Problem 36 - University Calculus: Early Transcendentals

Question

A particle moves along the parabola \(y=x^{2}\) in the first quadrant in such a way that its \(x\) -coordinate (measured in meters) increases at a steady \(10 \mathrm{m} / \mathrm{sec} .\) How fast is the angle of inclination \(\theta\) of the line joining the particle to the origin changing when \(x=3 \mathrm{m} ?\)

Step-by-Step Solution

Answer

The angle \( \theta \) changes at \( 0.1 \, \text{rad/s} \).

Step 1: Express the Problem Parameters

Given that the particle moves along the parabola \( y = x^2 \), the \( x \)-coordinate is increasing at \( 10 m/s \). We are asked to find out at what rate the angle of inclination \( \theta \) of the line joining the particle to the origin is changing when \( x = 3 \, m \).

Step 2: Setup the Relation Between \( \, \theta \, \) and \( \, x \, \)

We can express the angle \( \theta \) in terms of \( x \) and \( y \) using the relation:\[ \tan(\theta) = \frac{y}{x} = \frac{x^2}{x} = x \]So, \( \theta = \tan^{-1}(x) \).

Step 3: Differentiate with Respect to Time

To find the rate at which \( \theta \) changes with time, differentiate \( \theta = \tan^{-1}(x) \) with respect to time \( t \):\[ \frac{d\theta}{dt} = \frac{d}{dt} \left( \tan^{-1}(x) \right) = \frac{1}{1 + x^2} \cdot \frac{dx}{dt} \]

Step 4: Substitute Given Values

Substitute \( \frac{dx}{dt} = 10 \) and \( x = 3 \) into the differentiated equation:\[ \frac{d\theta}{dt} = \frac{1}{1 + 3^2} \cdot 10 = \frac{1}{10} \cdot 10 = 0.1 \, \text{rad/s} \]

Step 5: Conclusion

The rate at which the angle \( \theta \) is changing when \( x = 3 \) is \( 0.1 \) radians per second.

Key Concepts

Related Rates

The concept of related rates involves finding the rate at which one quantity changes with respect to another related quantity. This is a common type of problem in calculus. In the given exercise, we explore how the rate of change of a particle's horizontal position along a parabola affects the rate of change of an angle.

• The problem begins by outlining that the particle's location is defined by its position on a parabola, a common type of curve.
• The task is to understand how changes in the position ( x-coordinate) relate to the change in the angle of inclination from the origin.

By linking the change in the horizontal position with the change in the angle, we can use calculus to solve the problem efficiently. The key here is to differentiate the relation between the angle and the position with respect to time.

Parabolic Motion

Parabolic motion refers to the path traced by an object moving in a plane under constant gravity, such as a particle following a parabolic trajectory. It is characterized by a quadratic relationship, like the equation of a parabola: \(y = x^2\).In the exercise, the particle is moving along this specific parabola:

• The trajectory of this particle is entirely governed by the quadratic function, making it a classic example of parabolic motion.
• In this scenario, knowing the path allows us to relate the position along the path (x and y coordinates) to other rates of change.

This quadratic relationship is significant when applying calculus to understand how different aspects of motion, such as velocity and angle inclination, evolve over time.

Trigonometric Functions

Trigonometric functions often describe relationships between angles and sides in geometric contexts. In this problem, the function \( \tan(\theta) \) relates the angle \(\theta\) to the coordinates \(x\) and \(y\) of the particle along the parabola.

• The expression \( \tan(\theta) = \frac{y}{x} \) is used to calculate the angle based on the position.
• To isolate \(\theta\), we utilize the inverse function, resulting in \(\theta = \tan^{-1}(x)\).

This relationship is pivotal as it converts the problem from a spatial context to a form where calculus operations like differentiation can be applied to find the rate of change of \(\theta\). Trigonometric functions help in expressing and solving problems involving angles and their manipulation via calculus.

Derivatives

Derivatives represent the rate of change of a function with respect to a variable. In the context of the exercise, we differentiate the inverse tangent function to find how fast the angle \(\theta\) is changing with respect to time.

• Derivatives help in establishing the rate at which \(\theta\) changes due to its dependency on the changing \(x\) coordinate.
• The calculated derivative, \(\frac{d\theta}{dt} = \frac{1}{1 + x^2} \cdot \frac{dx}{dt}\), combines both the relationship between \(\theta\) and \(x\), as well as the input rate \(\frac{dx}{dt}\).
• By substituting the known values of \(x = 3\) and \(\frac{dx}{dt} = 10\), we find that the rate of change \(\frac{d\theta}{dt}\) is \(0.1\) radians per second.

Understanding derivatives in this context allows students to see the power of calculus in analyzing and predicting changes in dynamic systems. It is not only about computing rates but also interpreting what those rates mean in practical terms.