Question

Determine whether the following curves use arc length as a parameter. If not, find a description that uses arc length as a parameter. $$\mathbf{r}(t)=\langle 5 \cos t, 3 \sin t, 4 \sin t\rangle, \text { for } 0 \leq t \leq \pi$$

Short answer

If not, find a description of the curve that uses arc length as a parameter. **Solution:** The curve \(\mathbf{r}(t)=\langle 5 \cos t, 3 \sin t, 4 \sin t\rangle\) uses arc length as a parameter because the magnitude of its derivative is constant. Therefore, the curve's description in terms of arc length is already given by the equation \(\mathbf{r}(t)=\langle 5 \cos t, 3 \sin t, 4 \sin t\rangle\).

Step-by-step solution

Calculate the derivative of the given curve

We are given the vector equation \(\mathbf{r}(t)=\langle 5 \cos t, 3 \sin t, 4 \sin t\rangle\). To calculate its derivative, differentiate each component with respect to the parameter \(t\): $$\frac{d\mathbf{r}}{dt}=\frac{d}{dt}\langle 5 \cos t, 3 \sin t, 4 \sin t \rangle = \langle -5 \sin t, 3 \cos t, 4 \cos t \rangle$$

Find the magnitude of the derivative

To find the magnitude of the derivative, we apply the Pythagorean theorem for vectors: $$\left\lVert \frac{d\mathbf{r}}{dt} \right\rVert = \sqrt{(-5 \sin t)^2 + (3 \cos t)^2 + (4 \cos t)^2} = \sqrt{25 \sin^2 t + 9 \cos^2 t + 16 \cos^2 t}$$

Determine if the magnitude is constant

Now we want to check if the magnitude of the derivative is constant or not: $$\sqrt{25 \sin^2 t + 9 \cos^2 t + 16 \cos^2 t} = \sqrt{25\sin^2 t + 25 \cos^2 t} = \sqrt{25(\sin^2 t + \cos^2 t)} = \sqrt{25} = 5$$ The magnitude is constant, which means that the given curve uses arc length as a parameter.

Conclusion

As the magnitude of the derivative of the given curve is constant, this curve uses arc length as a parameter. Therefore, the curve's description in terms of arc length is already given by the equation \(\mathbf{r}(t)=\langle 5 \cos t, 3 \sin t, 4 \sin t\rangle\).

Key concepts

Vector-Valued Functions

Vector-valued functions are a crucial concept in multivariable calculus, connecting mathematics with real-world physical phenomena. Think of these functions as a bridge from the scalar world, where numbers stand alone, to a realm where direction and magnitude both play roles. A vector-valied function, like \( \mathbf{r}(t) \), maps each real number \( t \) to a vector in \( \mathbb{R}^n \). For example, \( \mathbf{r}(t)=\langle 5 \cos t, 3 \sin t, 4 \sin t\rangle \) is a 3-dimensional vector-valued function where \( t \) represents time or some other parameter, and the function outputs a point's location in space at that time.

Vector-valued functions can represent curves in space. In the given exercise, \( \mathbf{r}(t) \) describes a curve in 3D space as \( t \) varies, and the components of the vectors are the x, y, and z coordinates along the curve. The parameter \( t \) itself could represent anything from time to temperature, but when it represents the arc length of the curve, we're saying we can measure the curve by the distance traveled along it—a concept intertwined with physical contexts like the path of a particle.

Derivative of Vector Function

The derivative of a vector function is akin to velocity in physics; it reveals how the function's output is changing at any given moment. For a vector function \( \mathbf{r}(t) \) describing a position in space, its derivative \( \frac{d\mathbf{r}}{dt} \) gives us the velocity vector at time \( t \).

To calculate it, you take the derivative of each component function of \( \mathbf{r}(t) \) with respect to \( t \)—just like you'd take the derivative of a regular function. In the exercise, the derivative \( \frac{d\mathbf{r}}{dt} = \langle -5 \sin t, 3 \cos t, 4 \cos t \rangle \) tells us how fast and in what direction the curve's position is changing over time. A constant or zero derivative tells us a lot about the function's behavior, like whether it's moving at uniform speed or staying still.

Magnitude of a Vector

The magnitude of a vector reflects its 'size'. For a spatial analogy, if the vector were an arrow, the magnitude would be the length of the arrow. To find it, you can picture a right triangle where the vector is the hypotenuse, with its components along the axes forming the other sides.

In mathematical terms, the magnitude of vector \( \mathbf{v} \) is notated as \( \Vert \mathbf{v} \Vert \) and calculated using the square root of the sum of its components squared, \( \Vert \mathbf{v} \Vert = \sqrt{v_1^2 + v_2^2 + \dots + v_n^2} \) for \( \mathbf{v}=\langle v_1, v_2, ..., v_n \rangle \). In the given problem, the magnitude of the derivative gives us the speed of the curve’s movement through space. A constant magnitude indicates uniform speed, which relates to the concept of arc length parameterization.

Pythagorean Theorem in Vectors

The Pythagorean theorem is not just for triangles in high school geometry; it extends into the realm of vectors as well. This theorem helps compute the magnitude of a vector, affirming that the square of the length of the hypotenuse (the vector itself) equals the sum of the squares of the lengths of the other two sides (the vector's components).

In vector terms, for a 2D vector \( \mathbf{v} \) with components \( v_x \) and \( v_y \) we have \( \Vert \mathbf{v} \Vert^2 = v_x^2 + v_y^2 \). This scales up to three dimensions (or more) as in our exercise, where the magnitude of the derivative \( \frac{d\mathbf{r}}{dt} \) was found using the Pythagorean theorem, leading to the simplification \( \sqrt{25 \sin^2 t + 25 \cos^2 t} = \sqrt{25(\sin^2 t + \cos^2 t)} = 5 \), which tells us the curve in question moves at a consistent speed, confirming that its parameterization by \( t \) is indeed by arc length.