Question

Find the arc length of the given curve. \(x=t^{3 / 2}, y=3 t, z=4 t ; 1 \leq t \leq 4\)

Short answer

The arc length is approximately 16.82.

Step-by-step solution

Parameterize the Curve

The curve is already parameterized in terms of the variable \( t \): \( x(t) = t^{3/2} \), \( y(t) = 3t \), and \( z(t) = 4t \). These define the position of the curve in 3D space for different values of \( t \).

Determine the Derivatives

Find the derivatives of each component of the curve with respect to \( t \):\( x'(t) = \frac{d}{dt}[t^{3/2}] = \frac{3}{2} t^{1/2} \) , \( y'(t) = \frac{d}{dt}[3t] = 3 \), \( z'(t) = \frac{d}{dt}[4t] = 4 \).

Calculate the Magnitude of the Derivative Vector

The length of the derivative vector is given by \( \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \).That is:\( \sqrt{\left(\frac{3}{2} t^{1/2}\right)^2 + 3^2 + 4^2} = \sqrt{\frac{9}{4} t + 9 + 16} = \sqrt{\frac{9}{4} t + 25} \).

Set Up the Integral for Arc Length

The arc length \( L \) of a parametric curve from \( t = a \) to \( t = b \) is given by:\( L = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \, dt \).Substituting our expression, we have:\( L = \int_1^4 \sqrt{\frac{9}{4} t + 25} \, dt \).

Evaluate the Integral

To solve \( \int_1^4 \sqrt{\frac{9}{4} t + 25} \, dt \), perform a substitution.Let \( u = \frac{9}{4} t + 25 \). Then \( du = \frac{9}{4} \, dt \) or \( dt = \frac{4}{9} \, du \).When \( t = 1 \), \( u = \frac{9}{4}(1) + 25 = \frac{9}{4} + 25 = \frac{109}{4} \);when \( t = 4 \), \( u = \frac{9}{4}(4) + 25 = 9 + 25 = 34 \).Therefore, the integral becomes: \( L = \int_{109/4}^{34} \sqrt{u} \frac{4}{9} \, du \).Integrate to find: \( L = \frac{4}{9} \cdot \frac{2}{3} [u^{3/2}]^{109/4}_{34} = \frac{8}{27} [u^{3/2}]^{109/4}_{34} \).Plug in the limits:\( L = \frac{8}{27} \left(34^{3/2} - \left(\frac{109}{4}\right)^{3/2}\right) \).

Compute the Result

Calculate the following:\( 34^{3/2} = 34 \times \sqrt{34} \approx 198.57 \),\( \left(\frac{109}{4}\right)^{3/2} = \left(\frac{109}{4}\right) \times \sqrt{\frac{109}{4}} \approx 141.75 \).So your result will be:\( L = \frac{8}{27} (198.57 - 141.75) = \frac{8}{27} \times 56.82 \approx 16.82 \).

Key concepts

Parametric Equations

Parametric equations are a powerful way to describe curves in the coordinate plane and in three-dimensional space. Instead of representing a curve using a single equation, we use parameterized functions for each coordinate.
For example, in our exercise, the curve is defined by the parametric equations: \( x(t) = t^{3/2} \), \( y(t) = 3t \), and \( z(t) = 4t \). Here, \( t \) acts as the parameter, and varying \( t \) traces out the curve in 3D space.
Parametric equations are especially useful in representing curves that cannot be easily described by standard equations.

• They allow the expression of more complex paths, like loops and spirals.
• They often make it easier to compute properties of curves, such as tangents and normals, because the dependency on time (or another parameter) is explicit.

Integration Techniques

In this exercise, integration plays a crucial role in finding the arc length of the parametric curve. The general formula for computing arc length from \( t = a \) to \( t = b \) is given by the integral:
\[ L = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \, dt \]
To evaluate an integral like this, we often apply different techniques to simplify or directly solve it. One common technique, used here, is substitution.
We set \( u = \frac{9}{4} t + 25 \) and find the corresponding differential \( dt \) in terms of \( du \), which results in a more straightforward integral. Substitution is a handy skill as it can simplify many integration problems by making them resemble a familiar form.
Similarly, various techniques can be used to find integrals, such as partial fraction decomposition, trigonometric substitution, or integration by parts, each having its specific scenarios where applicable.

Derivatives

Derivatives are a fundamental tool in calculus, helping us understand rates of change. When working with parametric equations, you take the derivative of each component function separately.
The derivatives of the parametric functions provide a vector, which describes the instantaneous rate of change or velocity of the curve. For our problem:

• \( x'(t) = \frac{3}{2} t^{1/2} \)
• \( y'(t) = 3 \)
• \( z'(t) = 4 \)

These derivatives are combined to find the magnitude of the velocity vector, which is critical for computing the arc length.
The vector combines all three components of motion but only gives meaningful information when interpreted in the context of parametric curves, offering a way to measure how quickly the curve progresses through space.

3D Coordinate Geometry

3D coordinate geometry extends the principles of 2D coordinate geometry into three dimensions, adding a whole new layer to curve representation and analysis.
Using parametric equations, as in our given curve, each function component holds a specific function in the spatial geometry, making it possible to describe more complex motions and forms.
In 3D geometry, you often work with:

• **Vectors**, which represent points or directions in space.
• **Planes and lines**, which describe the orientation and intersections in 3D space.
• **Curves**, such as the one represented by \( x(t), y(t), z(t) \), which require careful computation to understand their spatial dynamics fully.

Understanding these concepts helps in visualizing the geometric aspects in computations like arc length, where each spatial direction contributes to the overall distance the curve covers in 3D space.