- #1

JJRKnights

- 53

- 0

What must the swimmer's speed, vs, be in order that he drifts only 1 mile downstream as he crosses the river?

The width of the river = 2 miles

The velocity of the river, vr = 3(1-x^2)

Relevant equations:

tanα=vr/vs

dy/dx = tanα

vr = v0(1-x^2/a^2)

Image:

So I started off with:

tanα=vr/vs

plugged in dy/dx for tanα, and vr = 3(1-x^2)

dy/dx = (3/vs)(1-x^2)

I assumed that vs was a constant.

∫dy/dx = (3/vs)∫(1-x^2)dx

y = (3/vs)(x - 1/3 x^3) + C

plugged in initial values y(-1) = 0,

0 = (3/vs)((-1) - 1/3(-1)^3) + C

((-1) - 1/3(-1)^3) = -2/3

0 = (3/vs)(-2/3) + C

C = 2/vs

now I have the equation all with 1 unknown:

y = (3/vs)(x - 1/3 x^3) + 2/vs

here is the part where I just improvised not really knowing what to do...

I figured that if I used the initial values, I wouldn't get my vs, so I used the final values:

y(1)=1

1 = (3/vs)(1 - 1/3) + 2/vs

and for this one I get that:

vs = 4 mi/h