Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

A bug is walking around the circle \(x^{2}+y^{2}=169 .\) At a certain instant the bug is at the point \((-5,12)\) and its \(y\) -coordinate is decreasing at a rate of 3 units per second. (a) Is the bug traveling the circle in a clockwise direction or a counterclockwise direction? (b) How fast is its \(x\) -coordinate changing at this instant?

Step-by-Step Solution

Answer

The bug is traveling in a clockwise direction. The rate of change of its \(x\) -coordinate is \(7.2\) units per second.

Step 1: Understand the Problem

At any point on the circular path, the x and y coordinates of the bug fulfill the equation \(x^{2} + y^{2} = 169\). Derived both sides of this equation with respect to the time \(t\), we can find the relationship between \(dx/dt\) and \(dy/dt\).

Step 2: Differentiate Circle Equation

Using the chain rule, the derivative of \(x^{2}\) is \(2x \cdot dx/dt\) and the derivative of \(y^{2}\) is \(2y \cdot dy/dt\). So, differentiate both sides of \(x^{2} + y^{2} = 169\), we get \(2x \cdot dx/dt + 2y \cdot dy/dt = 0\).

Step 3: Solve for dx/dt

We know that the bug is at the point \((-5, 12)\) and \(dy/dt\) is \(-3\) units per second. Plug in these values into the derived equation, we can solve for \(dx/dt\). So, \(-5 \cdot dx/dt + 12 \cdot (-3) = 0\), giving \(dx/dt = \frac{-12 \cdot (-3)}{-5} = 7.2\) units per second.

Step 4: Determine the Direction

Since \(dx/dt\) is positive while \(dy/dt\) is negative, we can conclude that the bug is moving in the clockwise direction. In a Cartesian plane, when \(x\) is increasing and \(y\) is decreasing, the direction of the movement is clockwise.

Key Concepts

Circle Equation

Circles are a fundamental concept in geometry and calculus, and they are often expressed algebraically through equations. A common form of a circle equation is given by

• \(x^{2} + y^{2} = r^{2}\)

where \(r\) is the radius of the circle. This equation represents all the points \((x, y)\) that are equidistant from the center of the circle, which in this case is the origin

• (0,0)

in the coordinate plane. The particular circle equation we are discussing, \(x^{2} + y^{2} = 169\), implies the circle has a radius of 13, since

• \(r = \sqrt{169} = 13.\)

The circle equation is vital for understanding the relationship between the \(x\) and \(y\) coordinates on the circle, which remains constant over time as the bug moves around the circle. This equation is essential for analyzing how these variables change relative to each other as time progresses.

Derivatives

Derivatives are a key concept in calculus, representing the rate at which one quantity changes with respect to another. In this exercise, we consider the derivatives of \(x\) and \(y\) with respect to time \(t\), denoted

• \(\frac{dx}{dt}\)
• \(\frac{dy}{dt}\).

These derivatives indicate how the \(x\) and \(y\) coordinates of the bug change as it travels around the circle. Differentiating gives us insight into the speed and direction of movement. For example, if

• \(\frac{dy}{dt} = -3\)

this means the \(y\)-coordinate is decreasing at a rate of 3 units per second. The process of taking these derivatives involves calculus techniques that tell us about instantaneous rates of change which is crucial in motion-related problems.

Chain Rule

The chain rule is an essential tool in differentiation, especially for problems involving related rates. It allows us to differentiate composite functions. When our functions depend on multiple variables, like \(x\) and \(y\) both being functions of time \(t\), the chain rule helps relate their derivatives.In this particular scenario, it is used to differentiate the circle equation:

• \(x^2 + y^2 = 169\).

To handle the differentiation of terms like \(x^2\), you must differentiate with respect to \(t\) using

• \(\frac{d}{dt}(x^2) = 2x \cdot \frac{dx}{dt}\)

and similarly for \(y^2\). This yields the expression:

• \(2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0\).

The chain rule efficiently connects variables that are not explicitly defined as functions of time, enabling us to solve for unknown rates of change, such as \(\frac{dx}{dt}\).

Direction of Movement

Understanding the direction of movement involves looking at the signs of the derivatives of the coordinates with respect to time. In our problem,

• \(\frac{dy}{dt} = -3\)

shows that the bug's \(y\) position is decreasing. After solving using the circle's equation, we find

• \(\frac{dx}{dt} = 7.2\),

which is positive, indicating that the bug's \(x\)-coordinate is increasing. When the \(x\)-coordinate is increasing and the \(y\)-coordinate is decreasing, the bug moves in a clockwise direction around the circle. This direction is intuitive; picture moving towards the right (positive \(x\)) and downwards (negative \(y\)) on a standard Cartesian plane. These dynamics illustrate how calculus can explain movement and orientation within a geometric shape.