Problem 37 - Calculus III

Question

Given \(\mathbf{r}(t)=\langle t+\cos t, t-\sin t\rangle\), find the velocity and the speed at any time.

Step-by-Step Solution

Answer

The velocity is \(\langle 1 - \sin t, 1 - \cos t \rangle\) and the speed is \(\sqrt{3 - 2(\sin t + \cos t)}\)."

Step 1: Define the Velocity Vector

The velocity vector is the derivative of the position vector with respect to time. Given \(\mathbf{r}(t)=\langle t+\cos t, t-\sin t\rangle\), we need to differentiate each component with respect to \(t\).

Step 2: Differentiate the Components

Differentiate the first component: \(t + \cos t\) gives \(1 - \sin t\). Differentiate the second component: \(t - \sin t\) gives \(1 - \cos t\). Thus, the velocity vector is \(\mathbf{v}(t) = \langle 1 - \sin t, 1 - \cos t \rangle\).

Step 3: Define Speed

The speed is the magnitude of the velocity vector. We use the formula for the magnitude of a vector: \(\|\mathbf{v}(t)\| = \sqrt{(v_1(t))^2 + (v_2(t))^2}\) where \(v_1(t)\) and \(v_2(t)\) are the components of the velocity vector.

Step 4: Calculate the Magnitude

Substitute the components of \(\mathbf{v}(t)\): \(\|\mathbf{v}(t)\| = \sqrt{(1 - \sin t)^2 + (1 - \cos t)^2}\). Expand the squared terms: \((1 - \sin t)^2 = 1 - 2\sin t + \sin^2 t\) and \((1 - \cos t)^2 = 1 - 2\cos t + \cos^2 t\).

Step 5: Simplify the Expression

Combine the terms: \(\|\mathbf{v}(t)\| = \sqrt{2 - 2(\sin t + \cos t) + \sin^2 t + \cos^2 t}\). Recall from trigonometric identities that \(\sin^2 t + \cos^2 t = 1\), so substitute it in: \(\|\mathbf{v}(t)\| = \sqrt{3 - 2(\sin t + \cos t)}\).

Key Concepts

Derivative of Position Vector

Understanding the concept of derivative is crucial in the study of motion. A position vector, denoted as \( \mathbf{r}(t) \), describes the location of a point in space as a function of time \( t \). When we differentiate this position vector with respect to time, we find the velocity vector. This tells us not only how fast but also in what direction the position is changing at any given moment.

In our exercise, the position vector is given by:

• \( \mathbf{r}(t) = \langle t + \cos t, t - \sin t \rangle \)

We take the derivative of each component of this vector:

• For \( t + \cos t \), the derivative is \( 1 - \sin t \).
• For \( t - \sin t \), the derivative is \( 1 - \cos t \).

Thus, the velocity vector \( \mathbf{v}(t) \) is composed of these derivatives:

• \( \mathbf{v}(t) = \langle 1 - \sin t, 1 - \cos t \rangle \)

This derivative process highlights how both linear and oscillatory (sine/cosine) movements are combined in the motion of the object.

Magnitude of Velocity

The magnitude of velocity, often referred to as speed, measures how fast an object is moving regardless of its direction. Once we have the velocity vector, we can find its magnitude using the formula:

• \( \|\mathbf{v}(t)\| = \sqrt{(v_1(t))^2 + (v_2(t))^2} \)

Substituting the components of our velocity vector, we get:

• \( \|\mathbf{v}(t)\| = \sqrt{(1 - \sin t)^2 + (1 - \cos t)^2} \)

Next, we expand and simplify each squared term:

• \( (1 - \sin t)^2 = 1 - 2\sin t + \sin^2 t \)
• \( (1 - \cos t)^2 = 1 - 2\cos t + \cos^2 t \)

Adding these, we use the identity \( \sin^2 t + \cos^2 t = 1 \), resulting in:

• \( \|\mathbf{v}(t)\| = \sqrt{3 - 2(\sin t + \cos t)} \)

Understanding this mathematical manipulation sheds light on how different types of movements contribute to the overall speed of an object.

Trigonometric Identities

Trigonometric identities are mathematical equations that relate the trigonometric functions sine, cosine, and tangent to one another. They can simplify complex mathematical expressions and are especially helpful in calculus and physics.

In the exercise, a key trigonometric identity used to find the speed of the object is:

• \( \sin^2 t + \cos^2 t = 1 \)

This identity comes from the Pythagorean Theorem, as it relates the squares of the sine and cosine of an angle \( t \) to 1. By substituting this identity, we reduced the complexity of our speed formula:
Instead of dealing with two squared terms, we combined them into a simpler expression that directly linked to the object's motion.

Trigonometric identities like this can also help:

• Simplify derivatives and integrals involving trigonometric functions
• Identify relationships between components of vectors in multidimensional space

Mastering these identities can greatly enhance your ability to solve a wide range of mathematical problems.