Problem 38 - Calculus

Question

Choose the number \(k\) so that \(x=\cos k t, y=\sin k t\) completes a rotation at \(t=1 .\) Find the speed and upward velocity.

Step-by-Step Solution

Answer

\( k = 2\pi \), speed is \( 2\pi \), upward velocity at \( t=1 \) is \( 2\pi \).

Step 1: Understand the Problem

We have the parametric equations for a circle: \( x = \cos(kt) \) and \( y = \sin(kt) \). This describes a circular path. We are asked to determine the value of \( k \) such that the curve completes one full rotation when \( t = 1 \).

Step 2: Complete a Full Rotation

A full rotation in terms of angle measurement occurs when the angle traversed equals \( 2\pi \) radians. For the given equation \( x = \cos(kt) \) and \( y = \sin(kt) \), the angle swept is \( kt \). We want this angle to be \( 2\pi \) when \( t = 1 \). Thus, we set up the equation \( k \times 1 = 2\pi \) and solve for \( k \).

Step 3: Solve for k

From \( k \times 1 = 2\pi \), we directly find that \( k = 2\pi \). This means that the parameter values \( k \) need to ensure one full rotation in one time unit.

Step 4: Find Speed

The speed \( v \) for parametric equations \( x = \cos(kt) \), \( y = \sin(kt) \) can be found using the formula \( v = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \). Calculate \( \frac{dx}{dt} = -k\sin(kt) \) and \( \frac{dy}{dt} = k\cos(kt) \). Then the speed is \( v = \sqrt{(-k\sin(kt))^2 + (k\cos(kt))^2} \= \sqrt{k^2(\sin^2(kt) + \cos^2(kt))} = k \). Since we have \( k = 2\pi \), the speed is \( 2\pi \).

Step 5: Calculate Upward Velocity

The upward (or vertical) velocity is given by \( \frac{dy}{dt} \). We previously calculated \( \frac{dy}{dt} = k\cos(kt) \). Substitute \( k = 2\pi \) to obtain \( \frac{dy}{dt} = 2\pi\cos(2\pi t) \). At \( t = 1 \), \( \cos(2\pi \times 1) = \cos(2\pi) = 1 \), hence the upward velocity is \( 2\pi \times 1 = 2\pi \).

Key Concepts

Full Rotation

Understanding what a full rotation entails is essential when dealing with circular motion problems involving parametric equations. A full rotation in the context of a circle is when a point travels around the circumference of the circle and returns to its starting position.
In terms of angular measurements, this is equivalent to traversing an angle of \(2\pi \) radians, which corresponds to 360 degrees.
In our exercise, we are using the equations \(x = \cos(kt)\) and \(y = \sin(kt)\), which describe a circular path. A metric of completion for a full rotation is when the angle \( kt \) becomes \( 2\pi \).
By setting \( kt = 2\pi \) when \( t = 1 \), we get the necessary condition that \( k = 2\pi \).

• This represents a full circular movement.
• It emphasizes the relation between linear time \(t\) and angular displacement \(kt\).

Speed Calculation

In parametric equations, the speed of a point moving along a curve can be determined using derivatives. When given \( x = \cos(kt) \) and \( y = \sin(kt) \), the speed \( v \) is calculated by evaluating
\[v = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\]Start by finding the derivatives:\( \frac{dx}{dt} = -k\sin(kt) \) and \( \frac{dy}{dt} = k\cos(kt) \).

Insert these into the speed formula:\[v = \sqrt{(-k\sin(kt))^2 + (k\cos(kt))^2}\]This simplifies to \( v = \sqrt{k^2(\sin^2(kt) + \cos^2(kt))} \). Since \( \sin^2(kt) + \cos^2(kt) = 1 \), simplify further to
\( v = k \).
Therefore, substituting \( k = 2\pi \) gives us a speed of \( 2\pi \).

• Speed is constant because the circle is uniform.
• We rely on trigonometric identities, notably \(\sin^2 \theta + \cos^2 \theta = 1\).

Upward Velocity

Upward velocity, in the context of circular motion, is specifically the change in the vertical component over time.
Given the parametric equations \( y = \sin(kt) \), we calculate the upward velocity by finding the derivative of \( y \) with respect to time \( t \).

Thus, \( \frac{dy}{dt} = k\cos(kt) \). The expression \( k\cos(kt) \) represents the rate of change of the vertical position component.
Substituting \( k = 2\pi \) gives \( \frac{dy}{dt} = 2\pi \cos(2\pi t) \).
Evaluate this expression at \( t = 1 \):\[\cos(2\pi \times 1) = \cos(2\pi) = 1\]Consequently, the upward velocity is \( 2\pi \times 1 = 2\pi \).

• The upward velocity is maximum when \( \cos(kt) \) is at its maximum value, which is 1.
• Understanding this component helps in analyzing oscillatory or wave-like motion climes.

Trigonometric Functions

Trigonometric functions, \(\cos(x)\) and \(\sin(x)\), are pivotal in describing circular motion and waves. In the context of our exercise, they help define the path traced by a point moving around a circle.
These functions oscillate between -1 and 1, which corresponds to the cyclical nature of angles.When you look at a unit circle, the radius is constantly 1, and both \(\cos(\theta)\) and \(\sin(\theta)\) define coordinates on this circle.

• \( \cos(x) \) is the horizontal (x-axis) component
• \( \sin(x) \) is the vertical (y-axis) component

A key identity to remember is \(\sin^2(x) + \cos^2(x) = 1\), fundamental in most trigonometric transformations and simplifications.
Using these functions with parametric equations lets us model not only circles but a wide array of periodic phenomena, from pendulum swings to alternating currents.

• Essential for solving problems that involve oscillations or rotations.
• Always relate trigonometric values with specific angles for better intuition.