Find a normal vector n𝑛 to the tangent plane to the surfacez=−3x2y3𝑧=−3𝑥2𝑦3 at the point (x,y,z)=(−3,4,−1728)(𝑥,𝑦,𝑧)=(−3,4,−1728) and enter your answer in the box below using Maple syntax.  n=𝑛=

To find the normal vector to the tangent plane of the surface z = -3x^2y^3 at the point (-3,4,-1728), we first need to find the gradient of the function f(x,y,z) = -3x^2y^3 - z.

The gradient of a function gives us a vector that is normal (perpendicular) to the level surface of the function at a given point. The gradient is found by taking the partial derivatives of the function with respect to each variable.

The partial derivative of f with respect to x is: fx = -6xy^3 The partial derivative of f with respect to y is: fy = -9x^2y^2 The partial derivative of f with respect to z is: fz = -1

So, the gradient of f is the vector <fx, fy, fz> = <-6xy^3, -9x^2y^2, -1>.

We then evaluate this vector at the point (-3,4,-1728) to get the normal vector at this point.

n = <-6*(-3)(4)^3, -9(-3)^2*(4)^2, -1> = <288, 432, -1>

So, the normal vector to the tangent plane at the point (-3,4,-1728) is n = <288, 432, -1>.