Greetings,

Attachment 30539I am almost done with this exercise, but I don't get a sqrt(5) in the denominator in my final answer, thus the correct answer is: 5/sqrt(5). How come I don't get a correct answer? Is the formula used incorrect?

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- Mar 28th 2014, 06:36 AMkaemperRate of change (partial differentiation)
Greetings,

Attachment 30539I am almost done with this exercise, but I don't get a sqrt(5) in the denominator in my final answer, thus the correct answer is: 5/sqrt(5). How come I don't get a correct answer? Is the formula used incorrect? - Mar 28th 2014, 08:19 AMhollywoodRe: Rate of change (partial differentiation)
When you calculate a directional derivative, you need to do the inner product with a unit vector. The unit vector in the direction is . So when you take the dot product, you should get . I assume your answer 5/sqrt(5) is a typo.

- Hollywood - Mar 28th 2014, 11:43 AMkaemperRe: Rate of change (partial differentiation)
Yes, it is certainly a mistake.

- Mar 29th 2014, 08:19 PMHallsofIvyRe: Rate of change (partial differentiation)
You also have the gradient itself wrong. The gradient of at (-1, -1) is NOT .

- Mar 30th 2014, 10:41 PMhollywoodRe: Rate of change (partial differentiation)
It's not? I calculated , and substituting gives .

What am I missing?

- Hollywood - Aug 13th 2014, 03:24 PMkaemperRe: Rate of change (partial differentiation)
- Aug 13th 2014, 03:45 PMCartesius24Re: Rate of change (partial differentiation)
I agree with previous (we say about unnormed vector)

- Aug 13th 2014, 04:16 PMkaemperRe: Rate of change (partial differentiation)
- Aug 13th 2014, 04:25 PMCartesius24Re: Rate of change (partial differentiation)
With You

- Aug 13th 2014, 04:32 PMHallsofIvyRe: Rate of change (partial differentiation)