Introduction
A radar antenna tracks a target by changing its pointing direction. Even when the target moves at a steady speed, the antenna does not generally turn at a steady angular rate: the rate depends on where the target is relative to the radar.
We will model a target moving along a straight path that stays a fixed distance from a stationary radar. The target’s position along that path and the antenna’s pointing angle are linked by a right triangle. Writing that geometric constraint before differentiating gives a direct way to find the antenna’s angular rate.
Set Up the Geometry
Place the radar at the origin. Let the target move along the horizontal line \(y=d\), where \(d>0\) is the fixed perpendicular distance between the radar and the target’s path. At time \(t\), let \(x(t)\) be the target’s horizontal position, so its coordinates are \((x(t),d)\).
Measure the antenna’s angle \(\theta(t)\) from the perpendicular to the target’s path, which is the positive \(y\)-direction. The right triangle has adjacent side \(d\) and opposite side \(x(t)\). For a target on the side where \(x\geq 0\), the angle is between \(0\) and \(\pi/2\). We can also use signed \(x\) and a correspondingly signed angle when the target crosses the point closest to the radar.
The quantity \(d\) is constant, while \(x\) and \(\theta\) change with time. This distinction matters when differentiating: the target’s position has a rate \(dx/dt\), but the fixed path distance does not.
Derive the Angular-Rate Formula
Differentiate the tangent constraint with respect to time. The chain rule gives the derivative of \(\tan\theta(t)\), and the derivative of \(x(t)/d\) is \(x'(t)/d\) because \(d\) is constant.
Since \(d>0\) and \(\sec^2\theta>0\), we can isolate the requested angular rate. The identity \(1+\tan^2\theta=\sec^2\theta\), together with \(\tan\theta=x/d\), gives a form involving the target’s position directly.
Here \(\theta\) is measured in radians, so the angular rate is expressed in radians per unit of time. The formula shows that the sign of the antenna’s angular rate is the same as the sign of the target’s horizontal velocity: as \(x\) increases, \(\theta\) increases; as \(x\) decreases, \(\theta\) decreases.
Why the Formula Works
The key is that the angle and the target position are not independent. At every instant, their values must form the right triangle described by \(\tan\theta=x/d\). Differentiation measures how quickly the two quantities must change together to keep that relationship true.
The triangle gives \(\tan\theta=x/d\), with \(d\) fixed.
The chain rule gives \(\sec^2\theta\,d\theta/dt=(1/d)\,dx/dt\).
Divide by \(\sec^2\theta\), or substitute \(\sec^2\theta=(d^2+x^2)/d^2\).
The simplified expression also explains why position matters. When \(|x|\) is large, the denominator \(d^2+x^2\) is large, so the same horizontal speed produces a smaller angular rate. Near the point directly across from the radar, the denominator is smaller and the antenna must turn more quickly for that same speed.
Worked Examples
Worked Example: A Target Moves Toward the Closest Point on Its Path
A target moves along a path \(300\) meters from a stationary radar. At the instant considered, its horizontal position is \(x=400\) meters and its horizontal velocity is \(dx/dt=-20\) meters per second. Find the antenna’s angular rate.
First, the constraint is \(\tan\theta=x/d\). Substituting the position and fixed distance gives
Therefore,
Differentiate the constraint and substitute the known values:
Now solve:
The negative sign means that the pointing angle is decreasing as the target moves toward the point on its path closest to the radar. Check with the simplified formula: \(d^2+x^2=300^2+400^2=90{,}000+160{,}000=250{,}000\), so
As a further check in the differentiated equation, \((25/9)(-0.024)=-0.0666\ldots=-1/15\ \text{s}^{-1}\), which agrees with \((1/300)(-20)=-1/15\ \text{s}^{-1}\).
Worked Example: The Target Crosses the Closest Point
The target moves along a path \(200\) meters from the radar. At the instant it crosses the point closest to the radar, its horizontal position is \(x=0\), and it is moving in the positive horizontal direction at \(30\) meters per second. Find the antenna’s angular rate.
