Finding specified partial derivatives

[dalarev]

Homework Statement

Homework Equations

The Attempt at a Solution

I follow the steps until I get to 2(x+y+z)(1 - sin(r+s) + cos(r+s))

The actual derivation process isn't the problem, I get lost in trying to figure out when to plug values for r and s.

 

[Dick]

Images don't load.

 

[paradigm]

images loaded fine for me...
anyhow, you have done everything correctly thus far. just plug the given values for s and r into the final expression 2(x+y+z)(1 - sin(r+s) + cos(r+s))

you will never go awry if you always plug in given values last

 

[dalarev]

images loaded fine for me...
anyhow, you have done everything correctly thus far. just plug the given values for s and r into the final expression 2(x+y+z)(1 - sin(r+s) + cos(r+s))

I'm guessing this is after I replace x,y,z with their respective functions, correct?

edit: Negative. I did that and got 0 as an answer. According to the answer guide (back of the book), the correct answer is 13.

 

[paradigm]

yes, I'm sorry, i should have specified... plug in x,y,z and then s, r

 

[dalarev]

yes, I'm sorry, i should have specified... plug in x,y,z and then s, r

That's not correct though, I get 6(0) = 0, correct answer should be 13.

 

[paradigm]

i'm currently unable to view the images you attached... what were x,y, and z again?

EDIT: i recall r and s being 1 and -1, right? if that's the case (1 - sin(r+s) + cos(r+s) should be equal to (1 - sin0 + cos0) = (1 - 0 +1) = 2, not zero..

 

[dalarev]

x = r - s
y = cos(r+s)
z = sin(r+s)

 

[paradigm]

so if r was 1 and s was -1, then you should have 2(2+1)(2) = 12... but you say the answer was 13?

 

[dalarev]

so if r was 1 and s was -1, then you should have 2(2+1)(2) = 12... but you say the answer was 13?

Yeah, that's what I got too. But then again, the answer guide has been known to be wrong on more than a couple of occasions. Disregard further posts, and thanks for your help.