Answer:
2.12 ft/s
Step-by-step explanation:
Given:
Let the fishing line be [tex]l[/tex] ft and horizontal distance between the bottom of the rod and the fish be [tex]b[/tex] ft.
Length of rod above water [tex]=8\textrm{ ft}[/tex].
Rate of reeling, [tex]\frac{dl}{dt}=2\textrm{ ft/s}[/tex]
Consider a triangle ABC as shown below.
AB is the length of the rod, BC is the distance between rod and fish and AC is the length of fish line.
For the triangle ABC,
[tex]AB^{2}+BC^{2}=AC^{2}[/tex]
[tex]8^{2}+b^{2}=l^{2}[/tex]
Differentiating both sides with respect to time t. This gives,
[tex]0+2b\frac{db}{dt}=2l\frac{dl}{dt}\\b\frac{db}{dt}=l\frac{dl}{dt}\\\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}[/tex].......... 1
For the given situation, AB = 8 ft, AC = 24 ft. Using Pythagoras theorem,
[tex]AB^{2}+BC^{2}=AC^{2}[/tex]
Plug in 8 ft for AB and 24 ft for AC, [tex]b[/tex] for BC.
[tex]8^{2}+b^{2}=24^{2}\\b^{2}=24^{2}-8^{2}\\b=\sqrt{576-64}=16\sqrt{2}\textem{ ft}[/tex]
Now, we have, [tex]b=16\sqrt{2},\frac{dl}{dt}=2,l=24[/tex]
Plug in these values in equation 1 and determine [tex]\frac{db}{dt}[/tex].
This gives,
[tex]\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}\\\frac{db}{dt}=\frac{24}{16\sqrt{2}}\times 2\\\frac{db}{dt}=\frac{3}{\sqrt{2}}=2.12\textrm{ ft/s}[/tex]
Therefore, the value of [tex]\frac{db}{dt}[/tex] is nothing but the rate at which the fish moves horizontally.
So, the rate at which the fish moving (horizontally) at the moment when the fishing line is exactly 24 ft long is 1.21 ft/s.
