calculus question

Question

A fish is being reeled in at a rate of 30cm/s from a bridge 4m above water. At what rate is the angle (in rad/s) between the line and the water changing when there is 8m of line out?

Answer 1

It looks like you have a right triangle with a constant height of 4, and a hypotenuse (the fishing line) shrinking at a rate of 0.30 m/s.

 

θ = angle between hypotenuse and water

r = length of hypotenuse

 

sinθ = 4 / r

 

Take d/dt on both sides.

 

cosθ dθ/dt = (-4/r²) dr/dt

 

dθ/dt = (1/cosθ)(-4/r²) dr/dt

 

dr/dt = -0.30 m/s  (negative because r is shrinking)

 

When the line is 8m out,

 

r = 8m

cosθ = √(1 - sin²θ) = √[1 - (4/r)²] = √[1 - (4/8)²] ≅ 0.866

 

dθ/dt = (1/0.866) [-4/(8²)] (-0.30) rad/sec = 0.0217 rad/sec

 

Note it makes sense to get a positive rate for θ, since it should increase to π/2 as the line is pulled in.

Answer 2

Yes, I agree. Arturo is correct. I'll second that! Arturo, I got your back!


It is a right triangle with constant height 4 m.
Let angle Theta = t and H be the hypotneuse which
in fact is decreasing at a rate of 0.3 m / s.
So dH/dt = -0.3

sin T = 4/H ---> h sin T = 4

Implicit differentiation:
h cos T dT/dt + dH/dt * sin T = 0

Now when h=8 then sin T = 4/8 = 1/2
This is a well known angle of 30 degrees or pi/6.


Substituting this and the given info. into the first derivative:
8 * cos(pi/6) * dT/dt + (-0.3)*sin(pi/6) = 0

Solving for dT/dt : dT/dt = 0.3*sin(pi/6) divided by 8 * cos(pi/6)

= 0.3 * 1/2 divided by 8 * rad3/2

= 0.15 divided by 4 * rad3

= 0.15 * rad3 divided by 12 <--- rationalizing the denominator

= 0.02165 radians per second
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Answer 3

It looks like you have a right triangle with a constant height of 4, and a hypotenuse (the fishing line) shrinking at a rate of 0.30 m/s.

θ = angle between hypotenuse and water
r = length of hypotenuse

sinθ = 4/r

Take d/dt on both sides.

cosθ dθ/dt = (-4/r²) dr/dt

dθ/dt = (1/cosθ)(-4/r²) dr/dt

dr/dt = -0.30 m/s  (negative because r is shrinking)

When the line is 8m out,

r = 8m
cosθ = √(1 - sin²θ) = √[1 - (4/r)²] = √[1 - (4/8)²] ≅ 0.866

dθ/dt = (1/0.866) [-4/(8²)] (-0.30) rad/sec = 0.0217 rad/sec

Note it makes sense to get a positive rate for θ, since it should increase to π/2 as the line is pulled in.