Answer:
(a) [tex]s=f(d)=\sqrt{d^2+25}[/tex].
(b) The distance traveled by the ship since noon time is [tex]d=g(t)=30t[/tex].
(c) [tex](f\circ g)(t)=5\sqrt{36t^2+1}[/tex]. It represents the distance between the lighthouse and the ship as a function of the time elapsed since noon.
Step-by-step explanation:
(a) To find the distance [tex]s[/tex] between the lighthouse and the ship as a function of [tex]d[/tex], the distance the ship has traveled since noon you must:
Observe the below figure.
From the Pythagorean theorem,
[tex]s^2=d^2+5^2\\\\s^2=d^2+25\\\\s=f(d)=\sqrt{d^2+25}[/tex]
(b) The relation between distance, speed, and time is, [tex]Speed = \frac{Distance}{Time}[/tex]. Therefore, we have
[tex]Distance = Speed \cdot Time[/tex].
We know that the ship is moving at a speed of 20 km/h.
[tex]d=g(t)[/tex] is the distance as a function of time.
So, the distance traveled by the ship since noon time is [tex]d=g(t)=30t[/tex]
(c) Given two functions [tex]f[/tex] and [tex]g[/tex] , the composite function [tex]f\circ g[/tex] is defined by
[tex](f\circ g)(x)=f(g(x))[/tex]
We have
[tex]s=f(d)=\sqrt{d^2+25}\\\\d=g(t)=30t[/tex]
So,
[tex](f\circ g)(t)=f(g(t))=\sqrt{(30t)^2+25}\\\\(f\circ g)(t)=5\sqrt{36t^2+1}[/tex]
It represents the distance between the lighthouse and the ship as a function of the time elapsed since noon.
Speed is the rate of change of distance over time
(a) The distance function s(d)
The relation between s and d can be illustrated by the following Pythagoras theorem (see attachment)
[tex]\mathbf{s^2 = d^2 + 5^2}[/tex]
Evaluate 5^2
[tex]\mathbf{s^2 = d^2 + 25}[/tex]
Take square roots of both sides
[tex]\mathbf{s = \sqrt{d^2 + 25}}[/tex]
Express s as a function of d
[tex]\mathbf{s(d) = \sqrt{d^2 + 25}}[/tex]
(b) The distance function d(t)
From the question, we have:
[tex]\mathbf{Speed = 20kmh^{-1}}[/tex]
Distance (d) is calculated as:
[tex]\mathbf{d = Speed \times Time}[/tex]
So, we have:
[tex]\mathbf{d = 20 \times t}[/tex]
[tex]\mathbf{d = 20t}[/tex]
Express as a function of t
[tex]\mathbf{d(t) = 20t}[/tex]
(c) Composite function f o g
In (a), we have:
[tex]\mathbf{s(d) = \sqrt{d^2 + 25}}[/tex]
In (b), we have:
[tex]\mathbf{d(t) = 20t}[/tex]
f o g represents s o d
So, we have:
[tex]\mathbf{s\ o\ d = s(d(t))}[/tex]
This gives
[tex]\mathbf{s\ o\ d = \sqrt{d(t)^2 + 25}}[/tex]
Substitute 20t for d(t)
[tex]\mathbf{s\ o\ d = \sqrt{(20t)^2 + 25}}[/tex]
[tex]\mathbf{s\ o\ d = \sqrt{400t^2 + 25}}[/tex]
Hence, f o g represents distance (s) as a function of time (t)
Read more about speed and distance at:
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