To reach the drowning person in the least amount of time, the lifeguard can swim some distance and run the remaining distance.
- [tex]Part \ 1 :\mathrm{The \ function \ is;} \ T(\theta) = \displaystyle \frac{ 43\times \cos(\theta)}{1.22} + \displaystyle \frac{43 \times \theta}{3.26}[/tex]
- Part 2: The angle at which the lifeguard should swim is approximately 21.98°
Reasons:
The given parameters are;
Diameter of the circular pool, AC = 43 meters
The rate at which the lifeguard swims = 1.22 m/s
Rate at which the lifeguard runs = 3.26 m/s
Part 1: The triangle ΔABC formed by the path AB = A right triangle
[tex]\displaystyle cos\left(\theta \right) = \frac{AB}{AC}[/tex]
AB = AC × cos(θ)
∠COB = 2 × θ (Angle at the center is twice angle on the circumference)
Length of arc BC = Radius of the circle × ∠COB = [tex]\displaystyle \frac{AC}{2} \times 2 \times \theta = AC \times \theta[/tex]
Therefore;
BC = AC × θ
The time it takes the lifeguard to swim, t₁, is given as follows;
[tex]\displaystyle t_1 = \mathbf{ \frac{AC \cdot \cos(\theta)}{1.22}}[/tex]
The time it takes the lifeguard to swim, t₂, is presented as follows;
[tex]\displaystyle t_2 = \frac{AC \times \theta}{3.26}[/tex]
Where AC = 43, we have;
Total time, T(θ) = t₁ + t₂
Which gives the function that measures the amount of time it takes to reach
the drowning person as a function of the swim angle θ as follows;
- [tex]\underline{T(\theta) = \displaystyle \frac{ 43\times \cos(\theta)}{1.22} + \displaystyle \frac{43 \times \theta}{3.26}}[/tex]
Part 2: The angle that gives the least amount of time is found as follows;
At the minimum time, we have;
[tex]T'(\theta) =0 = \mathbf{\displaystyle \frac{d}{d\theta} \left(\frac{ 43\times \cos(\theta)}{1.22} + \displaystyle \frac{43 \times \theta}{3.26} \right)} =\frac{2150}{163} -\frac{2150}{61} \cdot sin(\theta)[/tex]
Which gives;
[tex]\displaystyle \frac{2150}{163} =\frac{2150}{61} \cdot sin(\theta)[/tex]
[tex]\displaystyle \theta = arcsin\left(\frac{ \frac{2150}{163} }{\frac{2150}{61} } \right) = arcsin\left(\frac{ 61 }{163 } \right) \approx \mathbf{21.98^{\circ}}[/tex]
The angle at which the lifeguard should swim, θ ≈ 21.98°
Learn more here:
https://brainly.com/question/4950874
