Question

Find the gradient of the function and the maximum value of the directional derivative at the given point. $$ \frac{\text { Function }}{f(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}}} \frac{\text { Point }}{(1,4,2)} $$

Short answer

The gradient of the function at the given point is the vector calculated in step 2 and the maximum value of the directional derivative at this point is the magnitude of this gradient vector, calculated in step 3.

Step-by-step solution

Calculating the Gradient

The gradient of a function, \(f(x,y,z)\), is given by the vector: \(\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle\). This means we differentiate the function with respect to each of the variables to get partial derivatives and form the gradient vector. For the given function \(f(x, y, z) = \sqrt{x^{2}+y^{2}+z^{2}}\), this amounts to calculating \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\) and \(\frac{\partial f}{\partial z}\) and then combining the results into a vector.

Evaluating the Gradient at the Given Point

After calculating the gradient, evaluate it at the point (1,4,2). This gives the gradient vector at this particular point.

Finding the Maximum value of the Directional Derivative

The maximum value of the directional derivative is the magnitude of the gradient vector at the point. Therefore, calculate the magnitude of the gradient vector obtained in step 2. The magnitude of a vector \(\vec{v} = \langle v_1, v_2, v_3 \rangle\) is given by \(\|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}\).