It isn’t. Double check your computations.yecko said:
yecko said:
The OP thinks the directional derivative is ##\sqrt{3}## and ##\nabla f = \langle 1,1,1\rangle##, so he is not correct. This is why I suggested him to check his computation of the gradient in #2.Ray Vickson said:
The derivative of a directional vector is the rate of change of the vector with respect to a specific input variable. It measures how much the vector changes in magnitude and direction as the input variable changes.
To calculate the derivative of a directional vector, you first need to parametrize the vector in terms of a single variable. Then, you can use the chain rule and the product rule to find the derivative of each component of the vector. Finally, you can combine the derivatives to find the overall derivative of the directional vector.
The physical interpretation of the derivative of a directional vector is the instantaneous rate of change of the vector at a specific point. It tells us how fast and in what direction the vector is changing at that point.
The derivative of directional vector has many applications in physics, engineering, and economics. For example, it is used to calculate the velocity and acceleration of moving objects, the change in flow of fluids, and the rate of change of stock prices.
The derivative of directional vector is closely related to other mathematical concepts such as vectors, limits, and derivatives of multivariable functions. It is also a key concept in vector calculus, which studies the properties and behaviors of vector fields.
