Question

Prove that the curve \(\mathbf{r}(t)=\langle a \cos t, b \sin t, c \sin t\rangle\) is parameterized by arc length provided \(a^{2}=b^{2}+c^{2}=1\)

Short answer

Answer: Yes, the curve \(\mathbf{r}(t)=\langle a \cos t, b \sin t, c \sin t\rangle\) is parameterized by arc length because the magnitude of its derivative with respect to \(t\) is equal to \(1\) for all values of \(t\).

Step-by-step solution

Compute the derivative of the curve with respect to t

First, we will compute the derivative of the curve with respect to the parameter \(t\). The derivative is obtained by finding the derivative of each component of the vector function with respect to \(t\): \(\frac{d\mathbf{r}}{dt}=\frac{d}{dt}\langle a \cos t, b \sin t, c \sin t\rangle=\langle -a \sin t, b \cos t, c \cos t\rangle\)

Find the magnitude of the derivative

Next, we will find the magnitude of the derivative vector. This is needed to check if the curve is parameterized by arc length. The magnitude is given by the square root of the sum of the squares of each component: \(\left|\frac{d\mathbf{r}}{dt}\right| = \sqrt{(-a \sin t)^{2} + (b \cos t)^{2} + (c \cos t)^{2}}\)

Simplify the magnitude expression

Now, we will simplify the expression for the magnitude: \(\begin{aligned} \left|\frac{d\mathbf{r}}{dt}\right| &= \sqrt{a^{2}\sin^{2}t + b^{2}\cos^{2}t + c^{2}\cos^{2}t} \\ &= \sqrt{(b^{2}+c^{2})\cos^{2}t + a^{2}\sin^{2}t} \end{aligned}\) Remember that \(a^{2} = b^{2} + c^{2} = 1\), so we can substitute \(1\) for \((b^{2}+c^{2})\) in the expression for the magnitude: \(\left|\frac{d\mathbf{r}}{dt}\right| = \sqrt{1\cos^{2}t + 1\sin^{2}t} = \sqrt{\cos^{2}t + \sin^{2}t}\)

Show that the magnitude of the derivative is equal to 1

Finally, we need to show that the magnitude of the derivative is equal to \(1\). This is easily done by using the Pythagorean identity \(\sin^{2}t + \cos^{2}t = 1\): \(\left|\frac{d\mathbf{r}}{dt}\right| = \sqrt{1} = 1\) Since the magnitude of the derivative of \(\mathbf{r}(t)\) is equal to \(1\) for all values of \(t\), we can conclude that the curve is parameterized by arc length, as required.

Key concepts

Calculus in Motion

Calculus, the mathematical study of continuous change, plays an essential role in understanding the properties of objects when they are in motion. In the context of the problem, we used calculus to analyze the properties of a curve defined by a vector function. Calculus allows us to compute derivatives, which, in the case of motion, give us the velocity of the object at any point in time. The derivative tells us not only how fast an object is moving but also in which direction.

Vector Function Derivative

The derivative of a vector function works similarly to that of a scalar function in calculus. However, instead of resulting in just a single number, the derivative of a vector function results in another vector. This vector gives us the instantaneous rate of change of each component of the function. In our exercise, the derivative of the vector function \( \mathbf{r}(t) \) with respect to time \( t \) provided us with the instantaneous velocity vector of the curve's motion at any given time. Understanding the vector derivative is crucial for investigating how quantities such as position change over time.

Magnitude of a Vector

The magnitude of a vector represents its length or size and is essential for understanding the derivative's scale in the context of motion. In physics, when we speak about the velocity's magnitude, we are referring to speed. The magnitude is found by taking the square root of the sum of the squares of its components, as shown by the calculation \( |\frac{d\mathbf{r}}{dt}| \) in the solution. When the magnitude of our velocity vector is constant and equal to 1, it indicates the curve is parameterized by arc length, which means the curve's length corresponds exactly to the parameter \( t \) without any scaling.

Pythagorean Identity

The Pythagorean identity is a fundamental result in trigonometry that states \( \sin^{2}t + \cos^{2}t = 1 \). This identity derives from the Pythagorean theorem and the unit circle. It's especially useful in simplifying expressions involving trigonometric functions. In our exercise, it allowed us to prove that the magnitude of the velocity vector is 1 along the entire curve, thereby confirming that the vector function \( \mathbf{r}(t) \) was indeed parameterized by arc length. This identity reveals much about how trigonometric functions are interconnected and play a critical role in various areas of mathematics, including calculus.