Suppose you are climbing a hill whose shape is given by the equation z = 900 − 0.005x2 − 0.01y2, where x, y, and z are measured in meters, and you are standing at a point with coordinates (120, 80, 764). The positive x-axis points east and the positive y-axis points north. (a) If you walk due south, will you start to ascend or descend? ascend descend Correct: Your answer is correct.

Answer:AscendStep-by-step explanation:In order to solve this problem, we are going to use some principles of vector calculation. The concepts we are going to use are Partial derivatives, gradient vector, velocity vector, direction vector, and directional derivative. The gradient vector is a vector that describes how is the 'slope' in the space of a multivariable function at a specified point; it is built as a vector of partial derivatives. The vector velocity is a vector that describes the direction and speed of the movement of a body, if we make the velocity a unitary vector (a vector whose norm is 1), we obtain the direction vector (because we are not considering the real norm of the vector, just direction). Finally, the directional derivative is a quantity (a scalar) that describes the slope that we get on a function if we make a displacement from a particular point in a specific direction.  The problem we have here is a problem where we want to know how will be the slope of the hill if we stand in the point (120, 80, 764) and walk due south if the hill has a shape given by z=f(x,y). As you see, we have to find the directional derivative of z=f(x,y) at a specific point (120, 80, 764) in a given displacement direction; this directional derivative will give us the slope we need. The displacement direction 'u' is (0,-1): That is because 'y' axis points north and our displacement won't be to the east either west (zero for x component), just to south, which is the opposite direction of that which the y-axis is pointing (-1 for y component). Remember that the direction vector must be a unitary vector as u=(0,-1) is. Let's find the gradient vector: [tex]z=900-0.005x^2-0.01y^2\\\frac{\partial z}{\partial x}=-0.005*2*x=-0.01x\\\frac{\partial z}{\partial y}=-0.01*2*y=-0.02y\\ \nabla (z)=(-0.01x,-0.02y)[/tex] Evaluate the gradient vector at (120,80) (764 is z=f(120,80); you may confirm) [tex]\nabla (z(120,80))=(-0.01*120,-0.02*80)=(-1.2,-1.6)[/tex] Finally, find the directional derivative; if you don't remember, it can be found as a dot product of the gradient vector and the direction vector): [tex]D_{u,P_0}= \nabla (z)_{P_0}\cdot u\\D_{u,P_0}= (-1.2,-1.6)\cdot (0,-1)=1.6[/tex] As you see, the slope we find is positive, which means that we are ascending at that displacement direction.