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?

Printable View

- March 28th 2014, 05:36 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? - March 28th 2014, 07:19 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 - March 28th 2014, 10:43 AMkaemperRe: Rate of change (partial differentiation)
Yes, it is certainly a mistake.

- March 29th 2014, 07:19 PMHallsofIvyRe: Rate of change (partial differentiation)
You also have the gradient itself wrong. The gradient of at (-1, -1) is NOT .

- March 30th 2014, 09:41 PMhollywoodRe: Rate of change (partial differentiation)
It's not? I calculated , and substituting gives .

What am I missing?

- Hollywood