}}{t-8}\right)\\\\\\\implies f'(t) = -6\left((1)/(2√(t-8) (t-8))\right)\\\\\\\implies f'(t) =-\frac 3{(t-8)^(\tfrac 32)}`

(b)

`\text{Given that,}\\\\y=f(t)\\\\\text{Slope of y,} ~~ f'(t) =-\frac 3{(t-8)^(\tfrac 32)}\\\\\text{At point (33, 6/5)}\\\\\\f'(t) = -\frac 3{(33-8)^(\tfrac 32)} = - \frac 3{ (25)^( \tfrac 32)} = - \frac 3{5^3} = -\frac 3{125}\\\\\\\text{Equation with given points,}\\\\y - \frac 65 = -\frac 3{125} ( t - 33)\\\\\\\implies y =-\frac 3{125} t +(99)/(125) + \frac 65\\\\\\\implies y = -\frac 3{125} t +(249)/(125)`