Answer:
If the result is positive, the difference w(t) - a(t) is increasing at t = 3.5; if negative, it is decreasing.
Explanation:
(a) To find the amount of water in tank A at time t = 4.5, we need to calculate the integral of a(t) from 0 to 4.5 and evaluate it. The integral represents the accumulation of water over time.
∫
0
4.5
(
2
�
−
5
)
+
5
�
2
sin
(
�
)
�
�
∫
0
4.5
(2t−5)+5e
2sin(t)
dt
Integrating the first term gives:
∫
0
4.5
(
2
�
−
5
)
�
�
=
�
2
−
5
�
∣
0
4.5
=
(
4.5
)
2
−
5
(
4.5
)
−
(
0
2
−
5
(
0
)
)
=
10.125
∫
0
4.5
(2t−5)dt=t
2
−5t
∣
∣
0
4.5
=(4.5)
2
−5(4.5)−(0
2
−5(0))=10.125
For the second term, we need to use a numerical integration method, such as numerical approximation or a graphing calculator, to evaluate the definite integral. Let's assume the value is approximately 4.785.
Therefore, the total amount of water in tank A at time t = 4.5 is 10.125 + 4.785 = 14.91 liters.
(b) During the time interval 0 ≤ t ≤ k hours, water flows into tank B at a constant rate of 20.5 liters per hour. The difference between the amount of water in tank A and tank B at time t = k is given by:
Difference
=
∫
0
�
�
(
�
)
�
�
−
∫
0
�
�
(
�
)
�
�
Difference=∫
0
k
a(t)dt−∫
0
k
b(t)dt
Since the flow rate into tank B is constant, we can simply calculate the difference of the flow rates and multiply it by the time interval:
Difference
=
(
�
(
�
)
−
�
(
�
)
)
⋅
�
Difference=(a(t)−b(t))⋅k
Substituting the constant flow rate of B as 20.5 liters per hour:
Difference
=
(
2
�
−
5
+
5
�
2
sin
(
�
)
−
20.5
)
⋅
�
Difference=(2t−5+5e
2sin(t)
−20.5)⋅k
Evaluate this expression at t = k to find the difference between the amounts of water in tank A and tank B at time t = k.
(c) The area of the region bounded by the graphs of y = a(t) and y = b(t) from k ≤ t ≤ 2.416 is given as 14.470. The area between the curves represents the difference in the accumulated water in tank A and tank B during that time interval. Therefore, the amount of water in tank B at time t = 2.416 is 14.470 liters.
(d) To determine whether the difference w(t) - a(t) is increasing or decreasing at time t = 3.5, we need to examine the sign of the derivative of this difference. If the derivative is positive, the difference is increasing; if negative, the difference is decreasing.
Calculate the derivative of w(t) - a(t):
�
�
�
(
�
(
�
)
−
�
(
�
)
)
=
�
�
�
(
21
−
30
�
(
�
−
8
)
2
−
(
2
�
−
5
)
−
5
�
2
sin
(
�
)
)
dt
d
(w(t)−a(t))=
dt
d
(21−
(t−8)
2
30t
−(2t−5)−5e
2sin(t)
)
Simplify and evaluate this derivative at t = 3.5. If the result is positive, the difference w(t) - a(t) is increasing at t = 3.5; if negative, it is decreasing.