Question

Consider the following trajectories of moving objects. Find the tangential and normal components of the acceleration. $$\mathbf{r}(t)=\left\langle t, t^{2}+1\right\rangle$$

Short answer

Answer: The tangential component of the acceleration is $$a_\text{T}(t) = \frac{4t}{\sqrt{1+4t^2}}$$, and the normal component of the acceleration is $$a_\text{N}(t) =\sqrt{4-\frac{16t^2}{1+4t^2}}$$.

Step-by-step solution

Find the position vector\(\mathbf{r}(t)\)

The position vector is already given: $$\mathbf{r}(t)=\langle t, t^{2}+1\rangle$$

Calculate the velocity vector \(\mathbf{v}(t)\)

Taking the first derivative of \(\mathbf{r}(t)\) with respect to \(t\), we get the velocity vector: $$\mathbf{v}(t)=\frac{d\mathbf{r}}{dt}=\left\langle 1, 2t\right\rangle$$

Calculate the acceleration vector \(\mathbf{a}(t)\)

Taking the first derivative of \(\mathbf{v}(t)\) with respect to \(t\), we get the acceleration vector: $$\mathbf{a}(t)=\frac{d\mathbf{v}}{dt}=\left\langle 0, 2\right\rangle$$

Calculate the speed \(v(t)\)

The magnitude of the velocity vector \(\mathbf{v}(t)\) is the speed \(v(t)\). We calculate it as follows: $$v(t)=|\mathbf{v}(t)|=\sqrt{1^2+(2t)^2}=\sqrt{1+4t^2}$$

Calculate the unit tangent vector \(\mathbf{T}(t)\)

The unit tangent vector \(\mathbf{T}(t)\) is given by dividing the velocity vector by its magnitude: $$\mathbf{T}(t)=\frac{\mathbf{v}(t)}{v(t)}=\left\langle\frac{1}{\sqrt{1+4t^2}},\frac{2t}{\sqrt{1+4t^2}}\right\rangle$$

Find the tangential component of the acceleration

The tangential component of the acceleration, \(a_\text{T}(t)\), is given by the dot product of \(\mathbf{a}(t)\) and \(\mathbf{T}(t)\): $$a_\text{T}(t) = \mathbf{a}(t) \cdot \mathbf{T}(t)=\left\langle 0, 2\right\rangle \cdot \left\langle\frac{1}{\sqrt{1+4t^2}},\frac{2t}{\sqrt{1+4t^2}}\right\rangle = (0)(\frac{1}{\sqrt{1+4t^2}})+(2)(\frac{2t}{\sqrt{1+4t^2}})=\frac{4t}{\sqrt{1+4t^2}}$$

Find the normal component of the acceleration

Finally, the normal component of the acceleration, \(a_\text{N}(t)\), is given by the square root of the difference between the square of the magnitude of the acceleration vector and the square of the tangential component: $$a_\text{N}(t)=\sqrt{\left|\mathbf{a}(t)\right|^2-a_\text{T}^2(t)}=\sqrt{(0^2+2^2)-\left(\frac{4t}{\sqrt{1+4t^2}}\right)^2}=\sqrt{4-\frac{16t^2}{1+4t^2}}$$ Thus, the tangential component of the acceleration is \(a_\text{T}(t) = \frac{4t}{\sqrt{1+4t^2}}\) and the normal component of the acceleration is \(a_\text{N}(t) =\sqrt{4-\frac{16t^2}{1+4t^2}}\).