Question

Consider the following trajectories of moving objects. Find the tangential and normal components of the acceleration. $$\mathbf{r}(t)=\langle 20 \cos t, 20 \sin t, 30 t\rangle$$

Short answer

The tangential component of the acceleration is 0, and the normal component of the acceleration is 20.

Step-by-step solution

Find the velocity vector

Differentiate the position vector \(\mathbf{r}(t)\) with respect to time to get the velocity vector \(\mathbf{v}(t)\): $$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \langle-20\sin t, 20\cos t, 30\rangle$$

Find the acceleration vector

Differentiate the velocity vector \(\mathbf{v}(t)\) with respect to time to get the acceleration vector \(\mathbf{a}(t)\): $$\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = \langle-20\cos t, -20\sin t, 0\rangle$$

Find the unit tangent vector

To find the unit tangent vector \(\mathbf{T}(t)\), first find the magnitude of the velocity vector: $$||\mathbf{v}(t)|| = \sqrt{(-20\sin t)^2 + (20\cos t)^2 + 30^2} = \sqrt{400} = 20$$ Now, divide the velocity vector by its magnitude to get the unit tangent vector: $$\mathbf{T}(t) = \frac{\mathbf{v}(t)}{||\mathbf{v}(t)||} = \langle -\frac{\sin t}{\sqrt{2}},\frac{\cos t}{\sqrt{2}},\frac{3\sqrt{2}}{2}\rangle$$

Find the tangential component of the acceleration

The tangential component of the acceleration is the dot product of the acceleration vector and the unit tangent vector: $$a_T(t) = \mathbf{a}(t) \cdot\mathbf{T}(t) = (-20\cos t)(-\frac{\sin t}{\sqrt{2}}) + (-20\sin t)(\frac{\cos t}{\sqrt{2}}) + (0)(\frac{3\sqrt{2}}{2}) = 0$$

Find the normal component of the acceleration

The normal component of the acceleration is the vector difference between the total acceleration and the tangential component: $$\mathbf{a}_N = \mathbf{a}(t) - a_T(t)\mathbf{T}(t) = \langle -20\cos t, -20\sin t, 0\rangle - 0\langle -\frac{\sin t}{\sqrt{2}},\frac{\cos t}{\sqrt{2}},\frac{3\sqrt{2}}{2}\rangle = \langle -20\cos t, -20\sin t, 0\rangle$$ The magnitude of the normal component of the acceleration is: $$||\mathbf{a}_N|| = \sqrt{(-20\cos t)^2 + (-20\sin t)^2 + 0^2} = 20$$ So, the tangential component of the acceleration is \(0\), and the normal component of the acceleration is \(20\).