Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems 25-32, find the arc length of the given curve. } $$ $$ x=t^{2}, y=(4 / 3) t^{3 / 2}, z=t ; 0 \leq t \leq 8 $$

Step-by-Step Solution

Answer

The arc length of the curve is 72.

Step 1: Understand the Arc Length Formula

The arc length of a curve defined by parametric equations can be found using the formula \( L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt \). Given the parametric equations \( x=t^2 \), \( y=\frac{4}{3}t^{3/2} \), and \( z=t \), we will use this formula to find the arc length from \( t=0 \) to \( t=8 \).

Step 2: Calculate Derivatives of Parametric Equations

Find the derivatives of \( x \), \( y \), and \( z \) with respect to \( t \). Start with:- \( \frac{dx}{dt} = 2t \),- \( \frac{dy}{dt} = 2t^{1/2} \) (derived from \( y = \frac{4}{3}t^{3/2} \)),- \( \frac{dz}{dt} = 1 \).

Step 3: Formulate the Integral for Arc Length

Use the derivatives to set up the integral for arc length. Plug the derivatives into the arc length formula: \[ L = \int_0^8 \sqrt{(2t)^2 + (2t^{1/2})^2 + (1)^2} \, dt \].Simplify the expression inside the square root:\((2t)^2 = 4t^2\), \((2t^{1/2})^2 = 4t\), and \(1^2 = 1\).This results in \( \sqrt{4t^2 + 4t + 1} \).

Step 4: Evaluate the Integral

Evaluate the integral:\[ L = \int_0^8 \sqrt{4t^2 + 4t + 1} \, dt \].Note that \( \sqrt{4t^2 + 4t + 1} = \sqrt{(2t+1)^2} = |2t+1| \). Since \( t \) ranges from 0 to 8, \( 2t + 1 \) is always positive, so we can write it as \( \int_0^8 (2t+1) \, dt \).

Step 5: Solve the Integral

Now, solve the integral:\[ L = \int_0^8 (2t+1) \, dt = \left[ t^2 + t \right]_0^8 \].Evaluating the definite integral results in:\[ (8^2 + 8) - (0^2 + 0) = 64 + 8 - 0 = 72 \].

Step 6: Conclusion

The arc length of the curve from \( t=0 \) to \( t=8 \) is calculated to be 72.

Key Concepts

Parametric Equations

Parametric equations are a powerful tool in mathematics for describing curves. Whereas standard equations express one variable in terms of another, parametric equations allow us to use a third variable, often denoted by \( t \), to define both \( x \) and \( y \) (and sometimes \( z \)) as functions of \( t \). This is particularly useful when describing paths through space, where each coordinate of a point is dependent on time or another independent parameter. For example, in our exercise, the curve is defined as \( x = t^2 \), \( y = \frac{4}{3}t^{3/2} \), and \( z = t \).

• Each component is expressed independently, which makes them easier to differentiate or integrate.
• This method also allows the depiction of more complicated curves that can't be easily written in the form \( y = f(x) \).

Integral Calculus

Integral calculus plays a pivotal role in determining the arc length of parametric curves. In this context, the arc length is calculated using an integral that involves the derivatives of the parametric equations with respect to \( t \). The integral essentially sums up all the infinitesimally small line segments along the curve.

• We use the formula \( L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt \) to define the arc length \( L \).
• Taking derivatives \( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \) highlights changes in \( x, y, \) and \( z \) as \( t \) evolves.
• The integral then combines these changes over the interval \( [a, b] \) to determine total length.

This method allows us to account for curves in higher dimensions, significantly expanding the scope of problems we can solve with calculus.

Curve Evaluation

Evaluating the curve defined by parametric equations means calculating various properties of the curve, in this case, the arc length. This involves several steps:

• First, we find derivatives: the rate at which each coordinate changes with the parameter \( t \).
• The derivatives \( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \) are crucial for setting up the formula for arc length.
• Next, substitute these derivatives into the arc length integral.

Simplification is a key step. When we evaluate \( \sqrt{4t^2 + 4t + 1} \) and recognize it is a perfect square \( (2t + 1)^2 \), it becomes easy to integrate. Curve evaluation then culminates in integrating from \( t = 0 \) to \( t = 8 \), providing a comprehensive picture of the curve's characteristics and leading to the conclusion that the arc length is 72 units.