Mathematics, 24.04.2020 19:29 arom7470

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.
(A) Find W′(8) . Using correct units, explain the meaning of W′(8) in the context of the problem.
(B)The graph of W is concave down for 3≤t≤3.5 . Use the line tangent to the graph of W at t=3 to show that W(3.5)≤9
(C) in picture.

The depth of a river at a certain point is modeled by the function W defined above, where W(t) is me

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Answer from: justin20080

The depth of a river at a certain point is modeled by the function W. So, the value of W'(8) is -0.68 for t between 0 and 6 and by using the tangent it can be proved that the value of W(3.5) is less than or equal to 9.

Given :

The depth of a river at a certain point is modeled by the function W $\rm W(t) =\left \{ {{\dfrac{17}{2}-\dfrac{3}{2}cos \left(\dfrac{\pi t}{6}\right)\;\;\;\;\;\; for \;\;0\leq t \leq 6} \atop {10-\dfrac{1}{10}(t-6)^2\;\;\;\;\;\;for \;\; 6\leq t \leq 10}} \right.$ ---- (1) W(t) is measured in feet and time t is measured in hours.

A) Differentiate W(t) with respect to 't'.

$\rm W'(t) =\left \{ {{0+\dfrac{\pi}{6}\times \dfrac{3}{2} sin \left(\dfrac{\pi t}{6}\right)\;\;\;\;\;\; for \;\;0\leq t \leq 6} \atop {0-\dfrac{1}{5}(t-6)\;\;\;\;\;\;for \;\; 6\leq t \leq 10}} \right.$

Now, put the value of (t = 8) in order to evaluate W'(8).

$\rm W'(8) =\left \{ {{0+\dfrac{\pi}{6}\times \dfrac{3}{2} sin \left(\dfrac{4\pi}{3}\right)\;\;\;\;\;\; for \;\;0\leq t \leq 6} \atop {0-\dfrac{1}{5}(2)\;\;\;\;\;\;for \;\; 6\leq t \leq 10}} \right.$

$\rm W'(8) =\left \{ {{-\dfrac{\pi}{6}\times \dfrac{3}{2} \left(\dfrac{\sqrt{3} }{2}\right)\;\;\;\;\;\; for \;\;0\leq t \leq 6} \atop {0-\dfrac{1}{5}(2)\;\;\;\;\;\;for \;\; 6\leq t \leq 10}} \right.$

$\rm W'(8) =\left \{ {{-0.68 \;\;\;\;\;\; for \;\;0\leq t \leq 6} \atop {-0.4 \;\;\;\;\;\;for \;\; 6\leq t \leq 10}} \right.$

B) Differentiate W for 3 ≤ t ≤ 3.5, that is:

$\rm W'(t) =\dfrac{\pi}{6}\times \dfrac{3}{2} sin \left(\dfrac{\pi t}{6}\right)\;\;\;\;\;\; for \;\;0\leq t \leq 6$

Now, put (t = 3) in the above equation.

$\rm W'(3) = \dfrac{\pi}{6}\times \dfrac{3}{2}sin\dfrac{\pi}{2}$

$\rm W'(3) = \dfrac{\pi}{4}$

Now, put the value (t=3) in equation (1) only for 3 ≤ t ≤ 3.5.

$\rm W(3) = \dfrac{17}{2}-\dfrac{3}{2}cos\dfrac{\pi}{2}$

$\rm W(3) = \dfrac{17}{2}$

W(3) = 8.5

So, the tangent is given by the equation:

$y - \dfrac{17}{2}=\dfrac{\pi}{4}(x-3)$

Now, put x = 3.5 in above equation.

$y -\dfrac{17}{2}=\dfrac{\pi}{4}\times 0.5$

y = 8.8927

Therefore, W(3.5) $\leq$ 9

$\lim_{t \to \infty} \dfrac{W(t)-t^3+\frac{1}{4}}{t-2}$

$\rm A =\dfrac{ \dfrac{17}{2}-\dfrac{3}{2}cos\dfrac{\pi t}{6}-t^3+\dfrac{1}{4}}{t-2}$

Now, at (t=2) A tends towards 2.

$\rm \lim_{A \to 2} \dfrac{ \dfrac{17}{2}-\dfrac{3}{2}cos\dfrac{\pi t}{6}-t^3+\dfrac{1}{4}}{t-2}$

= $\rm \dfrac{ \dfrac{17}{2}-\dfrac{3}{2}cos\dfrac{\pi \times 2}{6}-2^3+\dfrac{1}{4}}{2-2}$

Thus the function is undefined at t = 2.

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Answer from: ZOE1000000

A) w'(8) = -4/5 ;w'(8) is less than 0, so it shows that w(t) function is a decreasing function and thus with the decrease in one function, the other will also decrease.

B) y = 8.8927 ;Thus we have proved that; w(3.5) ≤ 9

C) The function is undefined and thus has a vertical asymptote

Step-by-step explanation:

A) we are to find w'(8)

8 lies in the interval of 6 < t ≤ 10, we'll use;

w(t) = 10 - (1/5)(t - 6)²

Differentiating, we have;

w'(t) = -(2/5)(t - 6)

Thus, w'(8) = -(2/5)(8 - 6)

w'(8) = -4/5

As, w'(8) is less than 0,it shows that w(t) function is a decreasing function.

Thus, there will be a direct relation such that with the decrease in one function, the other will also decrease.