Question

The depth of a river at a certain point is modeled by the function W defined above, where W(t) is measured in feet and time T is measured in hours

Answer 1

Answer:

(a) The meaning of W'(8) is the rate of change of the depth of the water at time t = 8 hours is -0.8 ft/hr

(b) The tangent line equation is Y = 0.79×t +6.143

Therefore, W(3.5) ≤ 9 as 0.79×3.5 +6.143 = 8.908 < 9

(c) $\lim_{t \to 2 }\frac{W(t) - t^3 + \frac{1}{4} }{t -2}$ is $\frac{\sqrt{3} \pi -96 }{8}$

Step-by-step explanation:

Here we have

$W(t) = \begin{cases}\frac{17}{2}-\frac{3}{2}\cos \left (\frac{\pi t}{6} \right ) & \text{ if } 0\leq t\leq 6 \\ 10-\frac{1}{5}\left (t-6 \right )^{2} & \text{ if } 6< t\leq 10 \end{cases}$

(a) To find W'(8) we have

W(8) = $10-\frac{1}{5}\left (8-6 \right )^{2}$

Therefore, W'(8) given by the following relation;

$W'(t) = \frac{\mathrm{d} \left (10-\frac{1}{5}\left (t-6 \right )^{2} \right )}{\mathrm{d} t} = - \frac{2t-12}{5}$

∴$W'(8) =- \frac{2\times 8-12}{5} = -0.8 \ ft/hr$

The meaning of W'(8) is the rate of change of the depth of the water at time t = 8 hours = -0.8 ft/hr

b) Here we have the line tangent is given by the slope of the graph at the point t = 3, therefore we have

W'(t), t = 3 = $\frac{\pi \sin(\frac{\pi t}{6} )}{4}$

The tangent line equation is Y = 0.79×t +6.143

Therefore, W(3.5) ≤ 9 as 0.79×3.5 +6.143 = 8.908 < 9

c) $\lim_{t \to 2 }\frac{W(t) - t^3 + \frac{1}{4} }{t -2}$ where W(t) = $\frac{17}{2}-\frac{3}{2}\cos \left (\frac{\pi t}{6} \right )$

$\lim_{t \to 2 }\frac{W(t) - t^3 + \frac{1}{4} }{t -2}$  = $\frac{\sqrt{3} \pi -96 }{8}$.