Introduction
A rope pulled in toward a dock does not generally shorten at the same rate that the boat moves horizontally. If the rope runs from the boat to a point on the dock that is above the boat’s attachment point, the rope is the hypotenuse of a right triangle. As the boat approaches the dock, the horizontal side shortens, but the diagonal rope changes at a different rate.
The geometry gives a constraint between the rope length and the boat’s horizontal distance. Differentiating that constraint relates their instantaneous rates of change. The resulting factor depends on the boat’s current position, so the boat’s speed must be calculated using the geometry at the instant in question.
Set Up the Geometry
Assume the boat is connected to a fixed point on the dock by a straight, taut rope. Let \(x(t)\) be the boat’s horizontal distance from the dock, let \(h\) be the fixed vertical separation between the boat’s rope attachment point and the dock’s attachment point, and let \(L(t)\) be the rope’s length. The horizontal and vertical separations are perpendicular, so the rope is the hypotenuse of a right triangle.
Here \(x\), \(h\), and \(L\) are lengths measured in the same units. We take \(x>0\) while the boat is some positive horizontal distance from the dock, and \(h\geq 0\) is constant. The relation assumes the rope remains straight and taut, the dock attachment point stays fixed, and the boat moves along a horizontal line.
Likewise, if the rope is shortening, then \(\frac{dL}{dt}<0\). A statement that the rope is being pulled in at \(r\) units per unit time means \(\frac{dL}{dt}=-r\), not \(+r\). Keeping signed rates distinct from positive speed magnitudes prevents a common sign error.
Differentiate the Constraint
The Pythagorean relation describes the geometry at every instant. To relate the rates, differentiate both sides with respect to time, treating \(x\) and \(L\) as functions of time and \(h\) as a constant.
When \(x>0\), divide by \(2x\) to solve for the boat’s signed horizontal velocity:
The factor \(\frac{L}{x}\) is positive and greater than or equal to \(1\), because \(L^2=x^2+h^2\). Thus, when the rope shortens, the boat’s horizontal velocity is negative as well. The boat’s speed toward the dock is the positive value obtained by reversing the sign.
Why the Rate Equation Works
The differentiated equation follows directly from the geometric constraint, but its meaning is useful to make explicit. A small change in rope length does not have to match the horizontal change: the rope points along the hypotenuse, while the boat moves along the horizontal leg. The ratio of the corresponding instantaneous changes is determined by the triangle’s proportions.
Starting from \(L^2=x^2+h^2\), the difference in the squared lengths between two nearby instants satisfies the same constraint. In the instantaneous limit, differentiation gives \(2L\frac{dL}{dt}=2x\frac{dx}{dt}\), because the height does not change. Solving this equality yields the rate factor \(\frac{L}{x}\). Since the hypotenuse \(L\) is longer than the horizontal leg \(x\) whenever \(h>0\), the boat’s horizontal speed is then greater than the rope-shortening speed.
This is not a claim that the boat always moves faster than the rope shortens: if \(h=0\), then \(L=x\), and the rates have equal magnitudes. The exact relationship depends on the triangle at the instant being studied.
Worked Example: Find the Boat’s Speed from the Rope Rate
Worked Example: Find the Boat’s Speed from the Rope Rate
A boat is connected to a dock point \(12\) feet above its rope attachment point. At a particular instant, the boat is \(5\) feet horizontally from the dock, and the rope is being pulled in at \(2\) feet per second. Find the boat’s velocity and speed toward the dock.
Here \(h=12\) feet and \(x=5\) feet. First find the rope length using the Pythagorean relation:
The rope is shortening, so its signed rate is \(\frac{dL}{dt}=-2\) ft/s. Substitute the current values into the rate equation:
The negative sign means the horizontal distance is decreasing. Therefore the boat’s speed toward the dock is \(5.2\) ft/s.
Check by substituting the rates into the differentiated constraint. The left-hand side is \(2L\frac{dL}{dt}=2(13)(-2)=-52\text{ ft}^2/\text{s}\). The right-hand side is \(2x\frac{dx}{dt}=2(5)(-26/5)=-52\text{ ft}^2/\text{s}\). The two sides agree.
Worked Example: Use a Different Triangle Position
Worked Example: Use a Different Triangle Position
At another instant, a boat is \(24\) metres horizontally from the dock. The fixed vertical separation is \(7\) metres, and the rope is shortening at \(1.5\) m/s. Find the boat’s speed toward the dock.
First calculate the rope length:
Use the signed rope rate \(\frac{dL}{dt}=-1.5=-\frac{3}{2}\) m/s:
The boat moves toward the dock at \(1.5625\) m/s. The final decimal is exact here because \(\frac{25}{16}=1.5625\).
