Question

Find the curvature \(K\) of the curve. $$ \mathbf{r}(t)=a \cos \omega t \mathbf{i}+a \sin \omega t \mathbf{j} $$

Short answer

The curvature \(K\) of the curve defined by the given vector function is \(0\).

Step-by-step solution

Determine the first derivative (velocity)

To find the first derivative of the given vector function, use the chain rule of differentiation. For the vector function \( \mathbf{r}(t) = a \cos( \omega t) \mathbf{i} + a \sin( \omega t) \mathbf{j} \), its first derivative or velocity \( \mathbf{v}(t) \) can be found as: \( \mathbf{v}(t) = - a \omega \sin(\omega t) \mathbf{i} + a \omega \cos(\omega t) \mathbf{j} \).

Determine the second derivative (acceleration)

The next step is to find the second derivative of the vector function, also known as acceleration \( \mathbf{a}(t) \). To do this, again differentiate \( \mathbf{v}(t) \) using the chain rule. \( \mathbf{a}(t) \) can then be found as: \( \mathbf{a}(t) = - a \omega^2 \cos(\omega t) \mathbf{i} - a \omega^2 \sin(\omega t) \mathbf{j} \).

Find the magnitudes of velocity and acceleration

The magnitudes of the velocity \(v(t)\), and acceleration \(a(t)\) vectors are \(v(t) = \sqrt{(-a \omega \sin(\omega t))^2 +(a \omega \cos(\omega t))^2} = a \omega \) and \(a(t) = \sqrt{(-a \omega^2 \cos(\omega t))^2 +(- a \omega^2 \sin(\omega t) )^2} = a \omega^2\), respectively.

Find the curvature

Finally, using the formula for curvature \(K = \frac{\| \mathbf{v}(t) \times \mathbf{a}(t) \|}{\| \mathbf{v}(t) \|^3}\), where \(\times\) denotes the cross product and \(\|\) denotes the magnitude of the vector, we get \( K=\frac{0}{(a \omega)^3}=0\) as the velocity and acceleration are parallel and their cross product is zero.

Conclusion

So the curvature \(K\) of the curve defined by the given vector function is \(0\).

Key concepts

Vector Function Differentiation

Understanding vector function differentiation is essential for analyzing various physical phenomena, including the curvature of a curve.

When we speak about a vector function, like the one given by \( \mathbf{r}(t)=a \cos \omega t \mathbf{i}+a \sin \omega t \mathbf{j} \), it describes a curve in terms of components along the axises, dependent on a parameter, often time.

To differentiate a vector function, we apply derivative rules component-wise, just as we would for a scalar function. In the context of the given exercise, differentiating the cosine and sine functions within each vector component involves utilizing basic trigonometric derivatives as well as employing the chain rule of differentiation. The result of such differentiation provides the velocity vector \( \mathbf{v}(t) \).

Chain Rule of Differentiation

The chain rule is a fundamental tool in calculus, used whenever we differentiate a composition of functions.

In terms of vector functions, if a vector component includes a function of a function, like \( \cos(\omega t) \) or \( \sin(\omega t) \), we differentiate the outer function first and then multiply by the derivative of the inner function—this is the essence of the chain rule. In our exercise, this rule allows us to compute both the first derivative, yielding the velocity vector, and the second derivative, yielding the acceleration vector.

Applying the chain rule ensures that we accurately capture how changes in \( t \) affect the entire vector function. Without it, we would not be able to find the correct expressions for velocity and acceleration, critical factors when determining the curvature.

Magnitude of Vectors

The magnitude of a vector describes its length or size and is a measure of how far the vector reaches in its direction from the origin.

For a vector \( \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} \), the magnitude is calculated using the Pythagorean theorem resulting in \( \| \mathbf{v} \| = \sqrt{v_x^2 + v_y^2} \). This is why, in step 3 of our solution, despite having a trigonometric function in both components of \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \), we end up with magnitudes that are simply \( a \omega \) and \( a \omega^2 \) respectively—because the squares of the sine and cosine functions sum to 1.

Calculating the magnitude of vectors is fundamental when dealing with physical quantities like velocity and acceleration and is imperative for finding the curvature of a curve since curvature is influenced by the length of these vector quantities.