No. That is the unit vector in the direction from (0,0,0) to (0,-3,0). You are asked for the unit vector in the direction in which f(x,y,z) increases most rapidly. A function increases most rapidly in the direction of its gradient vector.brendan said:Homework Statement
Find a unit vector in the direction in which f(x,y,z) = (x-3y+4z)1/2 increases most rapidly at P(0,-3,0), and find the rate of increase of f in that direction
Homework Equations
The Attempt at a Solution
I've calculated the unit vector to be <0,-1,0>
That is correct. What is a unit vector in that direction?and the gradient to be <1/6,-1/2,2/3>
No, that's not at all right. Did you divide by the length or multiply? 1/36+ 1/4+ 4/9= 1/36+ 9/36+ 15/36= 26/36. The length of the gradient is [itex]\sqrt{26}/6[/itex]. Dividing by that will put the square root in the denominator.brendan said:So If I use the gradient <1/6 ,-1/2 ,2/3> and find its unit vector which is
<sqrt(26)/936, -sqrt(26)/312, sqrt(26)/234 >
Of course, that is wrong now, but in any case you did not need to do that product. If [itex]\vec{v}[/itex] is a vector of length [itex]||\vec{v}||[/itex], then the unit vector in that direction, [itex]\vec{u}[/itex], is [itex]\vec{v}/||/vec{v}||[/itex] and the inner product of the two vectors is [itex]\vec{v}\cdot\vec{v}/||\vec{v}}||= ||\vec{v}||^2/||\vec{v}||= ||\vec{v}||[/itex]. The derivative in the direction of the gradient is, as I said before, the length of the gradient.than find the dot product of them both <1/6 ,-1/2 ,2/3> . <sqrt(26)/936, -sqrt(26)/312, sqrt(26)/234 >
which is sqrt(26)/6 the magnitude of the gradient vector?
regards
Brendan
A directional derivative is a measure of how a function changes in a specific direction. It represents the rate of change of a function at a given point in the direction of the unit vector.
The directional derivative is calculated using the dot product of the gradient of the function and the unit vector in the desired direction.
Directional derivatives are important in multivariate calculus as they help us understand how a function changes in different directions. They are also used in optimization and vector calculus.
Yes, a directional derivative can be negative. It simply means that the function is decreasing in the direction of the unit vector.
Directional derivatives have various applications in fields such as engineering, physics, and economics. They are used to analyze the rate of change of physical quantities, optimize processes, and model real-world phenomena.