Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems 25-32, find the arc length of the given curve. } $$ $$ x=t, y=t, z=2 t ; 0 \leq t \leq 2 $$

Step-by-Step Solution

Answer

The arc length is \( 2 \sqrt{6} \).

Step 1: Parametric Equation Identification

Identify the parametric equations of the given curve. Here, we can directly see from the problem statement:\[ x(t) = t, \quad y(t) = t, \quad z(t) = 2t \] where the parameter \( t \) ranges from 0 to 2.

Step 2: Determine the Derivatives

Find the derivatives of each component with respect to \( t \):\[\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = 1, \quad \frac{dz}{dt} = 2\]

Step 3: Compute the Magnitude of the Velocity Vector

Use the derivatives to find the magnitude of the velocity vector \( \vec{v}(t) \):\[ \|\vec{v}(t)\| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2 }\]Substitute the derivatives:\[ \|\vec{v}(t)\| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}\]

Step 4: Integrate to Find the Arc Length

Now integrate the magnitude of the velocity vector over the interval given for \( t \), which is from 0 to 2:\[ L = \int_{0}^{2} \|\vec{v}(t)\| \, dt = \int_{0}^{2} \sqrt{6} \, dt\]Since \( \sqrt{6} \) is a constant, the integration becomes:\[L = \sqrt{6} \times [t]_{0}^{2} = \sqrt{6} \times (2 - 0) = 2 \sqrt{6}\]

Step 5: Conclude the Arc Length

Thus, the arc length of the curve from \( t = 0 \) to \( t = 2 \) is \( 2 \sqrt{6} \).

Key Concepts

Parametric Equations

Parametric equations are a fundamental part of understanding the geometry of curves in space. Instead of expressing curves in terms of a single variable like "x" or "y," we define each coordinate as a function of a common parameter often denoted as "t." In this exercise, the curve is defined with the parametric equations:

• \( x(t) = t \)
• \( y(t) = t \)
• \( z(t) = 2t \)

The parameter "t" essentially acts as a time variable showing how each component changes as "t" increases. This approach is particularly useful for describing motion in three-dimensional space, allowing us to capture the dynamics of movement rather than merely a static shape. By using parametric equations, we better understand the nature of curves and the paths of objects over time.
In the provided problem, the range for "t" is from 0 to 2, allowing us to evaluate how the coordinates evolve along this interval.

Velocity Vector

The velocity vector is crucial in determining how fast and in which direction a point is moving along a curve. When we have parametric equations, the velocity vector can be found by taking the derivative of each component with respect to the parameter "t."
For this specific curve, we calculate:

• \( \frac{dx}{dt} = 1 \)
• \( \frac{dy}{dt} = 1 \)
• \( \frac{dz}{dt} = 2 \)

The resulting velocity vector is \( \vec{v}(t) = (1, 1, 2) \). This vector indicates that for each unit increase in time, the point moves one unit in both x and y directions and two units in the z direction. Calculating the magnitude of this velocity vector helps us in computing the arc length of the curve, representing the rate of displacement of the point in space.

Integration

Integration is the mathematical process used to find total quantities, like areas under curves or lengths of curves. In our exercise, we use integration to calculate the arc length of the space curve defined by the parametric equations over a specific interval for the parameter "t."
The arc length formula for parametric curves involves integrating the magnitude of the velocity vector \( \|\vec{v}(t)\| \) over the desired range of "t."
Given that the magnitude is \( \sqrt{6} \), our integration task becomes simpler:\[L = \int_{0}^{2} \sqrt{6} \, dt\]Since \( \sqrt{6} \) is a constant, integration reduces to multiplying \( \sqrt{6} \) by the length of the interval from 0 to 2:\[L = \sqrt{6} \times (2 - 0) = 2 \sqrt{6}\]Through integration, we have efficiently calculated the arc length as \( 2 \sqrt{6} \), representing the complete path segment traced by the curve from its start to finish.

Derivative Calculation

Derivative calculation involves finding how a function changes as its input changes. For parametric equations, this means computing how each component of the curve changes with respect to the parameter "t."
In this problem, finding the derivatives of the parametric functions gives us the components of the velocity vector:

• Derivative of \( x(t) = t \) is \( \frac{dx}{dt} = 1 \).
• Derivative of \( y(t) = t \) is \( \frac{dy}{dt} = 1 \).
• Derivative of \( z(t) = 2t \) is \( \frac{dz}{dt} = 2 \).

These derivatives indicate how quickly the x, y, and z coordinates are changing at any instant. In general, derivatives help us understand the rate of change, which is vital for analyzing motion along curves and determining factors like speed and trajectory. In this context, the derivative calculations were pivotal for forming the velocity vector and subsequently for determining the curve's arc length.