Problem 41 - Calculus Volume 1

Question

For the following exercises, draw the situations and solve the related-rate problems. You are walking to a bus stop at a right-angle corner. You move north at a rate of 2 m/sec and are 20 m south of the intersection. The bus travels west at a rate of 10 m/sec away from the intersection – you have missed the bus! What is the rate at which the angle between you and the bus is changing when you are 20 m south of the intersection and the bus is 10 m west of the intersection?

Step-by-Step Solution

Answer

The angle changes at \(-0.44\) radians per second.

Step 1: Understand the Problem

You are moving north at 2 m/s and the bus is moving west at 10 m/s from a right-angle corner. We need to find the rate at which the angle between you and the bus is changing when you are 20 m south and the bus is 10 m west of the intersection.

Step 2: Draw the Situation

Visualize the scenario as a right-angled triangle. You are 20 m south and the bus is 10 m west of the intersection. The hypotenuse of the triangle is the line of sight between you and the bus. Denote your distance as 'y' and the bus's distance as 'x' with respect to the corner.

Step 3: Define the Known Rates

You are moving north, increasing 'y' at a rate of \(\frac{dy}{dt} = 2\, \text{m/s}\). The bus travels west, increasing 'x' at a rate of \(\frac{dx}{dt} = 10\, \text{m/s}\).

Step 4: Apply Trigonometric Relations

Consider the tangent function, \( \tan(\theta) = \frac{y}{x} \). Differentiate both sides with respect to time \( t \) to relate the angle \( \theta \) to \( x \) and \( y \).

Step 5: Differentiate the Tangent Equation

By the chain rule, we have \( \frac{d}{dt}[\tan(\theta)] = \sec^2(\theta)\frac{d\theta}{dt} = \frac{dx}{dt}\frac{y}{x^2} - \frac{dy}{dt}\frac{1}{x} \). This represents the rate of change of \( \theta \) with respect to time.

Step 6: Substitute Known Values

Substitute \( x = 10 \) m, \( y = 20 \) m, \( \frac{dx}{dt} = -10 \) m/s, \( \frac{dy}{dt} = 2 \) m/s into the differentiated equation: \( \sec^2(\theta)\frac{d\theta}{dt} = \left(-10\right)\frac{20}{10^2} - 2\left(\frac{1}{10}\right) \).

Step 7: Solve for \( \frac{d\theta}{dt} \)

Calculate \( \sec^2(\theta)\frac{d\theta}{dt} = -2 - 0.2\) leading to \(\sec^2(\theta)\) using right triangle relationships \(\sec^2(\theta) = \frac{x^2 + y^2}{x^2} = 5\). Hence, \(5\cdot \frac{d\theta}{dt} = -2.2\). Solving this gives \(\frac{d\theta}{dt} = -0.44\) radians per second.

Key Concepts

Trigonometric Functions

Trigonometric functions are crucial in relating angles to side lengths in triangles. In the context of this problem, we use the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right-angled triangle.
The scenario involving you and the bus forms a right triangle where you are represented by the vertical side (opposite), the bus by the horizontal side (adjacent), and the hypotenuse is the line of sight connecting both of you.
Mathematically, the tangent function is expressed as \( \tan(\theta) = \frac{y}{x} \), where \( \theta \) is the angle between your line of sight and the horizontal direction of the bus. Analyzing this helps us understand how changes in distances affect the angle between the two moving objects.

Differentiation

Differentiation allows us to understand how a function changes as its input changes. It is a fundamental concept in calculus that helps solve related rates problems. In this situation, we need to differentiate the tangent function with respect to time to find how the angle \( \theta \) changes as you and the bus move.
Using the chain rule, we differentiate \( \tan(\theta) \) to get its rate of change:
\[ \frac{d}{dt}[\tan(\theta)] = \sec^2(\theta)\frac{d\theta}{dt} \]
This formula shows that the rate of change of \( \theta \) can be found by multiplying the rate of change of the tangent by \( \sec^2(\theta) \), aiding us in solving the problem.

Right-Angled Triangles

Right-angled triangles are geometric shapes with one angle of 90 degrees. They are instrumental in solving problems involving relationships between different sides and angles through trigonometry.
In this problem, the situation with you and the bus forms a right-angled triangle, with you moving north and the bus moving west.

• The vertical side (y) is the distance you are from the intersection.
• The horizontal side (x) is the distance the bus is from the intersection.
• The hypotenuse represents the line of sight between you and the bus.

Understanding this configuration allows the application of trigonometric functions to relate the different components of the triangle accurately.

Rate of Change

The rate of change helps determine how quickly one quantity changes in relation to another. In this problem, you and the bus are moving at constant speeds in perpendicular directions, and we’re interested in how the angle \( \theta \) changes over time as you drift apart from each other.
The rate of change of \( \theta \), \( \frac{d\theta}{dt} \), is affected by how fast the distances \( x \) and \( y \) change, which in this problem are given by the speeds of you and the bus.
Through differentiation, we plugged in these known rates and distances to solve for \( \frac{d\theta}{dt} \). This process reveals that even with steady linear motions, the angular relationship undergoes dynamic change, resulting in \( \frac{d\theta}{dt} = -0.44 \) radians per second, indicating how quickly the angle is decreasing.