Question

Sketch the plane curve and find its length over the given interval. $$ \mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j}, \quad[0,2 \pi] $$

Short answer

The plane curve is a circle with radius \(a\) centered at the origin, and its length over the interval is \(2\pi a .\)

Step-by-step solution

Graph the curve

The provided equation represents a circle with radius 'a', centered at the origin. This is because it is in form of \(\mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j}\). The parameter t ranges from 0 to 2π, which covers the entire circle. Hence, the graph of this vector function is a circle of radius 'a' centered at origin.

Find the derivative

To find the length of the curve, the first step is to find the derivative of the vector function \(\mathbf{r}(t)\). The derivative of \(\mathbf{r}(t)\) is \(\mathbf{r}'(t) = -a \sin(t) \mathbf{i} + a \cos(t) \mathbf{j}\).

Calculate magnitude of the derivative

After finding the derivative, the next step is to calculate the magnitude of the derivative, which is given by \(\| \mathbf{r}'(t) \| = \sqrt{(-a \sin(t))^2 + (a \cos(t))^2} = \sqrt{a^2\sin^2(t) + a^2\cos^2(t)} = a \), since \(\sin^2(t) + \cos^2(t) = 1.\)

Compute the length

Now, we can calculate the length of the curve over the interval \([0, 2\pi]\). The length \(L\) of a curve from \(a\) to \(b\) is given by \(L = \int_a^b \| \mathbf{r}'(t) \| dt\). So, substituting into the given formula, we get \(L = \int_0^{2\pi} a dt = a \bigg[ t \bigg]_0^{2\pi} = 2\pi a . \)