If \( r \) varies direct as the square of \( t \) and \( r=216 \) when \( t=6 \), find \[ + \text { when } r=512 \]

Since \( r \) varies directly as the square of \( t \), we can express this relationship mathematically as: \[ r = k t^2 \] where \( k \) is a constant. Given that \( r = 216 \) when \( t = 6 \), we can substitute these values into the equation to find \( k \): \[ 216 = k (6^2) \] Calculating \( 6^2 \): \[ 6^2 = 36 \] Now substituting this back into the equation: \[ 216 = k \cdot 36 \] To solve for \( k \), we divide both sides by 36: \[ k = \frac{216}{36} = 6 \] Now that we have \( k \), we can express the relationship between \( r \) and \( t \) as: \[ r = 6 t^2 \] Next, we need to find \( t \) when \( r = 512 \): \[ 512 = 6 t^2 \] To isolate \( t^2 \), we divide both sides by 6: \[ t^2 = \frac{512}{6} = \frac{256}{3} \] Now, we take the square root of both sides to find \( t \): \[ t = \sqrt{\frac{256}{3}} = \frac{\sqrt{256}}{\sqrt{3}} = \frac{16}{\sqrt{3}} \] To rationalize the denominator, we multiply the numerator and denominator by \( \sqrt{3} \): \[ t = \frac{16 \sqrt{3}}{3} \] Thus, the value of \( t \) when \( r = 512 \) is: \[ \boxed{\frac{16 \sqrt{3}}{3}} \]