Questions: If you already know aN and v, then the formula aN = κv^2 gives a convenient way to find the curvature. Use it to find the curvature and radius of curvature of the curve r(t) = (cos t + t sin t) i + (sin t - t cos t) j, t>0. The curvature is

Solution Steps

The position vector is given by \(\mathbf{r}(t) = (\cos t + t \sin t) \mathbf{i} + (\sin t - t \cos t) \mathbf{j}\).

First, find the velocity vector \(\mathbf{v}(t)\) by differentiating \(\mathbf{r}(t)\) with respect to \(t\): \[ \mathbf{v}(t) = \frac{d}{dt}[(\cos t + t \sin t) \mathbf{i} + (\sin t - t \cos t) \mathbf{j}] = (-t \cos t) \mathbf{i} + (t \sin t) \mathbf{j} \]

Next, find the acceleration vector \(\mathbf{a}(t)\) by differentiating \(\mathbf{v}(t)\) with respect to \(t\): \[ \mathbf{a}(t) = \frac{d}{dt}[(-t \cos t) \mathbf{i} + (t \sin t) \mathbf{j}] = (-\cos t - t \sin t) \mathbf{i} + (\sin t + t \cos t) \mathbf{j} \]

The magnitude of the velocity vector \(|\mathbf{v}(t)|\) is: \[ |\mathbf{v}(t)| = \sqrt{(-t \cos t)^2 + (t \sin t)^2} = \sqrt{t^2 \cos^2 t + t^2 \sin^2 t} = \sqrt{t^2} = t \]

The normal component of the acceleration \(a_N\) is given by: \[ a_N = \sqrt{|\mathbf{a}(t)|^2 - \left(\frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{|\mathbf{v}(t)|}\right)^2} \]

First, calculate \(\mathbf{a}(t) \cdot \mathbf{v}(t)\): \[ \mathbf{a}(t) \cdot \mathbf{v}(t) = [(-\cos t - t \sin t)(-t \cos t) + (\sin t + t \cos t)(t \sin t)] \] \[ = t \cos^2 t + t^2 \sin t \cos t + t \sin^2 t + t^2 \sin t \cos t = t (\cos^2 t + \sin^2 t) + 2t^2 \sin t \cos t = t + 2t^2 \sin t \cos t \]

Now, calculate \(|\mathbf{a}(t)|^2\): \[ |\mathbf{a}(t)|^2 = (-\cos t - t \sin t)^2 + (\sin t + t \cos t)^2 \] \[ = \cos^2 t + 2t \sin t \cos t + t^2 \sin^2 t + \sin^2 t + 2t \sin t \cos t + t^2 \cos^2 t \] \[ = (\cos^2 t + \sin^2 t) + 2t \sin t \cos t + t^2 (\sin^2 t + \cos^2 t) + 2t \sin t \cos t \] \[ = 1 + 4t \sin t \cos t + t^2 \]

Substitute these into the formula for \(a_N\): \[ a_N = \sqrt{1 + 4t \sin t \cos t + t^2 - \left(\frac{t + 2t^2 \sin t \cos t}{t}\right)^2} \] \[ = \sqrt{1 + 4t \sin t \cos t + t^2 - (1 + 2t \sin t \cos t)^2} \] \[ = \sqrt{1 + 4t \sin t \cos t + t^2 - (1 + 4t \sin t \cos t + 4t^2 \sin^2 t \cos^2 t)} \] \[ = \sqrt{t^2 - 4t^2 \sin^2 t \cos^2 t} \] \[ = t \sqrt{1 - 4 \sin^2 t \cos^2 t} \]

The curvature \(\kappa\) is given by: \[ \kappa = \frac{a_N}{|\mathbf{v}(t)|^2} = \frac{t \sqrt{1 - 4 \sin^2 t \cos^2 t}}{t^2} = \frac{\sqrt{1 - 4 \sin^2 t \cos^2 t}}{t} \]

The radius of curvature \(R\) is the reciprocal of the curvature: \[ R = \frac{1}{\kappa} = \frac{t}{\sqrt{1 - 4 \sin^2 t \cos^2 t}} \]

Final Answer