- #1

- 420

- 0

Would it just be 1/sqrt((2^2+1))=1/sqrt(5), i am asking this because does the z=e^(xy+x-y) get used at all?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Punkyc7
- Start date

- #1

- 420

- 0

Would it just be 1/sqrt((2^2+1))=1/sqrt(5), i am asking this because does the z=e^(xy+x-y) get used at all?

- #2

- 1,255

- 3

1/sqrt(5) is how

Anytime you're looking for a rate of change, you should be thinking about 'derivatives'. In this case, because they're asking about a particular direction of motion, you should be thinking about 'directional derivatives'.

- #3

- 420

- 0

<y+1(e^(xy+x-y)), x-1(e^(xy+x-y))>

would you use the starting point or the ending point?

- #4

- 1,255

- 3

Great start Punkyc.

The key is taking the dot product of that, with a unit vector in the direction of interest.

**actually moving** from the origin to (1,2), but you are **at the origin** moving **in that direction**.

You have the gradient there, good; thats the first part of finding the http://en.wikipedia.org/wiki/Directional_derivative" [Broken].So e^(xy+x-y)

<y+1(e^(xy+x-y)), x-1(e^(xy+x-y))>

The key is taking the dot product of that, with a unit vector in the direction of interest.

Look at the wording of the question carefully. You're notwould you use the starting point or the ending point?

Last edited by a moderator:

- #5

- 420

- 0

So it woud be <1,-1> dot (1/sqrt(5)<1,2>

so we have -1/(sqrt(5))

so we have -1/(sqrt(5))

- #6

- 1,255

- 3

Keep in mind: even though this is almost the exact answer you guessed

- #7

- 420

- 0

Thanks for your help

Share: