To solve this problem, we need to determine when the particle’s velocity is zero, as that’s when it is closest to the origin.

Given Information

The position $p(t)$ of the particle is: $p(t) = \frac{t^4}{2} - 4t^2 + 10$ for $t > 0$ seconds.

Step 1: Find the Velocity Function

Velocity is the derivative of the position function $p(t)$: $v(t) = p'(t) = \frac{d}{dt} \left( \frac{t^4}{2} - 4t^2 + 10 \right)$

Calculating the derivative term by term:

1. $\frac{d}{dt} \left( \frac{t^4}{2} \right) = 2t^3$
2. $\frac{d}{dt} \left( -4t^2 \right) = -8t$
3. $\frac{d}{dt} (10) = 0$

Thus, $v(t) = 2t^3 - 8t$

Step 2: Set the Velocity to Zero

To find when the particle is closest to the origin, set $v(t) = 0$: $2t^3 - 8t = 0$

Factor out $2t$: $2t(t^2 - 4) = 0$

This gives: $t = 0 \quad \text{or} \quad t^2 = 4$

Since $t > 0$, we discard $t = 0$ and solve $t^2 = 4$: $t = 2$

Step 3: Find the Position at $t = 2$

Now, substitute $t = 2$ into $p(t)$ to find the position: $p(2) = \frac{(2)^4}{2} - 4 \cdot (2)^2 + 10$ $= \frac{16}{2} - 4 \cdot 4 + 10$ $= 8 - 16 + 10$ $= 2$

Conclusion

The particle is 2 meters from the origin when it is closest to it.

Answer: The closest distance to the origin is 2 meters.

Would you like more details on any step?

Here are five related questions for practice:

1. How would you find the maximum distance from the origin using this function?
2. If the position function was different, say $p(t) = t^3 - 3t + 5$, how would the solution change?
3. What does the second derivative tell us about the behavior of the particle's motion?
4. How would the solution change if we included negative values of $t$?
5. Can we determine if this point is a minimum using a second derivative test?

Tip: Always check that the value of $t$ falls within the specified domain when solving physics or calculus problems with constraints.