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

(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 )(W(t) - t^3 + (1)/(4) )/(t -2)` is
`(√(3) \pi -96 )/(8)`

Explanation:

Here we have

`W(t) = \begin{cases}(17)/(2)-(3)/(2)\cos \left ((\pi t)/(6) \right ) & \text{ if } 0\leq t\leq 6 \\ 10-(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-(1)/(5)\left (8-6 \right )^(2)`

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

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

∴
`W'(8) =- (2* 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 =
`(\pi \sin((\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 )(W(t) - t^3 + (1)/(4) )/(t -2)` where W(t) =
`(17)/(2)-(3)/(2)\cos \left ((\pi t)/(6) \right )`

`\lim_(t \to 2 )(W(t) - t^3 + (1)/(4) )/(t -2)` =
`(√(3) \pi -96 )/(8)`.