Let's solve the problem step by step.

Given: $f(t) = \frac{8}{\sqrt{t - 2}}$

Part (a): Find the derivative of $f(t)$

We will use the quotient rule or rewrite the function to a more manageable form. First, we rewrite $f(t)$ as: $f(t) = 8(t - 2)^{-\frac{1}{2}}$ Now, we apply the chain rule.

The derivative of $f(t)$ is: $f'(t) = 8 \cdot \left(-\frac{1}{2}\right) \cdot (t - 2)^{-\frac{3}{2}} \cdot 1$ Simplifying: $f'(t) = -4(t - 2)^{-\frac{3}{2}}$ Thus, the derivative is: $f'(t) = -\frac{4}{(t - 2)^{\frac{3}{2}}}$

Part (b): Equation of the tangent line

We are given the point $(t, y) = (11, \frac{8}{3})$. To find the equation of the tangent line, we need the slope of the tangent line at $t = 11$, which is $f'(11)$, and the point $(11, \frac{8}{3})$.

1. Find the slope $f'(11)$: $f'(11) = -\frac{4}{(11 - 2)^{\frac{3}{2}}} = -\frac{4}{9^{\frac{3}{2}}} = -\frac{4}{27}$

2. Use the point-slope form of a line: The point-slope form is: $y - y_1 = m(t - t_1)$ where $(t_1, y_1) = (11, \frac{8}{3})$ and $m = -\frac{4}{27}$.

Substitute the values: $y - \frac{8}{3} = -\frac{4}{27}(t - 11)$ Simplify the equation: $y = -\frac{4}{27}(t - 11) + \frac{8}{3}$

Thus, the equation of the tangent line is: $y = -\frac{4}{27}(t - 11) + \frac{8}{3}$

Would you like further clarification or details on any step?

Here are five related questions for further practice:

1. How would the derivative change if $f(t)$ had a different constant numerator?
2. How would you compute the second derivative of $f(t)$?
3. What is the geometric interpretation of the tangent line equation?
4. How would you solve the equation if the point of tangency were different?
5. What happens to the slope of the tangent line as $t \to 2$?

Tip: Remember that when differentiating square roots, converting them into fractional exponents makes applying the chain rule easier.