Question

Find the arc length of \(\vec{r}(t)\) on the indicated interval. $$ \vec{r}(t)=\langle 5 \cos t, 3 \sin t, 4 \sin t\rangle \text { on }[0,2 \pi] $$

Short answer

The arc length is \(10\pi\).

Step-by-step solution

Understand the Problem

We need to find the arc length of the vector function \( \vec{r}(t) = \langle 5 \cos t, 3 \sin t, 4 \sin t \rangle \) over the interval \([0, 2\pi]\). Arc length for a vector function is given by the integral of the magnitude of its derivative over the given interval.

Find the Derivative of \(\vec{r}(t)\)

First, compute the derivative of the vector function: \( \vec{r}'(t) = \left\langle \frac{d}{dt}(5\cos t), \frac{d}{dt}(3\sin t), \frac{d}{dt}(4\sin t) \right\rangle = \left\langle -5\sin t, 3\cos t, 4\cos t \right\rangle \).

Compute the Magnitude of \(\vec{r}'(t)\)

Calculate the magnitude of the derivative: \[ |\vec{r}'(t)| = \sqrt{(-5\sin t)^2 + (3\cos t)^2 + (4\cos t)^2} = \sqrt{25 \sin^2 t + 9 \cos^2 t + 16\cos^2 t} \].

Simplify the Expression

Simplify the expression inside the square root: \[ 25 \sin^2 t + 9 \cos^2 t + 16 \cos^2 t = 25 \sin^2 t + 25 \cos^2 t\]. This simplifies further to \( 25(\sin^2 t + \cos^2 t) = 25 \), since \( \sin^2 t + \cos^2 t = 1 \).

Evaluate the Arc Length Integral

Now, compute the arc length by integrating the magnitude from Step 4 over the interval \([0, 2\pi]\):\[ \int_0^{2\pi} |\vec{r}'(t)| dt = \int_0^{2\pi} 5 dt \].

Calculate the Final Integral

Evaluate the integral: \[ \int_0^{2\pi} 5 dt = 5t \Big|_0^{2\pi} = 5 (2\pi) - 5(0) = 10\pi \].

Conclusion

The arc length of the vector function \( \vec{r}(t) \) over the interval \([0, 2\pi]\) is \( 10\pi \).

Key concepts

Vector Calculus

Vector calculus is an essential branch of mathematics focusing on differentiation and integration of vector fields, mainly in 2D and 3D spaces. Here, we calculate properties like velocity, acceleration, and arc length. For any vector function, we can find its various derivatives and integrals, to understand the behavior of objects in motion. In our exercise, we deal with the vector function \[ \vec{r}(t) = \langle 5 \cos t, 3 \sin t, 4 \sin t \rangle \] moving in space over the interval \([0, 2\pi]\). Our goal is to find its arc length, a measure of the distance traveled by the point as the parameter \(t\) varies. Understanding this concept is crucial in physics, engineering, and computer graphics, where physical paths need to be quantified. Vector calculus tools such as divergence and curl also provide insights into fluid dynamics and electric fields that help in analyzing multidimensional fields and their flux.

Integration

Integration is a core concept in calculus, used for finding areas under curves, accumulated quantities, and in our case, arc lengths of curves represented by functions. For vector functions, integration is carried out over the parameter of the function, here denoted by \(t\). In the original exercise, we needed to integrate the magnitude of the derivative of the vector function over the interval \([0, 2\pi]\). The integral is given by \[ \int_0^{2\pi} \lvert \vec{r}'(t) \rvert \, dt \] which involves calculating the magnitude \(\lvert \vec{r}'(t) \rvert\) first. Integration helps in situations where direct measurement isn’t possible, and we use mathematical formulations to find the required values. In physics, this might relate to computing work done by force fields where direct computation is complex. Integration techniques ease the challenge by breaking these complex paths and journeys into simpler, manageable parts.

Derivative of Vector Functions

Derivatives play a pivotal role when it comes to understanding vector functions. They convey information about the rate of changes -- for instance, velocity is the derivative of position in physics. Finding the derivative of a vector function involves differentiating each component individually. In our exercise, the derivative of the vector function \[ \vec{r}(t) = \langle 5 \cos t, 3 \sin t, 4 \sin t \rangle \] is \[ \vec{r}'(t) = \langle -5 \sin t, 3 \cos t, 4 \cos t \rangle \]. This step is essential for computing the arc length because only after finding the derivative can we proceed to compute the magnitude of this derivative. The resulting magnitude then fed into the integral gives us the arc length. Notably, derivatives in vector calculus are used in optimizing problems, analyzing motion, and predicting future movement or states of systems. Mastery of finding and using derivatives of vector functions is a must for tackling more advanced problems in mathematics and applied sciences.