Question

An object moves on a trajectory given by $$\mathbf{r}(t)=\langle 10 \cos 2 t, 10 \sin 2 t\rangle, \text { for } 0 \leq t \leq \pi$$ How far does it travel?

Short answer

Answer: The object travels a total distance of \(20\pi\) units.

Step-by-step solution

Calculate the velocity vector

Differentiate the position vector, \(\mathbf{r}(t)\), with respect to time: $$\mathbf{v} (t) = \frac{d\mathbf{r}}{dt} = \left\langle\frac{d}{dt} (10 \cos 2t) , \frac{d}{dt} (10 \sin 2t) \right\rangle$$ Using chain rule for differentiation: $$\mathbf{v}(t) = \langle-20 \sin 2 t, 20 \cos 2 t\rangle.$$

Find the magnitude of the velocity vector

Calculate the magnitude of the velocity vector, \(|\mathbf{v}(t)|\), at any time \(t\): $$|\mathbf{v}(t)| = \sqrt{(-20 \sin 2 t)^2 + (20 \cos 2 t)^2}$$ Simplify the expression within the square root: $$|\mathbf{v}(t)| = \sqrt{400 (\sin^2 2t + \cos^2 2t)}$$ Since \(\sin^2 x + \cos^2 x = 1\), we can simplify the expression further: $$|\mathbf{v}(t)| = \sqrt{400} = 20$$

Calculate the total distance traveled

Integrate the magnitude of the velocity vector, \(|\mathbf{v}(t)|\), over the time interval \(0 \leq t \leq \pi\) to find the distance traveled: $$\text{Distance} = \int_0^{\pi} |\mathbf{v}(t)| dt = \int_0^{\pi} 20 dt$$ Evaluating the integral: $$\text{Distance} = 20t \Big|_0^{\pi} = 20\pi - 0 = 20\pi$$ So, the object travels a distance of \(20\pi\) units.

Key concepts

Velocity Vector Calculus

Understanding the velocity vector calculus concept is crucial when we aim to determine the movement of an object over time. Velocity is a vector quantity, which means it has both magnitude and direction. In calculus, when studying the motion of an object, the velocity vector, denoted as \textbf{v}(t), is the derivative of the position vector \textbf{r}(t) with respect to time (t).The velocity vector can be found by differentiating the components of the position vector. For instance, given the trajectory \textbf{r}(t)=\( \langle 10 \cos 2 t, 10 \sin 2 t\rangle \), we differentiate each component with respect to time using the chain rule. As a result, \textbf{v}(t) provides us with a vector that points in the direction of the object's instantaneous velocity and whose magnitude equals the speed at time t.

• Identifying the heading at a given instant is done by taking note of the direction of the velocity vector.
• The rate of change in both the x and y directions is given by the vector's individual components.
• Differentiating trigonometric functions often requires using the chain rule effectively.

Proper comprehension of the velocity vector's calculus lays the groundwork for exploring more complex motions and differentiating between speed and velocity in physical applications.

Magnitude of Velocity Vector

The magnitude of the velocity vector is often termed speed in physical contexts and represents how fast an object is moving regardless of its direction. It is denoted as |\textbf{v}(t)| and is calculated as the square root of the sum of the squares of the velocity vector's components.For example, using the velocity vector \textbf{v}(t) = \( \langle-20 \sin 2 t, 20 \cos 2 t\rangle \), the magnitude is computed as |\textbf{v}(t)|\( = \sqrt{(-20 \sin 2 t)^2 + (20 \cos 2 t)^2} \). This results in a simplified expression using the Pythagorean trigonometric identity \(\sin^2 x + \cos^2 x = 1\).When the magnitudes of the velocity components are squared, they sum up to a constant because of the specific trigonometric functions involved. Therefore, despite the motion's direction changing, the speed remains constant at \(20\) for this particular problem. This insight is particularly useful when dealing with circular or periodic motion where the speed may be constant, but the direction—and hence the vector velocity—changes.

Integration of Velocity

To find the total distance traveled by an object, we use the technqiue of integrating the velocity. This process involves finding the integral of the magnitude of the velocity vector over a given time period.In the context of the given problem, we are integrating |\textbf{v}(t)| over the time from \(0\) to \(\pi\). The integral of |\textbf{v}(t)| = 20 over this interval results in \(\text{Distance} = \int_0^{\pi} 20 dt\), which simplifies to \(20t\), evaluated from \(0\) to \(\pi\). The calculation yields \(20\pi\) units as the total distance traveled.

• Integration can be understood as the accumulation of quantities, such as distance over time.
• In scenarios where speed changes, the integral would not simplify so neatly, instead reflecting the area under the curve described by the speed-time graph.
• Even without a graph, knowing that the integrand is constant allows us to multiply the integrand by the width of the interval directly.

Integration is a powerful tool that extends well beyond constant velocity scenarios, allowing us to calculate distances for much more complex motions as long as we can describe velocity as a function of time.