For a rate check, \(2L\frac{dL}{dt}=2(25)(-3/2)=-75\text{ m}^2/\text{s}\). Also, \(2x\frac{dx}{dt}=2(24)(-25/16)=-75\text{ m}^2/\text{s}\). The differentiated constraint is satisfied.
Worked Example: Recover the Rope Rate from the Boat’s Motion
Worked Example: Recover the Rope Rate from the Boat’s Motion
Suppose the horizontal separation is \(20\) feet, the fixed vertical separation is \(15\) feet, and the boat moves toward the dock at \(4\) ft/s. How quickly is the rope shortening at that instant?
Since the boat is moving toward the dock, \(\frac{dx}{dt}=-4\) ft/s. First find the rope length:
Rearrange the differentiated constraint to solve for the rope rate:
The rope is shortening at \(3.2\) ft/s. Check the differentiated relation: \(2L\frac{dL}{dt}=2(25)(-16/5)=-160\text{ ft}^2/\text{s}\), while \(2x\frac{dx}{dt}=2(20)(-4)=-160\text{ ft}^2/\text{s}\). The rates are consistent.
A Common Mistake: Equating the Two Speeds
It is tempting to assume that pulling in \(2\) feet of rope per second makes the boat move horizontally at \(2\) ft/s. That is generally incorrect because the rope is diagonal. For example, suppose \(x=12\) feet and \(h=5\) feet. Then \(L=13\) feet. If the rope shortens at \(2\) ft/s, the incorrect calculation would be \(\frac{dx}{dt}=-2\) ft/s. Substituting that proposed rate into the differentiated constraint gives \(2L\frac{dL}{dt}=2(13)(-2)=-52\), but \(2x\frac{dx}{dt}=2(12)(-2)=-48\). The two sides do not match, so the proposed rate is wrong.
The correct calculation uses the geometric factor:
Now \(2L\frac{dL}{dt}=2(13)(-2)=-52\text{ ft}^2/\text{s}\), and \(2x\frac{dx}{dt}=2(12)(-13/6)=-52\text{ ft}^2/\text{s}\), so the corrected rate satisfies the equation. The boat’s speed is \(\frac{13}{6}\) ft/s, greater than the rope-shortening speed in this configuration.
A Reliable Procedure
For a boat connected to a fixed dock point by a taut rope, the calculation can be organized into a few steps. The order matters: the geometry determines the current rope length before the rate equation can be evaluated.
Let \(x\) be the horizontal separation, \(L\) the rope length, and \(h\) the constant vertical separation.
Use \(L^2=x^2+h^2\), because the rope and the two perpendicular separations form a right triangle.
Obtain \(2L\frac{dL}{dt}=2x\frac{dx}{dt}\). The fixed height contributes no rate.
If the rope is shortening at a stated positive rate, enter its derivative with a minus sign.
Use the sign to identify the direction, report speed as a positive magnitude, and verify the differentiated equation and units.
Conditions and Interpretation
The formula \(\frac{dx}{dt}=\frac{L}{x}\frac{dL}{dt}\) requires \(x>0\) so that division by \(x\) is valid. At the instant \(x=0\), the boat is directly below the dock attachment point, and this formula cannot be used to solve for \(\frac{dx}{dt}\) by division. The original differentiated equation, \(2L\frac{dL}{dt}=2x\frac{dx}{dt}\), remains the relevant relation, but a separate analysis of the instant may be needed.
Also, this model applies only while the rope is taut and the vertical separation stays fixed. If the rope is slack, its length is not necessarily the straight-line distance between the boat and dock point, so the right-triangle constraint does not describe the rope. If the dock point moves or the boat’s attachment height changes, then the corresponding changing quantity must also be included when differentiating.
Units offer a quick check. In \(\frac{L}{x}\frac{dL}{dt}\), the ratio \(\frac{L}{x}\) has no units, so the result has the same units as \(\frac{dL}{dt}\), namely length per time. The differentiated equation has units of length squared per time on both sides. A result with units of length, for instance, would signal an error.
The important idea is that the rope’s shortening rate and the boat’s horizontal velocity are linked by geometry, not assumed to be equal. Find the current triangle, use its side lengths in the rate equation, and then interpret the sign of the result.
Check Your Understanding
Use the right-triangle model, and keep signed rates separate from positive speeds.
- A boat is \(8\) metres from the dock horizontally, with a fixed vertical separation of \(15\) metres. What is the rope length?
- In the same position, the rope shortens at \(2\) m/s. Find \(\frac{dx}{dt}\) and state the boat’s speed toward the dock.
- Why is the signed rope rate negative when the rope is being pulled in?
- For what values of \(x\) can you use \(\frac{dx}{dt}=\frac{L}{x}\frac{dL}{dt}\) directly, and why?
- If \(h=0\), how do the magnitudes of the boat’s horizontal velocity and the rope’s rate of change compare?
