Gradient and Directional Derivatives

In summary, the previous exercise involved finding the gradient of f for a given equation and determining a vector normal to the surface at point P = (2, 2, 8) that points towards the xy-plane. The particle in question will pass through the point Q = (2, 2, 0) on the xy-plane and will take 1 second to reach this point if it travels at a constant speed of 8 cm/s along the path c(t) = (2, 2, 8) + t(-4, -2, -1).
  • #1
dmalwcc89
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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>.
 
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  • #2
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.
 

FAQ: Gradient and Directional Derivatives

What is a gradient?

A gradient is a mathematical concept that represents the direction and magnitude of the steepest slope of a function at a specific point.

What is a directional derivative?

A directional derivative is a measure of how a function changes in a specific direction, defined by a vector, at a given point.

How is the gradient related to directional derivatives?

The gradient is the vector that points in the direction of the steepest slope of a function at a specific point. The directional derivative is the rate of change of the function in that direction.

What is the formula for calculating a directional derivative?

The formula for calculating a directional derivative is Df(v) = ∇f · v, where ∇f is the gradient of the function f and v is the unit vector representing the direction in which the derivative is being calculated.

What are some real-world applications of gradient and directional derivatives?

Gradient and directional derivatives are used in fields such as physics, engineering, and economics to optimize functions and predict the behavior of systems. For example, they can be used to determine the direction in which a ball will roll on a sloped surface or to find the optimal direction for a plane to fly in order to minimize fuel consumption.

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