Modeling an Angle of Elevation
An angle of elevation changes when an object’s position changes relative to an observer. To find how fast that angle changes, first describe the situation with a right triangle. The key measurements are the horizontal distance from the observer to the point below the object and the object’s vertical separation above the observer.
For example, if an aircraft flies at a fixed altitude while approaching an observer, its horizontal distance decreases and its angle of elevation increases. A balloon directly above a fixed point can also change the angle: as it rises, its vertical separation increases, so the angle increases. Both situations can be analyzed using tangent and differentiation with respect to time.
The variables \(x\) and \(h\) may depend on time. Their rates, \(\frac{dx}{dt}\) and \(\frac{dh}{dt}\), describe signed changes in distance. For instance, if an object approaches while staying at the same height, then \(\frac{dx}{dt}<0\) and \(\frac{dh}{dt}=0\). If a balloon rises above a fixed horizontal position, then \(\frac{dx}{dt}=0\) and \(\frac{dh}{dt}>0\).
Deriving the Angular-Rate Formula
Start with the tangent relationship and differentiate both sides with respect to time. The chain rule applies because the angle \(\theta\) changes with time. The derivative of tangent, as used here, requires \(\theta\) to be measured in radians.
Differentiating gives
Because \(\tan\theta=\frac{h}{x}\), the identity \(\sec^2\theta=1+\tan^2\theta\) gives \(\sec^2\theta=\frac{x^2+h^2}{x^2}\). Substitute this expression into the differentiated equation and multiply both sides by \(x^2\). Since \(x>0\), the denominator \(x^2+h^2\) is positive, so we can solve for the angular rate.
This formula includes both vertical and horizontal motion. If the object stays at a fixed height, then \(\frac{dh}{dt}=0\), so
If the object remains above a fixed horizontal point, then \(\frac{dx}{dt}=0\), so
These special cases also make the signs clear. An approaching object has \(\frac{dx}{dt}<0\), making the angle increase when its height is fixed. A rising object has \(\frac{dh}{dt}>0\), making the angle increase when its horizontal distance is fixed.
Worked Examples
Worked Example: An Approaching Aircraft at Fixed Altitude
An aircraft is \(3\) km above an observer’s level ground and approaches horizontally at \(240\) km/h. Find the rate of change of its angle of elevation when it is \(4\) km horizontally from the observer.
Set \(h=3\) km, \(x=4\) km, \(\frac{dh}{dt}=0\) km/h, and \(\frac{dx}{dt}=-240\) km/h. The negative sign indicates that the horizontal distance is decreasing.
Use the general angular-rate formula:
Substitute the values, retaining the units:
The angle of elevation is increasing at \(28.8\) radians per hour at that instant. Check using \(\theta=\arctan(3/x)\): differentiating gives \(\frac{d\theta}{dt}=\frac{-3}{x^2+9}\frac{dx}{dt}\). At \(x=4\) and \(\frac{dx}{dt}=-240\), this is \(\frac{-3(-240)}{16+9}=28.8\) rad/h, as above.
Worked Example: A Rising Balloon Above a Fixed Point
An observer is \(120\) m horizontally from the point directly below a balloon. At a particular instant, the balloon is \(50\) m above the observer’s level and rising at \(4\) m/s. Find the rate of change of the angle of elevation at that instant.
The horizontal distance is constant, so \(x=120\) m and \(\frac{dx}{dt}=0\) m/s. The vertical separation is \(h=50\) m, and \(\frac{dh}{dt}=4\) m/s. Substitute into the formula:
The angle is increasing at \(\frac{24}{845}\) rad/s. Check directly from \(\theta=\arctan(h/120)\). Differentiating gives \(\frac{d\theta}{dt}=\frac{120}{120^2+h^2}\frac{dh}{dt}\). At \(h=50\) m and \(\frac{dh}{dt}=4\) m/s, this gives \(\frac{120(4)}{14400+2500}=\frac{24}{845}\) rad/s, matching the result.
Worked Example: An Aircraft’s Angle and Altitude Determine Its Distance
An aircraft flies horizontally at a constant height of \(2\) km above an observer. When its angle of elevation is \(30^\circ\), it is approaching at \(180\) km/h. Find the instantaneous rate of change of the angle.
First convert the angle to radians: \(30^\circ=\frac{\pi}{6}\). Since \(\tan\theta=\frac{h}{x}\), the horizontal distance at that instant is
The aircraft’s height is constant, so \(\frac{dh}{dt}=0\), and its approach gives \(\frac{dx}{dt}=-180\) km/h. Therefore,
The angular rate is positive because the aircraft is approaching, so the angle is increasing. Check using \(\frac{d\theta}{dt}=-\frac{h}{x^2+h^2}\frac{dx}{dt}\): substituting \(h=2\), \(x^2=12\), and \(\frac{dx}{dt}=-180\) gives \(-\frac{2(-180)}{12+4}=22.5\) rad/h.
Signs, Units, and a Common Mistake
Angular rates are signed. A positive value means the angle of elevation is increasing; a negative value means it is decreasing. Describe whether each distance is increasing or decreasing before assigning its rate. In particular, “approaching at \(240\) km/h” means \(\frac{dx}{dt}=-240\) km/h if \(x\) is the distance to the observer. The speed is positive, but the derivative of the decreasing distance is negative.
A common error is to use the positive speed as the derivative of the distance. For the aircraft in the first example, doing so would give
This answer incorrectly says the angle is decreasing. The correction is to use the signed distance rate, \(\frac{dx}{dt}=-240\) km/h:
There is also a unit condition: the derivative formula gives radians per unit of time because the derivative of \(\tan\theta\) has the form \(\sec^2\theta\) only when the angle is measured in radians. If a problem requests degrees per unit of time, first find the rate in radians per unit of time, then multiply by \(\frac{180}{\pi}\) degrees per radian.
Why the Relationship Is Useful
The tangent model is useful whenever a changing angle is determined by a right triangle whose horizontal and vertical dimensions can be measured or described. It converts a geometric question about sightlines into a rate equation involving distances and their rates. The formula also clarifies which motion matters: a change in vertical separation contributes through \(x\frac{dh}{dt}\), while a change in horizontal distance contributes through \(-h\frac{dx}{dt}\).
Before calculating, draw or imagine the right triangle, identify \(x\) and \(h\), and determine the signs of their rates. Then use the angle and distances at the specified instant. The rate equation describes how the quantities change at that instant; the distances themselves are not rates, and a speed should not be substituted for a signed derivative without considering direction.
Check Your Understanding
For each question, identify the signed rates before calculating.
- An aircraft remains \(4\) km above an observer and approaches horizontally. If its horizontal distance is \(3\) km and \(\frac{dx}{dt}=-120\) km/h, what is \(\frac{d\theta}{dt}\)?
- A balloon rises above a point \(80\) m from an observer. At an instant when it is \(60\) m high, it rises at \(3\) m/s. What is the angular rate?
- In the general formula, what does the sign of \(\frac{dx}{dt}\) indicate when \(x\) is the horizontal distance to an approaching aircraft?
- Why must the angle be measured in radians when differentiating \(\tan\theta\) with respect to time?
- If an object moves both horizontally and vertically, which two terms in the numerator account for those changes?
