Question

what force is required so that a particle of mass m has the position function r(t) = t³i + 3t²j + t³k? f(t) =

Explanation:

Step1: Find the velocity function

The velocity function \( \mathbf{v}(t)\) is the derivative of the position function \( \mathbf{r}(t)\).
If \( \mathbf{r}(t)=t^{3}\mathbf{i}+3t^{2}\mathbf{j}+t^{3}\mathbf{k}\), then using the power rule \((x^{n})^\prime = nx^{n - 1}\), we have:
\(\mathbf{v}(t)=\mathbf{r}^\prime(t)=3t^{2}\mathbf{i}+6t\mathbf{j}+3t^{2}\mathbf{k}\)

Step2: Find the acceleration function

The acceleration function \( \mathbf{a}(t)\) is the derivative of the velocity function \( \mathbf{v}(t)\).
Differentiating \( \mathbf{v}(t)=3t^{2}\mathbf{i}+6t\mathbf{j}+3t^{2}\mathbf{k}\) with respect to \(t\) using the power rule:
\(\mathbf{a}(t)=\mathbf{v}^\prime(t)=(3\times2t)\mathbf{i}+6\mathbf{j}+(3\times 2t)\mathbf{k}=6t\mathbf{i}+6\mathbf{j}+6t\mathbf{k}\)

Step3: Find the force function

By Newton's second law \( \mathbf{F}(t)=m\mathbf{a}(t)\).
Since \( \mathbf{a}(t)=6t\mathbf{i}+6\mathbf{j}+6t\mathbf{k}\) and \(m\) is the mass of the particle, then \( \mathbf{F}(t)=m(6t\mathbf{i}+6\mathbf{j}+6t\mathbf{k}) = 6mt\mathbf{i}+6m\mathbf{j}+6mt\mathbf{k}\)

Answer:

\(6mt\mathbf{i}+6m\mathbf{j}+6mt\mathbf{k}\)