Question

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 :)

Answer 1

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

Answer 2

i cant seem to receive the right answer.

Answer 3

I can't either.

Answer 4

Thanks for trying :)

Answer 5

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.