I have come at this a few different ways and I just can't seem to figure out how to get the correct answer for this.
The question is: A fisherman is reeling in a fish at a rate of 20 centimeters per second. If the tip of his fishing rod is 4.5 meters above the water, ad we are assuming that the fish is near the water surface throughout the process, how fast is it approaching when 7.5 meters of fishing line are still out? How fast is the angle θ between the fishing line and the water increasing at that instant?
As I see it, $$ {dl\over dt} = 20 cm/sec $$ $$ {dx\over dt} = -25 cm/sec $$ $$ 2x {dx\over dt} =2l {dl\over dt} $$ $$ {dx\over dt} = {l\over x} * {dl\over dt} $$
Then for θ $$ tanθ= {4.5\over x} $$ $$ sec^2θ * {dθ\over dt} = {-4.5\over x^2} * {dx\over dt} $$ $$ {dθ\over dt} = cos^2θ * {-4.5\over x^2} * {dx\over dt} $$ $$ cos^2θ = {6^2\over 7.5^2} $$
so $$ {dθ\over dt} = {6^2\over 7.5^2} * {-4.5\over 6^2} * (-25) = 2? $$ but my book says the answer should be .02 rad/s
I apologize for the format, this is my first time posting.
It appears that you just have a problem with units: You've used centimeters per second for velocities, but meters for lengths. To be consistent, you should stick to one unit for length, say centimeters. So the height of the fishing rod is $4.5$ meters, which is $450$ centimeters and the distance is $7.5$ meters, which is $750$ centimeters. The value $6$ looks like it should be $600$, but it cancels out anyway. These changes reduce your final answer by a factor of $100$, so it would match the expected result.