Directional derivative with angle

So the question is asking how the directional derivative of f(x,y,z) = xy + z^2 changes as you move in the direction of a vector that makes an angle of pi/4 with the gradient of f at point (2,2,3).In summary, the conversation discusses finding the directional derivative of a function in a specific direction, as well as clarifying the meaning of the direction given in the problem. It also touches on finding the unit vector along a given path and using it to calculate the directional derivative, as well as a question about finding the directional derivative in the direction of a given angle.
  • #1
-EquinoX-
564
1

Homework Statement



Find the directional derivative of the function given below as you arrive at (0,1,3) from the direction of (1,1,0). Give an exact answer.
f(x,y,z) = sin(5 x) + ln(y^2+1) + z^3

Homework Equations





The Attempt at a Solution


I know how to solve this if it were a 2 dimension, I am just confused on what should I multiply the k direction with? The i is multiplied with cos, the j is with sin and what about the k?
 
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  • #2
Use grad(f) again. The directional derivative of f in the direction n is grad(f).n (dot product, as usual).
 
  • #3
dot product with the vector? how do you get the vector here?
 
  • #4
You are GIVEN the direction vector. It's (1,1,0). Seems hard to read the problem any other way.
 
  • #5
and so what do I use the (0,1,3) for?
 
  • #6
Don't you need a point to evaluate the gradient at?
 
  • #7
ok so the derivative I found is 5*cos(5x)i + 2y/(y^2+1)j + (3z^2)k and after plugging in the points I have j + 27k, so I need to take the dot product of this with i + j ?
 
  • #8
Not to confuse anyone but maybe I'm confusing my self.. just to check

"Find the directional derivative of the function given below as you arrive at (0,1,3) from the direction of (1,1,0). Give an exact answer.
f(x,y,z) = sin(5 x) + ln(y^2+1) + z^3"it says FROM the direction of (1,1,0)

wouldn't that mean calculate the gradient at the point (1,1,0).

and the direction vector would be the in the direction so you would ARRIVE at (0,1,3).

in other words

V = (0 - 1, 1 - 1, 3 - 0) = (-1,0,3)
unit vector = [(-1,0,3)] /root(10)
so calculate

gradient at (1,1,0) dot [(-1,0,3)]/root(10)
 
  • #9
5*cos(5*0) isn't 0. You have a habit of asking questions that have already been answered. In post 2, I said find grad(f).n, and now you are asking if you should take the dot product again? Why?
 
  • #10
Therefore it is, 5i + j + 27k and I need to take a dot product of this with i + j and resulting in 6
 
  • #11
That's what I think. But I'm still scratching my head over salman213's suggestion. What do you think?
 
  • #12
And that's the reason why I am asking again and again, is to make sure that it is right.. as it seems like the other way around as it says from the direction
 
  • #13
looking at salman213 suggestion i get (-5cos(5))/sqrt(10), and I tried that but the result is wrong... I don't know if I miscalculate anything
 
  • #14
The question seems ambiguous but I do agree with Dick on this. it says from the direction of 1,1,0 not from the POINT 1,1,0. So when one is arriving at some place usually you arrive at a point. In this case you are arriving from a particular direction -as the statement says- and not another point.
 
  • #15
I tried 6 and it gives me a wrong result as well... weird
 
  • #16
salman213 said:
Not to confuse anyone but maybe I'm confusing my self.. just to check

"Find the directional derivative of the function given below as you arrive at (0,1,3) from the direction of (1,1,0). Give an exact answer.
f(x,y,z) = sin(5 x) + ln(y^2+1) + z^3"


it says FROM the direction of (1,1,0)

wouldn't that mean calculate the gradient at the point (1,1,0).

and the direction vector would be the in the direction so you would ARRIVE at (0,1,3).

in other words

V = (0 - 1, 1 - 1, 3 - 0) = (-1,0,3)
unit vector = [(-1,0,3)] /root(10)



so calculate

gradient at (1,1,0) dot [(-1,0,3)]/root(10)

Well, you are confusing people apparently. How can you 'arrive' at (0,1,3) after 'starting' at (1,1,0). That might make a little sense if that were on the same level curve, in which case the answer would be 0. But they aren't. How else do you 'get from' (1,1,0) to (0,1,3)? You might interpret it as 'find the directional derivative of f along the line from (1,1,0) to (0,1,3)', but I don't think so. What's the magnitude of the vector to find the directional derivative along, unit vector? I think you are throwing us into muddy water. I'd stick with the simple interpretation.
 
  • #17
The question it self is confusing and that's why I asked it here...
 
  • #18
-EquinoX- said:
I tried 6 and it gives me a wrong result as well... weird

Oh, just great. Do they mean a unit vector along the along the line connecting (1,1,0) to (0,1,3)?
 
  • #19
I guess it is, as 6 doesn't work...
 
  • #20
-EquinoX- said:
The question it self is confusing and that's why I asked it here...

I'll grant you that in retrospect.
 
  • #21
Did you try the result of dotting your gradient with the unit vector along the path from (1,1,0) to (0,1,3)? salman213 was wrong about where to take the gradient but was right about the unit vector.
 
  • #22
How do I get the unit vector along the path from (1,1,0) to (0,1,3)
 
  • #23
Assume the path is a straight line.
 
  • #24
(-1,0,3)/sqrt(10) is that the unit vector? if that's so doing a dot product of the gradient with this will give me -5/sqrt(10) + 81/sqrt(10) which results in 76/sqrt(10)
 
  • #25
It's worth a try. Seems ok to me.
 
  • #26
yay! it's correct thanks for helping out
 
  • #27
I have another question related to this though, if I am asked to find the directional derivative of f(x,y,z) in the direction of theta/2, how can I do this? it's kind of hard to represent theta/2 in terms of vectors
 
  • #28
It's kind of hard to figure out what you are talking about. What's the angle 'theta'? What direction is that?
 
  • #29
Say the question is like this:

Find the directional derivative of f(x,y,z) = xy + z2 at the point(2,2,3) in the direction of a vector making an angle of pi/4 with gradf (2,2,3). Give an exact answer.

Therefore my first step is to find the gradient and I need to something similar to what I've done with this question, however.. how do I find the vector if all I know is pi/4
 
  • #30
Yeah, but what angle? In three dimensions there is more than one angle.
 
  • #31
That's all the question provides, pi/4
 
  • #32
This is a conceptual question as it says "of a vector" so that means of any normal vector [tex]\vec{n}\cdot\vec{T}=0[/tex] (orthogonality condition)
 
  • #33
hmm..are you sure?
 
  • #34
djeitnstine said:
This is a conceptual question as it says "of a vector" so that means of any normal vector [tex]\vec{n}\cdot\vec{T}=0[/tex] (orthogonality condition)

I think I see what djeitnstine is saying. The directional derivative is grad(f).n. You also know an expression for the dot product in terms of the lengths of the vectors and the angle between them, I hope.
 
  • #35
-EquinoX- said:
Say the question is like this:

Find the directional derivative of f(x,y,z) = xy + z2 at the point(2,2,3) in the direction of a vector making an angle of pi/4 with gradf (2,2,3). Give an exact answer.

Therefore my first step is to find the gradient and I need to something similar to what I've done with this question, however.. how do I find the vector if all I know is pi/4
One major problem with this is that there are an infinite number of vectors "in the direction of a vector making an angle of pi/4 with gradf (2,2,3)". That is NOT a unique direction.
 

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