Problem 29 - Calculus

Question

A ball goes around a circle: \(x=\cos t, y=\sin t\). At \(t=3 \pi / 4\) the ball flies off on the tangent line. Find the equation of that line and the point where the ball hits the ground \((y=0)\).

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = x + \sqrt{2}\), and it hits the ground at \((-\sqrt{2}, 0)\).

Step 1: Identify the Position and Velocity

First, identify the position of the ball at time \(t = \frac{3\pi}{4}\). Given \(x = \cos t\) and \(y = \sin t\), plug in \(t = \frac{3\pi}{4}\) to get the coordinates: \(x = \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}\) and \(y = \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}\). The position of the ball at this time is \((-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\). Next, find the velocity (the derivative of the position): \(x'(t) = -\sin t\) and \(y'(t) = \cos t\). Evaluate at \(t = \frac{3\pi}{4}\) to find the velocity vector: \(x'(\frac{3\pi}{4}) = -\sin\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}\) and \(y'(\frac{3\pi}{4}) = \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}\).

Step 2: Find the Slope of the Tangent Line

To find the slope of the tangent line where the ball leaves the circle, use the velocity vector components: slope \(m = \frac{y'}{x'} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1\). The line is therefore in the direction with slope 1 passing through the point \((-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\).

Step 3: Write the Equation of the Tangent Line

The equation of the line is \(y - y_1 = m(x - x_1)\). With \(m = 1\), \(x_1 = -\frac{\sqrt{2}}{2}\), and \(y_1 = \frac{\sqrt{2}}{2}\), plug these into the line equation: \(y - \frac{\sqrt{2}}{2} = 1(x + \frac{\sqrt{2}}{2})\). Simplifying yields \(y = x + \sqrt{2}\).

Step 4: Determine the Intersection with the Ground

The point where the ball hits the ground is when \(y = 0\). Set \(y = x + \sqrt{2}\) to zero: \(0 = x + \sqrt{2}\). Solving for \(x\) gives \(x = -\sqrt{2}\). Therefore, the point where the line hits the ground is \((-\sqrt{2}, 0)\).

Key Concepts

Equation of Motion

In physics, the equation of motion gives a mathematical description of an object's motion. In our problem, we deal with a ball moving around a circle defined by the parametric equations \(x = \cos t\) and \(y = \sin t\). These equations describe the circle's movement as a function of time \(t\).

By plugging in specific values for \(t\), such as \(t = \frac{3\pi}{4}\), we find the corresponding position of the ball on the circle. This process allows us to understand how the ball's coordinates change over time, illustrating the concept of motion in a circular path.

Derivative

The derivative is a powerful mathematical tool used to determine the rate at which something changes. In this task, derivatives help find the ball's velocity as it moves around the circle. The position given by \(x = \cos t\) and \(y = \sin t\) can be differentiated to find velocity.

Compute the derivative to obtain: \(x'(t) = -\sin t\) and \(y'(t) = \cos t\). These derivatives represent the velocity components in the \(x\) and \(y\) directions, respectively. At \(t = \frac{3\pi}{4}\), calculating these values provides crucial information about how fast and in what direction the ball is moving at that specific point in time.

Circle Coordinates

In our scenario, we use parametric equations to describe the circle coordinates. These equations, \(x = \cos t\) and \(y = \sin t\), map out every point on the circle as \(t\) varies from 0 to \(2\pi\).

At \(t = \frac{3\pi}{4}\), substituting gives the coordinates \(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\). These coordinates represent the exact position of the ball on the circle at this time. Understanding these coordinates is essential to grasping the ball's path and predicting future positions.

Velocity

Velocity is the measure of both speed and direction of an object's movement. In this context, the velocity of the ball at any point on the circle is the derivative of its position with respect to time.

For \(t = \frac{3\pi}{4}\), we found the velocity components by differentiating \(x = \cos t\) and \(y = \sin t\) to get \(x'(t) = -\sin t\) and \(y'(t) = \cos t\). For this specific \(t\), both components equal \(-\frac{\sqrt{2}}{2}\), indicating that the ball is moving at a steady pace diagonally across the plane, which helps us find the slope of the tangent line.

Intersection Point

The intersection point in this exercise is crucial for determining where the ball will hit the ground. Once the ball leaves the circular path at \(t = \frac{3\pi}{4}\), it travels along a tangent line.

The tangent line equation derived from the velocity vector is \(y = x + \sqrt{2}\). To find where the ball hits the ground, set \(y = 0\). Solving \(0 = x + \sqrt{2}\) reveals the \(x\)-coordinate of the intersection point, \(-\sqrt{2}\). Thus, the point \((-\sqrt{2}, 0)\) is where the ball meets the ground.