# The position function of a particle is given by r(t) = (t^2, 9t, t^2 - 16t). When is the speed a...

## Question:

The position function of a particle is given by {eq}r(t) = \left \langle t^2, 9t, t^2 - 16t \right \rangle {/eq}. When is the speed a minimum?

## Motion:

Recall that the velocity is the time derivative of the position, and that the speed is the magnitude of the velocity. To minimize it we will use differentiation by getting the derivative, setting it equal to 0, and solving.

## Answer and Explanation:

We differentiate to find the velocity:

{eq}\begin{align*} \vec v (t) &= \frac{d}{dt} \left< t^2, 9t, t^2 - 16t \right> \\ &= \left< 2t, 9, 2t - 16 \right> \end{align*} {/eq}

And so the speed is

{eq}\begin{align*} | \vec v (t) | &= \sqrt{\left( 2t \right)^2+\left( 9 \right)^2+\left( 2t-16 \right)^2} \\ &= \sqrt{4t^2 + 81 + 4t^2 - 64t + 256} \\ &= \sqrt{8t^2 - 64t + 337} \end{align*} {/eq}

Minimizing the speed is the same as minimizing the square of the speed, so we can simply minimize

{eq}\begin{align*} 8t^2 - 64t + 337 \end{align*} {/eq}

which we instantly recognize as an upward open parabola, which means its only extremum is a minimum, and thus the minimum we seek. We differentiate and set equal to 0 to find the minimum:

{eq}\begin{align*} \frac{d}{dt} \left( 8t^2 - 64t + 337 \right) &= 0 \\ 16t - 64 &= 0 \\ 16t &= 64 \\ t &= 4 \end{align*} {/eq}

And so the minimum speed is

{eq}\begin{align*} | \vec v (4) | &= \sqrt{8(4)^2 - 64(4) + 337} \\ &\approx 14.46 \end{align*} {/eq}

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