Find the normal vector to the tangent plane of \( z=3 e^{x^{2}-4 y} \) at the point \( (8,16,3) \) \( x \) component \( = \) \( y \) component \( = \) \( z \) component \( =-1 \)

To find the normal vector to the tangent plane, we first need to determine the gradient vector of the function at the given point. The function is \( z = 3 e^{x^2 - 4y} \). To find the components of the normal vector, we calculate the partial derivatives: 1. \(\frac{\partial z}{\partial x} = 6x e^{x^2 - 4y}\) 2. \(\frac{\partial z}{\partial y} = -12 e^{x^2 - 4y}\) Now, evaluating these derivatives at the point \((8, 16)\): 1. \( \frac{\partial z}{\partial x} \bigg|_{(8, 16)} = 6 \cdot 8 \cdot e^{8^2 - 4 \cdot 16} = 48 e^{64 - 64} = 48 \) 2. \( \frac{\partial z}{\partial y} \bigg|_{(8, 16)} = -12 e^{64 - 64} = -12 \) Thus, the gradient vector (which is perpendicular to the tangent plane) is given by: \[ \nabla = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, -1 \right) = (48, -12, -1) \] So the components of the normal vector are \( x \) component \( = 48 \), \( y \) component \( = -12 \), \( z \) component \( = -1 \).