Why Perpendicular Motion Creates a Changing Separation
When two boats or vehicles move along perpendicular paths, their separation is the hypotenuse of a right triangle. The two legs of that triangle change as the objects move, so the separation changes too. The Pythagorean theorem gives a constraint connecting these quantities; differentiating it with respect to time relates their instantaneous rates.
The signs of the component rates matter. One part of the separation can be shrinking while another is growing. The total distance may therefore increase, decrease, or momentarily stay constant, depending on how the two changes combine.
Set Up Perpendicular Position Components
Choose perpendicular coordinate directions along the paths or across the relevant displacement. Let \(x(t)\) and \(y(t)\) be the signed components of the displacement from one object to the other. Their values and rates must use the same coordinate directions throughout the calculation. The distance \(D(t)\) between the objects is nonnegative.
Equivalently, the Pythagorean constraint is \(D^2=x^2+y^2\). Using signed components is helpful even when an object passes a chosen origin: the sign records which side of that origin it is on. If instead \(x\) and \(y\) represent positive gaps that remain in the same quadrant, they can be treated as those gaps and their rates must be signed according to whether each gap is growing or shrinking.
Deriving the Separation-Rate Formula
Differentiate the Pythagorean constraint with respect to time. Each position component depends on time, so the chain rule applies to all three squared quantities.
When \(D>0\), divide by \(D\) to obtain the formula for the rate of separation.
The numerator combines the contribution of each changing component. If a component is positive and shrinking, its contribution to \(xx'+yy'\) is negative. A growing positive component contributes positively. Thus, it is not enough to notice that one object is moving quickly: the direction of each component change matters.
Worked Example: Two Boats Depart on Perpendicular Courses
Worked Example: Two Boats Depart on Perpendicular Courses
Two boats leave the same point at the same time. One travels east at \(3\) kilometers per hour; the other travels north at \(4\) kilometers per hour. Find their separation and the rate at which it is changing \(2\) hours after departure.
Let \(x(t)=3t\) kilometers and \(y(t)=4t\) kilometers, where \(t\) is measured in hours. Their component rates are \(x'=3\) kilometers per hour and \(y'=4\) kilometers per hour.
At \(t=2\) hours, the positions along the two perpendicular directions are
The distance is
Apply the separation-rate formula:
Check the result by differentiating the full distance function. Since \(D(t)=\sqrt{(3t)^2+(4t)^2}=\sqrt{25t^2}=5t\) for \(t\geq 0\), it follows that \(D'(t)=5\) kilometers per hour. This agrees with the rate computed at \(t=2\).
Signs Determine Whether the Distance Grows
The formula uses signed component rates, not just their magnitudes. A positive rate does not automatically mean that each component is increasing; its sign depends on the chosen coordinates. In a setup where both components are positive gaps, a negative component rate means that gap is shrinking.
Both contributions are added before dividing by \(D\). As a result, one component can be closing while the other opens, and the total separation rate reflects their combined effect.
Worked Example: One Gap Shrinks While the Other Grows
Worked Example: One Gap Shrinks While the Other Grows
At a particular instant, two vehicles have perpendicular separation components of \(300\) meters and \(400\) meters. The first gap is shrinking at \(6\) meters per second, while the second is growing at \(8\) meters per second. Find the instantaneous rate of change of their separation.
Use positive gaps for the two components. At the instant in question, \(x=300\) meters, \(y=400\) meters, \(x'=-6\) meters per second, and \(y'=8\) meters per second. First find the distance:
Then calculate the separation rate:
The positive result means that the vehicles are moving farther apart at \(2.8\) meters per second at this instant, even though the first gap is shrinking. Check the calculation in the undivided equation \(DD'=xx'+yy'\): the left side is \((500)(2.8)=1400\), and the right side is \((300)(-6)+(400)(8)=-1800+3200=1400\). The two sides agree.
Worked Example: Two Vehicles Approach Along Perpendicular Directions
Worked Example: Two Vehicles Approach Along Perpendicular Directions
Two vehicles move along perpendicular paths. Let their separation components, in meters, be \(x(t)=600-30t\) and \(y(t)=800-20t\), where \(t\) is measured in seconds. Find their separation rate at \(t=10\) seconds.
The component rates are constant:
At \(t=10\) seconds, the component values are
The distance at this instant is
Now substitute into the rate formula:
To check, multiply this rate by the distance: \((300\sqrt{5})(-14\sqrt{5})=-4200(5)=-21000\text{ m}^2/\text{s}\). The other side of the differentiated constraint is \(xx'+yy'=(300)(-30)+(600)(-20)=-9000-12000=-21000\text{ m}^2/\text{s}\), as required. The negative rate means the separation is decreasing.
A Common Mistake: Adding Component Rates
A tempting but incorrect approach is to add \(x'\) and \(y'\) and call the result the rate of separation. The distance is not the sum \(x+y\); it is the hypotenuse \(\sqrt{x^2+y^2}\). For the \(300\)-meter and \(400\)-meter gaps in the previous example, the incorrect calculation would be
That happens to be positive, but it is not the separation rate. The correct calculation weights each component rate by its current component and divides by the current distance:
The numerical difference illustrates why the Pythagorean constraint must be differentiated rather than replaced with an informal addition rule. Always identify the geometric relationship first, then differentiate that relationship.
What the Formula Tells You
The numerator \(xx'+yy'\) determines the direction of change whenever \(D>0\). If it is positive, \(D'>0\) and the objects are separating. If it is negative, \(D'<0\) and they are approaching. If it is zero, \(D'=0\): their separation is momentarily neither increasing nor decreasing, even though one or both component rates may be nonzero.
Units provide another useful check. Each product \(xx'\) or \(yy'\) has units of distance squared per time. Dividing by \(D\), which has units of distance, gives distance per time, as a separation rate should.
Check Your Understanding
For each question, keep track of the coordinate directions and the units.
- Two boats travel east and north from the same point at \(5\) kilometers per hour and \(12\) kilometers per hour. What is their separation rate after \(1\) hour?
- At an instant, perpendicular gaps are \(5\) meters and \(12\) meters. The first is growing at \(2\) meters per second and the second is shrinking at \(1\) meter per second. Find the separation rate.
- Why must a shrinking positive gap have a negative rate in the formula?
- What information does the sign of \(D'\) give about the motion?
- Why can the formula \(D'=(xx'+yy')/D\) not be used when \(D=0\)?
