I am having difficulty with this math question in related rates. can anyone help?

Question

A fishing line is reeled in at a rate of 1 foot per second from a bridge 15 feet above the water. At what rate is the angle between the line and the water changing when 25 feet of line is out?

Answer 1

I'm having fun answering your questions...but you're only hurting yourself by not doing them yourself.

The basic equation here is the Pythagorean Theorem. 25 is the hypotenuse, and 15 is a side.

a² + b² = c²
a² + 15² = 25²
a² + 225 = 625
a² = 400
a = 20

Then, the angle between the line and the water is given by the tangent:

tan θ = b/a = baˉ¹

So at that time,

θ = tanˉ¹(15/20) ≈ 36.87°

Differentiating both sides with respect to time, you get:

θ'sec²θ = b(-aˉ²)a' + b'(aˉ¹)

So

θ' = cos²θ[b'/a - ba'/a²]

Plugging in the values you know:

θ' = cos²(36.87°)[b'/20 - 15a'/20²]

Ah, but you still don't know b' and a'....

Let's go back to the Pythagorean Theorem:

a² + b² = c²

Differentiating:

2aa' + 2bb' = 2cc'
aa' + bb' = cc'

We've got a, b, c, and c':

20a' + 15b' = 25(1)

I believe (and this part I'm not 100% sure about) we also know, that the ratio of a' to b' is the same as the ratio of a to b:

a'/b' = 3/4
a' = 3/4 b'

Plugging in:

20(3/4 b') + 15b' = 25
15b' + 15b' = 25
30b' = 25
b' = 25/30 = 5/6

Thus

a' = 3/4b' = (3/4)(5/6) = 5/8

So back to:

θ' = cos²(36.87°)[b'/20 - 15a'/20²]
θ' = cos²(36.87°)[(5/6)/20 - 15(5/8)/20²]

I get:

θ' = 0.0116666°/s

If I didn't screw something up.

Answer 2

Differentiate implicitly. rg tan Q= v² rg(sec² Q)(dQ/dt) = 2v(dv/dt) dQ/dt = 2v(dv/dt)/{rg(sec² Q)} (dQ/dt)/(dv/dt) = 2v/{rg(sec² Q)} Plugging interior the value for g we get (dQ/dt)/(dv/dt) = 2v/{32r(sec² Q)} (dQ/dt)/(dv/dt) = v/{16r(sec² Q)}