Question

If r(t) = (2t3)i + (5t)j- (t2)k is a position vector, at

what time the velocity and acceleration vectors are
orthogonal?

Answer 1

Only at
`t = 0`.

Explanation:

Differentiate
`\vec{r}(t)` (the position vector at time
`t`) with respect to
`t\!` to find expressions for the velocity vector
`\vec{v}(t)` and the acceleration vector
`\vec{a}(t)`.

`\begin{aligned} & \vec{v}(t) \\ &= (d)/(dt)\left[\vec{r}(t)\right] \\ &= (d)/(dt)\left[\left(2\, t^(3)\right)\, \vec{i} + (5\, t) \, \vec{j} - \left(t^(2)\right)\, \vec{k}\right] \\ &= \left((d)/(dt)\left[2\, t^3\right]\right)\, \vec{i} + \left((d)/(dt)\left[5\, t\right]\right)\, \vec{j} - \left((d)/(dt)\left[t^2\right]\right)\, \vec{k} \\ &= \left(6\, t^2\right)\, \vec{i} + 5\, \vec{j} - 2\, t\, \vec{k} \end{aligned}`.

`\begin{aligned} & \vec{a}(t) \\ &= (d)/(dt)\left[\vec{v}(t)\right] \\ &= (d)/(dt)\left[\left(6\, t^(2)\right)\, \vec{i} + 5 \, \vec{j} - \left(2\, t\right)\, \vec{k}\right] \\ &= \left((d)/(dt)\left[6\, t^2\right]\right)\, \vec{i} + \left((d)/(dt)\left[5\right]\right)\, \vec{j} - \left((d)/(dt)\left[2\, t\right]\right)\, \vec{k} \\ &= \left(12\, t\right)\, \vec{i} + 0\, \vec{j} - 2\, \vec{k} \end{aligned}`.

Two vectors are orthogonal to one another when their dot product is zero.

Calculate the dot product of
`\vec{v}(t)` and
`\vec{a}(t)`:

`\begin{aligned}& \left(\vec{v}(t)\right) \cdot \left(\vec{a}(t)\right) \\ &= \left(6 \, t^2\right) \cdot (12\, t) + 5 * 0 + (-2\, t) \cdot (-2) \\ &= 72\, t^3 + 4\, t\end{aligned}`.

Set the dot product of
`\vec{v}(t)` and
`\vec{a}(t)` to zero and solve for
`t`:

`18\, t^3 + t = 0`.

`t\, \left(18\,t ^2 + 1\right) = 0`

The only real root of this equation is
`t = 0`.

Therefore,
`\vec{v}(t)` and
`\vec{a}(t)` are orthogonal to one another only at
`t = 0`.