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Irodov 1.9Hardkinematicsrelative-motionriver-crossingoptimization

Minimum Drift River Crossing

Problem 1.9
A boat moves relative to water with a velocity which is n=2.0n = 2.0 times less than the river flow velocity. At what angle to the stream direction must the boat move to minimize the drift?
Answer: θ=120∘\theta = 120^\circ from stream direction
Solution Path
When a boat is slower than the river current (v < u = nv, n = 2), it cannot avoid downstream drift. The minimum drift occurs at angle alpha = arcsin(1/n) = 30 degrees upstream from the perpendicular, giving theta = 120 degrees from the stream direction. The trade-off: more upstream aim reduces drift speed but increases crossing time. The optimal angle balances these effects.
01Problem Restatement
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A boat can move through still water at speed vv, but the river flows at speed u=nvu = nv where n=2.0n = 2.0. This means the river is twice as fast as the boat. The boat needs to cross the river, but because it is slower than the current, it will always be carried downstream to some extent.Given:
- vv = boat speed relative to water
- u=nvu = nv = river flow speed, with n=2.0n = 2.0
- v<uv < u (boat cannot overcome the current)We need to find: the angle θ\theta (measured from the stream direction) at which the boat must move to minimize the drift downstream when crossing the river.
Find θ\theta to minimize drift (boat slower than river)
02Physical Picture and Strategy
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Picture a river flowing from left to right at speed uu. The boat starts at point AA on one bank and needs to reach the opposite bank. Since v<uv < u, the boat can never aim upstream enough to cancel the current completely, so there will always be some drift downstream.We define α\alpha as the angle between the boat's velocity and the perpendicular to the banks (the boat aims upstream at angle α\alpha from the straight-across direction).Cross-river speed: vcos⁡αv\cos\alpha (this gets the boat across)Crossing time: t=dvcos⁡αt = \dfrac{d}{v\cos\alpha}, where dd is the river widthNet downstream speed: u−vsin⁡αu - v\sin\alpha (current minus the upstream component of boat velocity)Strategy: We will write the total drift DD as a function of α\alpha, then minimize it using calculus. The river width dd will factor out, so the optimal angle is independent of how wide the river is.
Drift speed = u−vsin⁡αu - v\sin\alpha, crossing time = d/(vcos⁡α)d/(v\cos\alpha)
03Step 1: Derive the Drift Formula
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The drift distance is the downstream displacement during the crossing. Using displacement = velocity ×\times time:D=(u−vsin⁡α)⋅dvcos⁡αD = (u - v\sin\alpha) \cdot \frac{d}{v\cos\alpha}Substituting u=nvu = nv:D=d(nv−vsin⁡α)vcos⁡α=d⋅v(n−sin⁡α)vcos⁡αD = \frac{d(nv - v\sin\alpha)}{v\cos\alpha} = \frac{d \cdot v(n - \sin\alpha)}{v\cos\alpha}The boat speed vv cancels from numerator and denominator:D=d(ncos⁡α−tan⁡α)=d(nsec⁡α−tan⁡α)D = d\left(\frac{n}{\cos\alpha} - \tan\alpha\right) = d(n\sec\alpha - \tan\alpha)This means: the drift depends only on the ratio n=u/vn = u/v and the aiming angle α\alpha, not on the actual speeds individually. The river width dd is just a scaling factor.
D=d(nsec⁡α−tan⁡α)D = d(n\sec\alpha - \tan\alpha)
04Step 2: Why Drift Can Never Be ZeroKEY INSIGHT
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For the drift to be zero, we would need u−vsin⁡α=0u - v\sin\alpha = 0, which gives sin⁡α=u/v=n=2.0\sin\alpha = u/v = n = 2.0. But sin⁡α\sin\alpha can never exceed 1, so zero drift is impossible when n>1n > 1.Physically: the boat is too slow to fully compensate for the river current. Even if it aimed directly upstream (α=90∘\alpha = 90^\circ), the current would still carry it downstream because u>vu > v.Comparing drift at different angles:- α=0∘\alpha = 0^\circ (straight across): D/d=2.00D/d = 2.00 (all current, no fighting)
- α=15∘\alpha = 15^\circ: D/d=1.81D/d = 1.81
- α=30∘\alpha = 30^\circ: D/d=1.73D/d = 1.73 (minimum!)
