Problem 4 - Essential Calculus Early Transcendentals

Question

Find the length of the curve. $$\mathbf{r}(t)=12 t \mathbf{i}+8 t^{3 / 2} \mathbf{j}+3 t^{2} \mathbf{k}, \quad 0 \leqslant t \leqslant 1$$

Step-by-Step Solution

Answer

The length of the curve is 15.

Step 1: Understand the Problem

We need to find the length of the curve described by the vector function \( \mathbf{r}(t)=12 t \mathbf{i}+8 t^{\frac{3}{2}} \mathbf{j}+3 t^{2} \mathbf{k} \) over the interval \( 0 \leqslant t \leqslant 1 \). The curve length formula for a vector function \( \mathbf{r}(t) \) is given by \( L = \int_a^b \| \mathbf{r}'(t) \| \, dt \).

Step 2: Find the Derivative of \( \mathbf{r}(t) \)

The derivative \( \mathbf{r}'(t) \) is obtained by differentiating each component: \( \mathbf{r}'(t) = \frac{d}{dt}(12t) \mathbf{i} + \frac{d}{dt}(8t^{3/2}) \mathbf{j} + \frac{d}{dt}(3t^2) \mathbf{k} \). This results in \( \mathbf{r}'(t) = 12 \mathbf{i} + 12t^{1/2} \mathbf{j} + 6t \mathbf{k} \).

Step 3: Calculate the Magnitude of \( \mathbf{r}'(t) \)

The magnitude \( \| \mathbf{r}'(t) \| \) is calculated as \( \sqrt{(12)^2 + (12t^{1/2})^2 + (6t)^2} \). This simplifies to \( \sqrt{144 + 144t + 36t^2} \).

Step 4: Set Up the Integral for the Length

The length \( L \) of the curve is given by \( L = \int_0^1 \sqrt{144 + 144t + 36t^2} \, dt \).

Step 5: Simplify the Integrand

Observe that \( \sqrt{144 + 144t + 36t^2} = \sqrt{36(4 + 4t + t^2)} = 6\sqrt{(t+2)^2} = 6(t+2) \) since \( t+2 > 0 \) over \( [0,1] \).

Step 6: Evaluate the Integral

Now, integrate: \( L = 6 \int_0^1 (t+2) \, dt = 6 \left[ \frac{t^2}{2} + 2t \right]_0^1 = 6 \left( \frac{1}{2} + 2 \right) = 6 \times \frac{5}{2} = 15 \).

Step 7: Conclusion

The length of the curve is \( 15 \).

Key Concepts

Curve Length

The curve length in vector calculus is a measure of how long a curve extends over a particular interval. Think of it like unfurling a tangled string and measuring it from one end to the other. For a vector function, which assigns a position vector for each time point within an interval, the length of the curve can be computed from its derivatives. The formula to determine the length, denoted as \(L\), of a curve represented by a vector function \( \mathbf{r}(t) \) over a given interval \([a, b]\) is given by:

• \(L = \int_a^b \| \mathbf{r}'(t) \| \, dt\)

This integral uses the magnitude of the derivative of \( \mathbf{r}(t) \) to calculate the arc length over the interval. In simpler terms, it involves integrating, or summing up, the small line segments obtained from the rate of change along the curve. This formula works for space curves not defined in a simple Cartesian way but through vector functions.

Vector Functions

Vector functions play an essential role in describing paths and curves in a space with intuitive and flexible components. They take a real-valued input and return a vector output. For example, in our problem:

• \( \mathbf{r}(t) = 12t \mathbf{i} + 8t^{3/2} \mathbf{j} + 3t^2 \mathbf{k} \)

This representation is vital because it describes how an object's position changes in three-dimensional space as a function of time \(t\). Each vector function component, associated with \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), corresponds to movement along the x, y, and z axes. They're instrumental in physics for describing various motion dynamics, including projectile paths, fluid flows, or anything else moving through space over time.

Integration

Integration is the process of evaluating the area under a curve or discovering accumulated quantities, such as the length of a curve or area under a graph. Here, we applied it to find the curve length using the definite integral formula. When integrating the magnitude of a vector's derivative over a given interval, we're essentially adding up infinitesimally small elements to determine a total measure. In this context:

• We calculated \(L = \int_0^1 6(t+2) \, dt\)

The definite integral gives us a single numerical output, representing the total curve span from \(t=0\) to \(t=1\). Integration here acts as a summative process over the interval, breaking down a potentially complex form into a computable length.

Derivatives

Derivatives in vector calculus let us understand how fast and in which direction components of a vector function change with a parameter, usually time. Essentially, it's about looking at how a vector function morphs as \(t\) varies. For the vector function \( \mathbf{r}(t)\):

• \( \mathbf{r}'(t) = 12 \mathbf{i} + 12t^{1/2} \mathbf{j} + 6t \mathbf{k} \)

The derivative, represented as \( \mathbf{r}'(t) \), reveals the instantaneous velocity vector at any point \(t\). By calculating the derivative, we got the rate at which the curve's position vector is changing. It's crucial for understanding motion dynamics and was necessary here for determining the curve length as we derived each component to find how it changes concerning \(t\). This step provides the groundwork for subsequent length integration.