3
$\begingroup$

$$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)$$

These components are the rate of increase along the $x$, $y$ and $z$ directions respectively and according to gradient operation maximum increase is along the axis.

Why does it have to be along the axis? Why does maximum increase have to be along the $x$, $y$ and $z$ direction?

$\endgroup$
1
$\begingroup$

Let's say we calculate the gradients at point $(x_0,y_0,z_0)$. We are interested in the rate of increase of $f$ in the direction of some vector $(x_1, y_1, z_1)$, and let's say that the vector has norm $1$.

Let's have a single variable function

$$g(t)=f(x_0 + tx_1, y_0 + ty_1, z_0 + tz_1)$$

and let's find $g'(t)$.

Using the chain rule, you can verify that

$$\frac{dg}{dt}(t) = \nabla f(x_0 + tx_1, y_0 + ty_1, z_0 + tz_1) \cdot (x_1, y_1, z_1)$$

So the rate of change at $t=0$ is

$$\nabla f(x_0 , y_0 , z_0 ) \cdot (x_1, y_1, z_1)$$ This inner product is maximum (if the norm of the second vector is $1$) when the two vectors are paralel, so the conclusion is that the rate of change is biggest in the direction of $\nabla f(x_0,y_0,z_0)$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks 5xum. I have seen that several times. But this explanation is just an explanation for the result. You are looking the result and says: "It must be that way". And I still don't understand why taking derivative in vector calculus shows the biggest increase. Can you please tell us not looking result.I mean in single variable calculus, taking derivative doesn't represent biggest increase, but in vector calculus it does. Can you please explain not by looking result for inner product property. $\endgroup$ – user50322 Sep 3 '15 at 15:30
  • $\begingroup$ @user50322 Actually, in single variable calculus, taking derivative does represent biggest increase. If the derivative is positive, then the biggest decrease is in the direction of "$+1$" (which is a vector in $\mathbb R^2$), because $+1$ is the normed gradient of the (single variable) function. $\endgroup$ – 5xum Sep 3 '15 at 15:33
  • $\begingroup$ @user50322 Also, your statement "why does maximum increase have to be in the $x$, $y$ and $z$ direction" is non-sensical. The maximum increase happens in the direction of $\nabla d$, and $\nabla f$ can be calculated by taking partial derivatives over $x$, $y$ and $z$. But the maximum change does not hapen in the direction of $x$, it happens in the direction of $\nabla f$! $\endgroup$ – 5xum Sep 3 '15 at 15:35
  • $\begingroup$ You said:"The maximum increase happens in the direction of ∇f, and ∇f can be calculated by taking partial derivatives over x, y and z." Doesn't taking partial derivative over x,y and z mean: the rate of increase along the x,y and z directions respectively? So if we want to maximize increase, shouldn't move in ratio to the rate of increase along the axis? $\endgroup$ – user50322 Sep 3 '15 at 21:10
  • $\begingroup$ @user50322 If we want to maximize increase, we need to move in the direction of $\nabla f$. $\nabla f$ can be expressed in our traditional $xyz$ coordinate system as the vector of partial derivatives over $x,y$ and $z$. I don't see that the confusion is... $\endgroup$ – 5xum Sep 4 '15 at 5:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.