Question

A position function \(\vec{r}(t)\) of an object is given. Find the speed of the object in terms of \(t,\) and find where the speed is minimized/maximized on the indicated interval. $$ \vec{r}(t)=\langle 12 t, 5 \cos t, 5 \sin t\rangle \text { on }[0,4 \pi] $$

Short answer

Speed is constant at 13, minimized/maximized for all \( t \).

Step-by-step solution

Find the Velocity Vector

The velocity vector \( \vec{v}(t) \) is the derivative of the position function \( \vec{r}(t) \). Compute \( \vec{v}(t) = \frac{d}{dt} \langle 12t, 5\cos t, 5\sin t \rangle \). The resulting components are \( \langle 12, -5\sin t, 5\cos t \rangle \).

Calculate the Speed

The speed is the magnitude of the velocity vector. Calculate the speed as \( \|\vec{v}(t)\| = \sqrt{12^2 + (-5\sin t)^2 + (5\cos t)^2} \). Simplify to get \( \|\vec{v}(t)\| = \sqrt{144 + 25(\sin^2 t + \cos^2 t)} = \sqrt{169} = 13 \).

Determine Interval-specific Speed Characteristics

Here, the speed function is constant and equal to 13 for all \( t \) in the interval \( [0, 4\pi] \). A constant speed indicates it is both minimized and maximized over the interval, as there are no changes in speed.

Key concepts

Position Function

The position function of an object, often denoted as \( \vec{r}(t) \), describes the object's position in space at any given time \( t \). It acts as a map that provides the object's coordinates based on time inputs. This function can have different components depending on how many dimensions it covers:

• For motion along a line, the position function is usually a single-variable function.
• For planar motion, it could have two components, like \( \langle x(t), y(t) \rangle \).
• For motion in three dimensions, it would typically be of the form \( \langle x(t), y(t), z(t) \rangle \).

In our exercise, the position function is given by \( \vec{r}(t) = \langle 12t, 5\cos t, 5\sin t \rangle \). This indicates that the object moves in three-dimensional space. The expression suggests linear motion in the first component and circular motion in the second and third components.

Velocity Vector

The velocity vector of an object represents its speed and direction of motion at any moment. It is found by taking the derivative of the position function with respect to time. This derivative essentially provides the rate of change of the object's position:

• The velocity vector points in the direction the object is moving.
• The magnitude of this vector gives you the object's speed.

For our specific exercise, we derived the velocity vector \( \vec{v}(t) \) by differentiating each component of the position function \( \vec{r}(t) \):

• First component: \( \frac{d}{dt}[12t] = 12 \)
• Second component: \( \frac{d}{dt}[5\cos t] = -5\sin t \)
• Third component: \( \frac{d}{dt}[5\sin t] = 5\cos t \)

Thus, the velocity vector becomes \( \langle 12, -5\sin t, 5\cos t \rangle \). Consistent application of derivatives here tells us how the object's directional motion changes over time.

Speed

Speed is a scalar quantity that measures how fast an object is moving. Unlike velocity, speed does not have a direction associated with it. It is simply the magnitude (or length) of the velocity vector:

• Speed is always a non-negative value.
• It shows the rate of position change but not the direction.

To find the speed from the velocity vector \( \vec{v}(t) \), you calculate its magnitude:\[\|\vec{v}(t)\| = \sqrt{12^2 + (-5\sin t)^2 + (5\cos t)^2}\]Substituting values into the equation gives us:\[\|\vec{v}(t)\| = \sqrt{144 + 25(\sin^2 t + \cos^2 t)} = \sqrt{169} = 13\]In this exercise, the speed is a constant 13 across the given interval \([0, 4\pi]\). This consistent speed indicates steady movement without acceleration or deceleration.

Derivative

The derivative is a fundamental concept in calculus that represents the rate of change of a function with respect to a variable. In simple terms, it measures how a function varies as its input varies:

• In the context of motion, the derivative of the position function gives the velocity vector.
• Derivatives allow us to understand dynamic changes, like acceleration, by taking further derivatives of the velocity (yielding acceleration).

When working with derivatives, it's essential to apply rules like the power rule, product rule, and the chain rule, depending on the type of functions you're dealing with. In our exercise, deriving the position function \( \vec{r}(t) = \langle 12t, 5\cos t, 5\sin t \rangle \) resulted in the velocity vector \( \langle 12, -5\sin t, 5\cos t \rangle \). Each component derivative gives insight into how each directional motion changes over time.