Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems 25-32, find the arc length of the given curve. } $$ $$ x=t^{3 / 2}, y=3 t, z=4 t ; 1 \leq t \leq 4 $$

Step-by-Step Solution

Answer

The arc length is approximately 16.68 units.

Step 1: Recall the formula for arc length in parametric form

The arc length of a vector-valued function \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) from \( t = a \) to \( t = b \) is given by \[L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt.\] This provides the total length traced by the parameterized curve over the interval.

Step 2: Find the derivatives

First, we need to find the derivatives of the parametric equations with respect to \( t \):- \( \frac{dx}{dt} = \frac{d}{dt}(t^{3/2}) = \frac{3}{2}t^{1/2} \).- \( \frac{dy}{dt} = \frac{d}{dt}(3t) = 3 \).- \( \frac{dz}{dt} = \frac{d}{dt}(4t) = 4 \).

Step 3: Substitute derivatives into the arc length formula

Substitute these expressions into the arc length integral formula:\[L = \int_1^4 \sqrt{ \left( \frac{3}{2}t^{1/2} \right)^2 + (3)^2 + (4)^2 } \, dt.\]

Step 4: Simplify inside the square root

Simplify the expression under the square root:\[\left( \frac{3}{2}t^{1/2} \right)^2 = \frac{9}{4}t,\]\[L = \int_1^4 \sqrt{ \frac{9}{4}t + 9 + 16 } \, dt,\]\[L = \int_1^4 \sqrt{ \frac{9}{4}t + 25 } \, dt.\]

Step 5: Evaluate the integral

To solve this integral, first recognize that it takes the form of a standard integration problem that can be solved by substitution. Let \( u = \frac{9}{4}t + 25 \), then \( du = \frac{9}{4} dt \). So, \( dt = \frac{4}{9} du \). The limits of integration change accordingly as \( t = 1 \) gives \( u = \frac{9}{4} \times 1 + 25 = 27.25 \), and as \( t = 4 \) gives \( u = \frac{9}{4} \times 4 + 25 = 34 \).We substitute back and change limits:\[L = \int_{27.25}^{34} \sqrt{u} \frac{4}{9} \, du = \frac{4}{9} \int_{27.25}^{34} u^{1/2} \, du.\]Integrate:\[L = \frac{4}{9} \left[ \frac{2}{3} u^{3/2} \right]_{27.25}^{34} = \frac{8}{27} \left( (34^{3/2}) - (27.25^{3/2}) \right).\]

Step 6: Calculate the definite integral

Compute the values:\[34^{3/2} = 34 \times \sqrt{34} \approx 197.403\]\[27.25^{3/2} = 27.25 \times \sqrt{27.25} \approx 141.097\]Substitute these values back into the evaluated integral:\[L = \frac{8}{27} \times (197.403 - 141.097) \approx \frac{8}{27} \times 56.306 \approx 16.68.\]

Step 7: Final Answer

Hence, the arc length of the curve from \( t=1 \) to \( t=4 \) is approximately 16.68 units.

Key Concepts

Parametric Equations

Parametric equations are a set of equations that express the coordinates of the points of a curve as functions of a variable, often denoted as \( t \). These equations allow for describing a path in a more flexible way, as opposed to using a single function like \( y = f(x) \).
In the context of our exercise, the parametric equations are given as:

• \( x = t^{3/2} \)
• \( y = 3t \)
• \( z = 4t \)

These equations help us describe the trajectory of the curve in three-dimensional space by using a parameter \( t \), which varies within a given interval \( 1 \leq t \leq 4 \).
This form of representation is particularly useful in vector calculus, as it provides a clear picture of movement through space.

Integral Calculus

Integral calculus is the branch of mathematics concerned with the concepts of integrals. It deals with finding the quantity where the rate of change is known.
An integral, generally speaking, is the mathematical tool used to calculate areas, volumes, central points, and arc lengths, among others.
The fundamental idea in our exercise is applying integral calculus to determine the arc length of a curve described by parametric equations.
To find the arc length \( L \) of a curve, we use the formula:

• \[ L = \int_a^b \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 } \, dt. \]

This expression includes differentiation, followed by integration, which are key operations of calculus that help in determining the overall length of the curve traced by the parameter \( t \).

Vector-Valued Functions

Vector-valued functions extend the concept of parametric equations by encapsulating them into a vector form. Each component of the vector represents a function of \( t \), which are coordinates in a multidimensional space.
For this exercise, the vector-valued function \( \mathbf{r}(t) \) is:

• \( \mathbf{r}(t) = \langle t^{3/2}, 3t, 4t \rangle \)

This means that the function uniquely determines a point in space for every value of \( t \).
The beauty of using vector-valued functions lies in their ability to convey information about an object's position in physics or other multidimensional movements succinctly.
When applied to computing arc lengths, we take the derivative of each component, and integrate their resultant magnitudes over the specified interval.

Definite Integrals

Definite integrals are a concept in calculus used to find the exact net area under a curve over a specific interval.
They are crucial in solving problems that involve a summation of continuous data points, like finding distances or areas.
In the exercise, we're tasked with calculating the arc length between \( t = 1 \) and \( t = 4 \). This is done by integrating a function from one limit to another, giving us a scalar value as a result.
In mathematical terms, the solution to our problem involves evaluating:

• \[ L = \int_{27.25}^{34} \sqrt{u} \frac{4}{9} \, du \]

To evaluate this integral, we adjust the function and limits using substitution, transforming the problem into one solvable by standard calculus techniques.
This definite integral provides a precise calculation of the total arc length, accounting for every tiny segment of the curve within the chosen interval of \( t \).