- α=45∘\alpha = 45^\circ: D/d=1.83D/d = 1.83 (angling too much reduces crossing speed)This means: there is a sweet spot. Angling upstream a little reduces drift, but angling too much slows the crossing so much that the current has more time to push the boat sideways. The minimum occurs at α=30∘\alpha = 30^\circ.
Drift is never zero when v<uv < u. An optimal angle balances upstream aim against crossing speed.
05Step 3: Minimize Drift with Calculus
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We minimize D=d(nsec⁡α−tan⁡α)D = d(n\sec\alpha - \tan\alpha) by differentiating with respect to α\alpha and setting the derivative to zero.Using the chain rule (derivative of sec⁡α\sec\alpha is sec⁡αtan⁡α\sec\alpha\tan\alpha, derivative of tan⁡α\tan\alpha is sec⁡2α\sec^2\alpha):dDdα=d(nsin⁡αcos⁡2α−1cos⁡2α)=0\frac{dD}{d\alpha} = d\left(\frac{n\sin\alpha}{\cos^2\alpha} - \frac{1}{\cos^2\alpha}\right) = 0Since cos⁡2α≠0\cos^2\alpha \neq 0, we can multiply through:nsin⁡α−1=0n\sin\alpha - 1 = 0sin⁡α=1n=12\sin\alpha = \frac{1}{n} = \frac{1}{2}α=30∘\alpha = 30^\circConverting to the angle from the stream direction: θ=90∘+α=120∘\theta = 90^\circ + \alpha = 120^\circ.This means: the boat should aim 30 degrees upstream from the perpendicular, or equivalently, at 120 degrees from the downstream direction.
sin⁡α=1/n=1/2\sin\alpha = 1/n = 1/2, so α=30∘\alpha = 30^\circ, θ=120∘\theta = 120^\circ
06Verification and Summary
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Verification: At α=30∘\alpha = 30^\circ with n=2n = 2:Dmin⁡=d(2cos⁡30∘−tan⁡30∘)=d(23/2−13)=d(43−13)=3d3=d3D_{\min} = d\left(\frac{2}{\cos 30^\circ} - \tan 30^\circ\right) = d\left(\frac{2}{\sqrt{3}/2} - \frac{1}{\sqrt{3}}\right) = d\left(\frac{4}{\sqrt{3}} - \frac{1}{\sqrt{3}}\right) = \frac{3d}{\sqrt{3}} = d\sqrt{3}We can verify this is a minimum (not maximum) by checking the second derivative or noting that D→∞D \to \infty as α→90∘\alpha \to 90^\circ and D=2dD = 2d at α=0\alpha = 0, while D=d3≈1.73dD = d\sqrt{3} \approx 1.73d at α=30∘\alpha = 30^\circ, which is indeed less. ✓\checkmarkThe formula sin⁡α=1/n\sin\alpha = 1/n is general: for any n>1n > 1, the optimal upstream angle satisfies this relation. ✓\checkmarkSummary: When a boat is slower than the river current (v<uv < u), it cannot avoid drifting downstream. The minimum drift occurs when the boat aims upstream at angle α=arcsin⁡(1/n)\alpha = \arcsin(1/n) from the perpendicular. For n=2n = 2, this gives α=30∘\alpha = 30^\circ, or θ=120∘\theta = 120^\circ from the stream direction. The key insight is the trade-off: angling upstream reduces the drift speed but increases the crossing time.
θ=120∘ from stream direction\boxed{\theta = 120^\circ \text{ from stream direction}}
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