Ytukyg

Consider a function $g: mathbb{R^3} to mathbb{R}$ defined by $g(x,y,z) = z^2 -x^3 + 2z + 3y^3$

Find the gradient of g at the point (2,1,-1)

Is the gradient $nabla g(2,1,-1)$ given by a vector? i.e. $nabla g(2,1,-1) = -12i + 9j$ If so then what does the directional derivative mean?

The gradient is a vector; it points in the direction of steepest ascent.

The directional derivative is a number; it is the rate of change when your point in $Bbb R^3$ moves in that direction. (You can imagine "reducing" your function to a function of a single variable, say $t$, by "slicing" the curve in that direction; the directional derivative is just then the 1-D derivative of that "sliced" function.)

Be careful that directional derivative of a function is a scalar while gradient is a vector.

The only difference between derivative and directional derivative is the definition of those terms. Remember:

• Directional derivative is the instantaneous rate of change (which is a scalar) of $f(x,y)$ in the direction of the unit vector $u$.


• Derivative is the rate of change of $f(x,y)$, which can be thought of the slope of the function at a point $(x_0,y_0)$.

Yes, the gradient is given by the row vector whose elements are the partial derivatives of $g$ with respect to $x$, $y$, and $z$, respectively. In your case the gradient at $(x,y,z)$ is hence $[-3x^2,9y^2,2z+2]$. The gradient at $(2,1,-1)$ is therefore $[-12,9,0]$.

The directional derivative at a point $(x,y,z)$ in direction $(u,v,w)$ is the gradient multiplied by the direction divided by its length. So if $u^2+v^2+w^2=1$ then the directional derivative at $(x,y,z)$ in direction $(u,v,w)$ is just $-3x^2 u + 9y^2v + (2z+2)w$.

If $u^2+v^2+w^2neq 1$ then you should divide the number above by $sqrt{u^2+v^2+w^2}$.

In sum, the gradient is a vector with the slope of the function along each of the coordinate axes whereas the directional derivative is the slope in an arbitrary specified direction.

In simple words, directional derivative can be visualized as slope of the function at the given point along a particular direction. For example partial derivative w.r.t x of a function can also be written as directional derivative of that function along x direction.

Gradient is a vector and for a given direction, directional derivative can be written as projection of gradient along that direction.

Please go through this link to have a clear idea:
http://omega.albany.edu:8008/calc3/directional-derivatives-dir/define-m2h.html

• 
  A Directional Derivative is a value which represents a rate of change
  


• 
  A Gradient is an angle/vector which points to the direction of the steepest ascent of a curve.
  

Let us take a look at the plot of the following function:

$$ bbox[lightgray] {f(x) = -x^2+4}qquad (1)$$

The 1st derivative of the function is:

$$ bbox[lightgray] {frac{dy}{dx} = -2x}qquad (2)$$

Putting $x=-1$ in $(2)$ we obtain,

$implies frac{dy}{dx} = frac{rise}{run}= 2 ... ... ...qquad (3)$

Also,

$tan theta = 2$
$implies theta = tan^{-1}(2)$
$implies theta = 0.964 $ radian

So, $theta = 55.23 ^circ ... ... ...qquad (4)$

.

Similarly, putting $x=-2$ in $(2)$ we obtain,

$implies frac{dy}{dx} = frac{rise}{run}= 4 ... ... ...qquad (5)$

So, $theta = 57.25 ^circ ... ... ...qquad (6)$

So, the answer is:

• (3) and (5) are Directional Derivatives.


• Directional Derivatives are scalar values.

And,

• (4) and (6) are Gradients.


• Gradients are vector values.