Gradient and finding the direction of maximum rate of change

[Taylor_1989]

Homework Statement

Hi guys, it a very simple question, but it causing me a great deal of confusion. The questions are as follows:

So I worked out the ans for one which I have displayed below. But what I don't understand is what they want from the second question. Because the way I see it, the direction of maximum rate of change is the gradient itself. Am I missing something here because I am really lost.

Homework Equations

The Attempt at a Solution

1: $\nabla f=\frac{-xi}{(x^2+y^2+z^2)^3} -\frac{yj}{(x^2+y^2+z^2)^3}-\frac{zk}{(x^2+y^2+z^2)^3}$

$\nabla f = le^{lx+my+nz}i+me^{lx+my+nz}j+ne^{lx+my+nz}k$

2: I make the direction as follows: $(-xi-yj-zk)$ & $li+mj+nk$

Is this correct or have I miss understood the question?

 

[BvU]

Check the exponent in the denominator in the first f in part 1).
For part 2: I agree with you : $-\bf\hat r$ and $(l,m,n)$.

[edit] perhaps for 'direction' the exercise composer wants to see $\bf \hat r$ ?

 

[PeroK]

But what I don't understand is what they want from the second question. Because the way I see it, the direction of maximum rate of change is the gradient itself.

?

The direction of maximum rate of change is the direction of the gradient. That generally means a unit vector,

 

[Taylor_1989]

Check the exponent in the denominator in the first f in part 1).
For part 2: I agree with you : $-\bf\hat r$ and $(l,m,n)$.

[edit] perhaps for 'direction' the exercise composer wants to see $\bf \hat r$ ?

Thank you I did not put the 3/2 in thanks.

 

[Taylor_1989]

The direction of maximum rate of change is the direction of the gradient. That generally means a unit vector,

when u say unit vector you mean $\frac{\nabla f}{|\nabla f|}$?

 

[PeroK]

when u say unit vector you mean $\frac{\nabla f}{|\nabla f|}$?

In this case, yes. If you look at the difference between the gradient and the direction if the gradient for the second function - the exponential - you'll see the point.