The Growing Tetrahedron

Calculus Level pending

The vectors $<3, 2, -1>, <1, 0, 2>,$ and $<t^2, -t^2, 2t^2>$ define a tetrahedron, where $t$ is the time elapsed in seconds. In cubic units per second, at what rate does the volume of this tetrahedron changing when $t = 3$ ?

98 14 7 42

1 solution

Andrew Ellinor
Aug 27, 2015

The determinant of the matrix below gives the volume of the tetrahedron formed by the three vectors.

$V = \frac{\left| \begin{array}{ccc} 3 & 2 & -1 \\ 1 & 0 & 2 \\ t^2 & -t^2 & 2t^2 \end{array} \right|}{6} = \frac{7t^2}{6}$

We take the derivative of the volume with respect to time and then evaluate at $t = 3$ to find the desired answer.

$\frac{dV}{dt} = \frac{7t}{3} |_{t=3} = \boxed{7}$

The Growing Tetrahedron

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