Question

A television camera at ground level is filming the lift-off of a space shuttle at a point 750 meters from the launch pad. Let \(\theta\) be the angle of elevation of the shuttle and let \(s\) be the distance between the camera and the shuttle (as shown in the figure). Write \(\theta\) as a function of \(s\) for the period of time when the shuttle is moving vertically. Differentiate the result to find \(d \theta / d t\) in terms of \(s\) and \(d s / d t\).

Short answer

\(\frac{d\theta}{dt} = \frac{1}{1 + (s/750)^2} \cdot \frac{ds}{dt}\)

Step-by-step solution

Setting up the Equation

This problem involves a right triangle formed by the camera on the ground, the shuttle in the air at height \(s\), and the point on the ground below the shuttle (forming a length of 750m). According to the trigonometric principle, the tangent of the angle \(\theta\) is the ratio of the height of the shuttle to the distance between the launch pad and the camera. Therefore, we have:\[tan(\theta) = \frac{s}{750}\]However, we need to express \(\theta\) as a function of \(s\). This requires finding the inverse tangent (also known as the arctangent) on both sides to solve for \(\theta\):\[\theta = arctan \left(\frac{s}{750}\right). \]

Differentiating the Equation

In this step, we differentiate \(\theta\) with respect to \(t\). Using the chain rule, we get:\[\frac{d\theta}{dt} = \frac{d\theta}{ds} \cdot \frac{ds}{dt}.\]The derivative of \(arctan(x)\) is \(1 / (1 + x^2)\). So, \[\frac{d\theta}{ds} = \frac{1}{1 + (s/750)^2}.\]Substituting this into the chain rule expression, we get:\[\frac{d\theta}{dt} = \frac{1}{1 + (s/750)^2} \cdot \frac{ds}{dt},\]which is the expression for the rate of change of the angle of elevation in terms of \(s\) and \(ds/dt\).

Key concepts

Angle of Elevation

The term 'angle of elevation' refers to the angle between the horizontal plane and the line of sight from an observer's eye to an object above the horizontal plane. Imagine you are standing on the ground and you look up to see a bird in the sky; the angle your line of sight makes with the ground is the angle of elevation.

In the context of the given exercise, a television camera is observing the lift-off of a space shuttle. As the shuttle ascends, the angle of elevation, denoted by \(\theta\), changes. This angle is crucial in trigonometry and navigation, as it helps determine an object's altitude or positioning with respect to the observer's point. Calculating the angle of elevation involves understanding right-angled triangles and the trigonometric function tangent, which correlates to the ratio of the opposite side to the adjacent side in a right-angled triangle.

Trigonometric Functions

Trigonometric functions are mathematical functions that relate the angles of a triangle to the lengths of its sides. They are fundamental in the study of geometry, waves, oscillations, and many areas of science and engineering.

These functions are sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot). In a right-angled triangle, for a given angle \(\theta\), the sine function represents the ratio of the length of the side opposite to \(\theta\) to the length of the hypotenuse. Similarly, cosine gives the ratio of the adjacent side to the hypotenuse, and tangent is the ratio of the opposite to the adjacent side.

In our exercise, we use the tangent function, defined as \(tan(\theta) = \frac{opposite}{adjacent}\), where 'opposite' is the vertical distance from the camera to the shuttle and 'adjacent' is the horizontal distance from the camera to the launch pad.

Differentiation

Differentiation is a central concept in calculus that deals with rates of change. It provides a mathematical way of finding the rate at which a quantity changes instantaneously. In real-world applications, differentiation helps us understand motion, optimization, and changing systems.

When we differentiate a function, we are finding its derivative - a new function that gives the slope of the tangent line to the original function at any point. The derivative represents how the function's value changes as its input changes. With the trigonometric functions, differentiation can be a bit more complex, as these functions behave differently from linear or polynomial functions.

In the exercise, we differentiate the angle of elevation with respect to time, which gives us the rate of change of the angle as the shuttle ascends. This requires using the chain rule, a method in calculus where we take the derivative of the outside function and multiply it by the derivative of the inside function.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find an angle when given the ratio of sides of a right-angled triangle. They are the inverses of the basic trigonometric functions and include arcsine (\(arcsin\)), arccosine (\(arccos\)), and arctangent (\(arctan\)), among others.

These functions are particularly useful in situations where we have the lengths of sides of a triangle and need to find the measure of an angle. For instance, if we know the tangent of an angle, we can use arctangent to find the actual angle.

In our exercise scenario, the arctangent is used to express the angle of elevation \(\theta\) as a function of the shuttle's distance \(s\). We then differentiated this function with respect to time to find the rate at which the angle of elevation changes as the shuttle moves away from the camera. Understanding the application of inverse trigonometric functions is essential for solving many problems in trigonometry and calculus.