Introduction
When two objects move away from the same point, their separation depends not only on how far each has travelled but also on the angle between their directions. If the paths are not perpendicular, the right-triangle distance formula does not apply directly. The Law of Cosines gives the needed geometric relationship.
To find how quickly the separation changes, we let the distances and, when necessary, the angle vary with time. Differentiating the geometric relationship then connects their rates. The angle may stay fixed, or it may change and contribute its own term.
Set Up the Distance Relationship
Suppose two objects are observed from a common point. Let \(r(t)\) and \(s(t)\) be their distances from that point, let \(\theta(t)\) be the angle between the two rays from the common point to the objects, and let \(d(t)\) be the distance between the objects. We take \(r\), \(s\), and \(d\) to be nonnegative distances, and measure \(\theta\) in radians.
At each time, the three lengths form a triangle, so the Law of Cosines gives a relation between them. The angle in this relation is the angle between the two rays from the common point, not an angle measured from either object's direction of motion.
All these quantities may depend on time. A dot, as in \(\dot r\), will denote a rate with respect to time; equivalently, \(\dot r=dr/dt\). A positive distance rate means that distance is increasing at the instant considered, while a negative rate means it is decreasing.
Differentiate the Constraint
Differentiate the Law of Cosines with respect to time. The chain rule applies to each time-dependent quantity, including the angle if it changes. In particular, differentiating \(\cos\theta(t)\) gives \(-\sin\theta\,\dot\theta\).
When \(d>0\), divide by \(2d\) to solve for the separation rate:
This formula requires that the quantities are differentiable at the instant in question and that \(d\ne0\). If the angle is fixed, then \(\dot\theta=0\), so the angular term vanishes. If the angle changes, its rate must be included. When using a numerical angle rate, it must be expressed in radians per unit time.
Why the Differentiation Gives the Rate
The Law of Cosines is a constraint: at every instant, the three lengths and the included angle must satisfy the same geometric equation. Differentiating that equation describes how its two sides change together as time passes. The steps below show where each term comes from.
Since \(d=d(t)\), the chain rule gives \(\frac{d}{dt}(d^2)=2d\dot d\).
The terms \(r^2\) and \(s^2\) give \(2r\dot r\) and \(2s\dot s\).
The product \(rs\) contributes \(\dot r s+r\dot s\), and differentiating \(\cos\theta\) contributes \(-\sin\theta\,\dot\theta\). Including the original factor \(-2\) produces both terms in the differentiated relation.
When \(d>0\), divide by \(2d\). The resulting equation relates the instantaneous separation rate to the distances, their rates, and the angle and its rate.
This reasoning also explains why knowing only the two distance rates is not always enough: if the angle varies, the last term depends on how quickly it varies.
Worked Examples
Worked Example: Fixed Angle of 60 Degrees
Two objects are \(3\) km and \(4\) km from a common point. Their included angle is fixed at \(60^\circ\). At the instant considered, the first distance is increasing at \(2\) km/h and the second is decreasing at \(1\) km/h. Find the rate of change of their separation.
First, \(60^\circ=\pi/3\) radians and \(\cos(\pi/3)=1/2\). The Law of Cosines gives
The angle is fixed, so \(\dot\theta=0\). The given rates are \(\dot r=2\) km/h and \(\dot s=-1\) km/h. Substitute into the differentiated constraint:
Since \(d=\sqrt{13}\), divide by \(2\sqrt{13}\):
Check by substitution into the differentiated equation: \(2d\dot d=2\sqrt{13}\left(-1/(2\sqrt{13})\right)=-1\) km²/h, matching the right-hand side. The negative rate means the separation is decreasing at this instant.
Worked Example: A Changing Included Angle
Two objects are \(2\) m and \(3\) m from a common point, and their included angle is \(\pi/3\) radians. At this instant, their distances are increasing at \(1\) m/s and \(2\) m/s, respectively. The angle is increasing at \(\pi/6\) rad/s. Find their separation rate.
Find the current separation first:
Now use the differentiated relation with \(\dot r=1\), \(\dot s=2\), \(\dot\theta=\pi/6\), and \(\sin(\pi/3)=\sqrt3/2\):
Therefore,
Check: substituting this value into the left-hand side gives \(2\sqrt7\,(9+\pi\sqrt3)/(2\sqrt7)=9+\pi\sqrt3\), exactly the value calculated on the right-hand side. The positive result means the separation is increasing.
Worked Example: One Distance Decreases at a Fixed Obtuse Angle
Two objects are \(10\) km and \(6\) km from a common point, with a fixed included angle of \(120^\circ\). The first distance is decreasing at \(1\) km/h, while the second is increasing at \(0.5\) km/h. Find the rate of change of their separation.
Since \(\cos(120^\circ)=-1/2\), the current separation is
The angle is fixed, and the distance rates are \(\dot r=-1\) km/h and \(\dot s=0.5\) km/h. Substituting gives
Thus
Check: \(2d\dot d=2(14)(-15/28)=-15\) km²/h, equal to the differentiated right-hand side. The separation is decreasing, even though one of the objects is moving farther from the common point.
Common Pitfall: Omitting the Angle Rate
A frequent error is to differentiate the Law of Cosines as though the angle were fixed when it is actually changing. In the second worked example, omitting the angle term would give
But the actual angle-rate contribution to \(2d\dot d\) is \(2rs\sin\theta\,\dot\theta=\pi\sqrt3\), which is not zero. The correct rate is \((9+\pi\sqrt3)/(2\sqrt7)\) m/s. The omitted calculation would only be valid if the angle were fixed at that instant, meaning \(\dot\theta=0\).
Another useful check is to keep the signs of the distance rates. “Decreasing at \(1\) km/h” means \(\dot r=-1\) km/h, not \(+1\) km/h. A negative separation rate is not an error: it says the distance between the objects is shrinking.
A Reliable Problem-Solving Sequence
For a distance-rate problem involving two paths meeting at an angle, organize the calculation before substituting the instantaneous values:
- Draw or identify the triangle, label the two distances \(r\) and \(s\), the included angle \(\theta\), and the separation \(d\).
- Write \(d^2=r^2+s^2-2rs\cos\theta\) as a relation that holds over time.
- Differentiate with respect to time, retaining the \(\dot\theta\) term unless the angle is specified to be fixed.
- Substitute the values and signed rates for the instant, including the current value of \(d\).
- Solve for \(\dot d\), attach the appropriate units, and check the result in the differentiated equation.
Keeping the relation symbolic until after differentiation matters because the quantities are linked throughout the motion. Substituting a single instant too early can hide which quantities vary and which derivative terms are needed.
Check Your Understanding
Use the Law of Cosines and its time derivative to answer these questions. For rate calculations, include units and interpret the sign.
- In the distance relation, what does the included angle measure?
- If the angle is fixed, which term in the differentiated equation vanishes, and why?
- Two distances are momentarily constant, but the angle between them is changing. Can the separation still change? Explain using the angle-rate term.
- Why must an angular rate used in the differentiated formula be expressed in radians per unit time?
- In the third worked example, what does the negative value of \(\dot d\) say about the objects' separation?
