Johan14th
  • Johan14th
A fish is being reeled in at the rate of 2 meters/ second (that is the fishing line is being shortened by 2m/s) by a fisher woman at Mill brook. If the fisher is sitting on a dock 30 meters above the water, how fast is the fish moving through the water when the line is 50 meters long? Thank you :)
Calculus1
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SOLVED
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chestercat
  • chestercat
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Johan14th
  • Johan14th
a^2+b^2=c^2 db/dt =? h= 30 b= 50 dc/dt= 2 2a+ 2b (db/dt)= 2c (dc/dt) the answer should be -1.855
Johan14th
  • Johan14th
i cant seem to receive the right answer.
retirEEd
  • retirEEd
I can't either.

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Johan14th
  • Johan14th
Thanks for trying :)
retirEEd
  • retirEEd
Your mixing of variables is confusing me. a^2+b^2=c^2 so b^2 + h^2 = fl^2 I'll use h for a, b is good for the base and fl for the fishing line (hypotenuse db/dt =? This is good h= 30 true enough b= 50 dc/dt= 2 I like fl = 50 so d(fl)/dt = -2 neg getting shorter 2a+ 2b (db/dt)= 2c (dc/dt) b^2 + h^2 = fl^2 respect to dt 2b db/dt + 2h (dh/dt) = 2fl (dc/dt) the 2s cancel b db/dt + h (dh/dt) = fl (dc/dt) db/dt = { fl (dc/dt) - h (dh/dt) } / b again fl = 50 dc/dt = -2 m/s h = 30 dh/dt = 0 since height is not changing b = sqrt ( fl^2 - h^2) or 40 db/dt = ((50)(-2) - 0) / 40 -2.5 is not equal to -1.855 What did you get? I did it another was and got the same answer -2.5. I think the answer is wrong.

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