Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

A Ferris wheel of radius 20 feet is rotating counterclockwise with an angular velocity of 1 radian per second. One seat on the rim is at \((20,0)\) at time \(t=0\). (a) What are its coordinates at \(t=\pi / 6\) ? (b) How fast is it rising (vertically) at \(t=\pi / 6\) ? (c) How fast is it rising when it is rising at the fastest rate? 25\. Find the equation of the tangent line to \(y=\tan x\) at \(x=0\).

Step-by-Step Solution

Answer

(a) Coordinates: (10√3, 10); (b) Rising at 10√3 ft/s; (c) Max rising speed: 20 ft/s; Tangent line: y = x.

Step 1: Determine Coordinates at t=π/6

To find the coordinates at \( t = \pi / 6 \), we use the parametric equations for a circle: \( x(t) = 20 \cos(t) \) and \( y(t) = 20 \sin(t) \). At \( t=\pi/6 \), we calculate:\[ x = 20 \cos(\pi/6) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \]\[ y = 20 \sin(\pi/6) = 20 \times \frac{1}{2} = 10 \]Thus, the coordinates are \((10\sqrt{3}, 10)\).

Step 2: Calculate Vertical Velocity at t=π/6

The vertical velocity at any time \( t \) is given by the derivative of \( y(t) \), which is \( y'(t) = 20 \cos(t) \cdot \frac{d}{dt}(t) = 20 \cos(t) \times 1 = 20 \cos(t) \). At \( t = \pi/6 \):\[ y'(\pi/6) = 20 \cos(\pi/6) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \]So, the seat is rising at a rate of \( 10\sqrt{3} \) feet per second.

Step 3: Find Maximum Vertical Velocity

The maximum vertical velocity occurs when \( \cos(t) = 1 \), which is when \( t = 0 \). Therefore, the maximum vertical velocity is:\[ y'(t) = 20 \cos(t) = 20 \]This indicates the maximum rate of rising is 20 feet per second.

Step 4: Equation of the Tangent Line to y=tan x at x=0

To find the equation of the tangent line, first determine the slope by differentiating \( y = \tan x \), so \( y' = \sec^2(x) \). At \( x = 0 \):\[ y'(0) = \sec^2(0) = 1 \]The value of \( y \) at \( x = 0 \) is \( \tan(0) = 0 \). Thus, the tangent line has a slope of 1 and goes through \((0, 0)\), giving the equation \( y = x \).

Key Concepts

Angular Velocity

Angular velocity is a key concept in rotational motion. It describes how quickly an object rotates or spins around a specific axis. In our example, the Ferris wheel completes one radian of rotation per second. This unit, radians per second, is a standard way to measure angular velocity.

• One radian equals approximately 57.3 degrees.
• Angular velocity emphasizes how fast the wheel rotates, not its linear speed.

Understanding angular velocity helps in determining positions and velocities in scenarios involving rotation. With a constant angular velocity of 1 radian/second, we can predict the wheel's movements at any given time, such as finding the position of a seat on the Ferris wheel.

Parametric Equations

Parametric equations are used to express the coordinates of points on a curve as functions of a parameter, in this case, time. For the Ferris wheel, the parametric equations for the x and y positions of a point on the wheel's rim are given by:

• \( x(t) = 20 \cos(t) \)
• \( y(t) = 20 \sin(t) \)

These equations allow us to calculate the exact position of the seat at any time \( t \). They are especially useful in describing cyclic or circular motion where the coordinates change with the rotation. By using these parametric functions, we determined the position of the Ferris wheel seat at \( t = \pi/6 \). This highlights the coordinate system that rotates in a circular path, based on trigonometric functions.

Vertical Velocity

Vertical velocity refers to how quickly something moves up or down. In the case of the Ferris wheel, we're interested in how fast the seat rises vertically. To find this, we calculate the derivative of the vertical (y) position with respect to time:

• The formula is \( y'(t) = 20 \cos(t) \).

At \( t = \pi/6 \), this derivative tells us that the seat's vertical speed at that moment is \( 10\sqrt{3} \) feet per second. This mathematical process of differentiation highlights how calculus plays a key role in analyzing motion, especially in providing rates of change at precise moments.

Tangent Line

The tangent line is a straight line that touches a curve at a single point, matching the curve's slope at that point. It provides a linear approximation of the curve near that point. In the exercise, we're finding the tangent line to \( y = \tan x \) at \( x = 0 \).

• We first find the derivative \( y' = \sec^2(x) \).
• Evaluating at \( x = 0 \), we find the slope is 1.

Next, knowing the point \((0, 0)\) lies on the tangent, we construct the equation of the tangent line: \( y = x \). Understanding tangent lines is crucial in calculus as they help approximate and analyze complex curves with simpler linear equations.