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. 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 :

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

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

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

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

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

W(3) = 8.5

So, the tangent is given by the equation:

Now, put x = 3.5 in above equation.

y = 8.8927

Therefore, W(3.5) 9

C)

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

=

Thus the function is undefined at t = 2.

For more information, refer to the link given below:

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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.

B) To find the line tangent to the graph of W at t=3; 3 falls into the interval of 0 ≤ t < 6,

Thus;

w(t) = 17/2 - (3/2)(cos(πt/6))

Now let's differentiate so we can find w'(3)

So, w'(t) = -(3/2 x -π/6)(sin(πt/6))

w'(t) = (π/4)(sin(πt/6))

So; w'(3) = (π/4)(sin(π*3/6))

w'(3) = (π/4)(sin(π/2))

sin (π/2) in radians = 1

Thus,w'(3) = (π/4)

Now, let's find the y coordinate which is at w(3);

w(3) = 17/2 - (3/2)(cos(π x 3/6))

w(3) = 17/2 - (3/2)(cos(π/2))

(cos(π/2)) in radians = 0

Thus;

w(3) = 17/2

So the tangent line will be;

y - 17/2 = (π/4)(x - 3)

y = (π/4)(x - 3) + 17/2

Now, at t = 3.5, putting 3. 5for x;

y = (π/4)(3.5 - 3) + 17/2

y = 0.5(π/4) + 8.5

y = 0.3927 + 8.5

y = 8.8927

Thus; w(3.5) ≤ 9

C) we want to find the limit as t approaches 2.

Now, 2 falls into the interval of 0 ≤ t < 6,

Thus;

w(t) = 17/2 - (3/2)(cos(πt/6))

So, z = [17/2 - (3/2)(cos(πt/6)) - t³ + ¼]/(t - 2)

At t = 2,we have;

Lim z → 2 = [17/2 - (3/2)(cos(π x 2/6)) - 2³ + ¼]/(2 - 2)

Lim z → 2 = [17/2 - (3/4) - 8 + ¼]/0

The denominator is zero and thus the function is undefined.

So, at t = 2, the function has a vertical asymptote