Find the positive \( t \) when the vector \( r(t)=\left\langle 5 t, 2 t^{2}, 7 t^{2}-8\right\rangle \) is perpendicular to \( r^{\prime}(t) \) \( t=\square \)

To find the positive \( t \) when the vector \( r(t)=\left\langle 5 t, 2 t^{2}, 7 t^{2}-8\right\rangle \) is perpendicular to \( r^{\prime}(t) \), we first need to compute the derivative \( r^{\prime}(t) \). Calculating the derivative, we get: \[ r^{\prime}(t) = \left\langle 5, 4t, 14t \right\rangle. \] Now, for the vectors \( r(t) \) and \( r^{\prime}(t) \) to be perpendicular, their dot product must equal zero: \[ r(t) \cdot r^{\prime}(t) = 5t \cdot 5 + 2t^{2} \cdot 4t + (7t^{2}-8) \cdot 14t = 0. \] Calculating the dot product: \[ 25t + 8t^{3} + 14t(7t^{2}-8) = 25t + 8t^{3} + 98t^{3} - 112t = 0, \] which simplifies to: \[ (8t^{3} + 98t^{3}) + (25t - 112t) = 0, \] or \[ 106t^{3} - 87t = 0. \] Factoring out \( t \), we have: \[ t(106t^{2} - 87) = 0. \] Setting the factor equal to zero gives: 1. \( t = 0 \) or 2. \( 106t^{2} - 87 = 0 \). Solving for \( t \) when \( 106t^{2} = 87 \): \[ t^{2} = \frac{87}{106}, \quad t = \sqrt{\frac{87}{106}}. \] To find the positive \( t \): \[ t = \frac{\sqrt{87}}{\sqrt{106}}. \] Thus, the positive \( t \) when \( r(t) \) is perpendicular to \( r^{\prime}(t) \) is: \[ t = \frac{\sqrt{87}}{\sqrt{106}} \approx 0.866. \]