At \(x=0\), the tangent constraint gives \(\tan\theta=0/200=0\), so \(\theta=0\) and \(\sec^2\theta=1\). The differentiated constraint becomes
Thus the antenna is turning in the positive direction at \(0.15\) radians per second. Check with the simplified formula: \(d^2+x^2=200^2+0^2=40{,}000\), and
Substitution into the differentiated constraint also checks: \(\sec^2\theta\,d\theta/dt=1(3/20)=3/20\), while \((1/d)\,dx/dt=30/200=3/20\). The two sides agree.
Worked Example: A Target Farther Along the Path
A target moves along a path \(600\) meters from a stationary radar. At an instant, the target is \(800\) meters along the path from the closest point, and its horizontal velocity is \(-25\) meters per second. Find the antenna’s angular rate.
The constraint gives \(\tan\theta=800/600=4/3\). Hence, \(\sec^2\theta=1+(4/3)^2=25/9\). Using the differentiated constraint,
Therefore,
The target is moving back toward the closest point, so a decreasing angle is consistent with the geometry. For the independent check, \(d^2+x^2=600^2+800^2=360{,}000+640{,}000=1{,}000{,}000\), and
Finally, \((25/9)(-3/200)=-75/1800=-1/24\), matching \((-25)/600=-1/24\) in the differentiated constraint.
Units and Signs
If \(x\) and \(d\) are measured in meters and \(dx/dt\) in meters per second, then \(d/(d^2+x^2)\) has units of inverse meters. Multiplying by meters per second leaves inverse seconds, interpreted here as radians per second. Radians are dimensionless in the equations, but writing \(\text{rad/s}\) makes clear that the result is an angular rate.
The sign comes from the chosen coordinate direction. With positive \(x\) pointing along the target’s path away from the closest point, \(dx/dt>0\) means the target moves farther along that direction and the angle increases. A negative \(dx/dt\) means the angle decreases. If a problem describes the target’s speed but not its direction, the speed alone is not enough to determine the sign; identify the direction of motion first.
Common Pitfall: Differentiating the Wrong Triangle Ratio
In this setup, the angle is measured from the perpendicular to the target’s path. Thus the opposite side is \(x\) and the adjacent side is \(d\), so the correct relationship is \(\tan\theta=x/d\). Reversing the sides would give \(d/x\), which is the tangent of a different angle. It would produce a different derivative and an incorrect rate for the antenna angle defined here.
A second common error is treating the fixed distance \(d\) as if it changes with time. It does not: the target stays on the same straight path, so \(d\) is constant and \(d(d)/dt=0\). The right-hand side differentiates to \((1/d)\,dx/dt\), not to a quotient rule expression with an additional rate for \(d\).
It is also important not to substitute numerical values into the constraint before differentiating if that would remove the variable whose rate is being sought. First differentiate the relation involving \(x(t)\) and \(\theta(t)\); then substitute the values at the stated instant. This preserves the link between the two changing quantities.
Check Your Understanding
Use the constraint for a target moving along a path at fixed distance from a stationary radar. Keep track of the angle’s reference direction and the sign of the target’s motion.
- A target’s path is \(150\) meters from the radar. Write the tangent constraint when \(\theta\) is measured from the perpendicular to the path.
- Starting from the constraint, derive the equation relating \(\sec^2\theta\), \(d\theta/dt\), \(d\), and \(dx/dt\).
- At \(d=300\) meters and \(x=400\) meters, is \(\sec^2\theta\) equal to \(25/9\) or \(25/16\)? Explain using the tangent constraint.
- For \(d=500\) meters, \(x=0\), and \(dx/dt=-10\) meters per second, calculate the signed angular rate.
- Why does the simplified formula predict a smaller angular rate at a large \(|x|\) than near \(x=0\), for the same horizontal speed?
