Gradient and Directional Derivatives

[dmalwcc89]

Homework Statement

Suppose, in the previous exercise, that a particle located at the point P = (2, 2, 8) travels towards the xy-plane in the direction normal to the surface.

a) Through which point Q on the xy-plane will the particle pass?

b) Suppose the axes are calibrated in centimeters. Determine the path c(t) of the particle if it travels at a constant speed 8 cm/s. How long will it take the particle to reach Q?

Homework Equations

Gradient of F: <dF/dx, dF/dy, dF/dz>

The Attempt at a Solution

I completed the "previous exercise:" I found the gradient of f after given the equation z^2 - 2x^4 - y^4 = 16 and asked to find vector n normal to this surface at P = (2, 2, 8) that points in the direction of the xy-plane. After normalizing a vector and finding the gradient, I was left with 1/(sqrt. 21)<-4, -2, 1>. The option was either + or - this value, and since (2, 2, 8) lies above the xy-plane, I needed the negative value. My answer was finally -1/(sqrt. 21)<-4, -2, 1>.

 

[lippyka]

a) Through which point Q on the xy-plane will the particle pass? The particle travels in the direction of -1/(sqrt. 21)<-4, -2, 1>, so it will move in the negative z direction in the xy-plane. The x and y coordinates of P are (2, 2), so the point Q is at (2, 2, 0). b) Suppose the axes are calibrated in centimeters. Determine the path c(t) of the particle if it travels at a constant speed 8 cm/s. How long will it take the particle to reach Q? The path c(t) of the particle is c(t) = (2, 2, 8) + t(-4, -2, -1) where t is the time the particle travels. The particle travels at a constant speed of 8 cm/s, and the distance between P and Q is 8 cm, so it will take the particle 1 second to reach Q.