for 0<=t<=21 the rate of change of the number of black flies on a coastal island at time t days is modeled by R(t)=3sqrt(t)cos(t/3) flies per day. There are 500 flies on the island at the time t=0. To the nearest whole #, what is the max # of flies for 0<=t<=21?
The maximum occurs when R(t) = 0
That would be when t/3 = pi/2
t = 4.712 days
To get the number of flies at that time, you have to integrate R(t) from t=0 to t= 4.712 days, with the number of flies being 500 at t=0
wait...dont u have to integrate it from 0 to 21 or something?? n how would you integrate this equation?
To find the maximum number of flies on the island for 0 ≤ t ≤ 21, we need to find the maximum value of the rate of change function R(t).
Given: R(t) = 3√t * cos(t/3)
Step 1: Determine the critical points.
Critical points occur when the derivative of R(t) is equal to zero or does not exist.
To find the critical points, let's first find the derivative of R(t).
R'(t) = d/dt (3√t * cos(t/3))
= 3/2 * t^(-1/2) * cos(t/3) - 1/2 * √t * sin(t/3)
Now, set R'(t) = 0 and solve for t:
3/2 * t^(-1/2) * cos(t/3) - 1/2 * √t * sin(t/3) = 0
Simplifying this expression is quite challenging as it involves trigonometric and square root functions. It cannot be easily solved analytically. Instead, we can use numerical methods or graphing calculators to estimate the critical points.
Step 2: Use numerical methods or graphing calculators to estimate the critical points.
By using numerical methods or graphing calculators, we can determine the critical points to be around t ≈ 5.2 and t ≈ 18.9.
Step 3: Evaluate R(t) at the endpoints and critical points.
To find the maximum number of flies, we need to evaluate R(t) at the endpoints of the interval [0, 21] and the critical points.
R(0) ≈ 3√0 * cos(0/3) = 3 * 1 = 3 (there are 3 flies at t = 0)
R(21) ≈ 3√21 * cos(21/3) ≈ 3√21 * cos(7) ≈ 3 * 4.58 * (-0.80) ≈ -46.55 (negative value doesn't make sense in this context)
From the calculations above, we can see that the maximum number of flies occurs at one of the critical points.
Step 4: Determine the maximum number of flies.
To find the maximum number of flies, we need to evaluate R(t) at the critical points we estimated earlier.
R(5.2) ≈ 3√5.2 * cos(5.2/3) ≈ 3 * 2.28 * 0.73 ≈ 15.14
R(18.9) ≈ 3√18.9 * cos(18.9/3) ≈ 3 * 3.57 * (-0.98) ≈ -33.13
Since we are looking for the maximum number of flies, we take the larger value of R(5.2) and R(18.9), which is approximately 15.14.
Therefore, the maximum number of flies, to the nearest whole number, is 15.
To find the maximum number of flies on the island for the given time interval, we need to find the maximum value of the rate function R(t).
Since R(t) is given by R(t) = 3√(t)cos(t/3), we can find the maximum by finding the critical points of this function.
To find the critical points, we need to find where the derivative of R(t) is equal to zero.
First, let's find the derivative of R(t):
R'(t) = d/dt [3√(t)cos(t/3)]
= 3cos(t/3) * d/dt [√(t)]
To find the derivative of √(t), we can use the power rule:
d/dt [√(t)] = (1/2)t^(-1/2)
= 1/(2√(t))
Now, let's substitute this derivative into R'(t):
R'(t) = 3cos(t/3) * 1/(2√(t))
= 3cos(t/3)/(2√(t))
To find the critical points, we set R'(t) equal to zero and solve for t:
3cos(t/3)/(2√(t)) = 0
Since the numerator, 3cos(t/3), cannot be zero for any value of t, the only possible solution is when the denominator, 2√(t), is equal to zero.
Setting 2√(t) = 0 and solving for t, we get:
√(t) = 0
Squaring both sides, we get:
t = 0
Therefore, the only critical point is t = 0.
Next, we need to check the endpoints of the interval [0, 21] to see if there are any maximum or minimum values at those points.
At t = 0, there are 500 flies on the island, which is the initial number.
Next, we need to evaluate R(t) at t = 21:
R(21) = 3√(21)cos(21/3)
≈ 3(4.5826)(0.8260)
≈ 45.29 flies per day
Since the number of flies cannot be negative, the maximum number of flies is either at t = 0 or t = 21.
Comparing the values, we see that R(21) ≈ 45.29 is greater than the initial number of 500 flies at t = 0.
Therefore, the maximum number of flies for 0 <= t <= 21, to the nearest whole number, is 500 flies.