3v_1 - (11 + 3v_1) = 5 \implies -11 = 5, \text{ which is a contradiction.}
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- From the third equation, $v_2 = 6 + 2v_1$. Substituting into the first equation:
- 2v_3 - 3(6 + 2v_1) = 4 \implies 2v_3 - 18 - 6v_1 = 4 \implies 2v_3 = 22 + 6v_1 \implies v_3 = 11 + 3v_1.
- Substitute $v_3$ into the second equation:
- However, the condition $\mathbf{v} \cdot \mathbf{a} = 0$ implies $v_1 + 2v_2 + 3v_3 = 0$. Substituting $v_2 = 6 + 2v_1$ and $v_3 = 11 + 3v_1$:
- v_1 + 2(6 + 2v_1) + 3(11 + 3v_1) = 0 \implies v_1 + 12 + 4v_1 + 33 + 9v_1 = 0 \implies 14v_1 + 45 = 0 \implies v_1 = -\frac{45}{14}.
- Then $v_2 = 6 + 2(-\frac{45}{14}) = \frac{84 - 90}{14} = -\frac{6}{14} = -\frac{3}{7}$, and $v_3 = 11 + 3(-\frac{45}{14}) = \frac{154 - 135}{14} = \frac{19}{14}$.
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Now substitute $ t = 5 $ into $ E(t) $:
So, the maximum elevation change is:
This is an upward-opening parabola, so the minimum occurs at the vertex. The $ t $-coordinate of the vertex is:
So, the minimum storage usage occurs in:
This is a quadratic equation, and using the quadratic formula, we find:
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