1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this use of the grad function correct?

  1. Feb 9, 2013 #1
    I am currently studying vector calculus, and to try and understand it I have created a question for myself, and it would be greatly appreciated if someone could verify that I have answered my question correctly, and I have done everything right.

    Okay here it is; Point Q in the room is a heat source, how does the temperature vary as you move away from the origin (0,0,0)

    http://img571.imageshack.us/img571/4250/gascube.png [Broken]
    http://imageshack.us/a/img841/4250/gascube.png [Broken]

    .The further away from point Q you are the less heat there is. This can be modelled by:

    [itex] dt = \nabla t .dr[/itex]

    .(dt) is a scalar, it is a change in temperature
    .(dr) is a vector, it is a change in distance of x, y and z [itex](dr=dx\hat{x} +dy\hat{y} +dz\hat{z} )[/itex]
    .(del t) is a vector, it is the amount t changes for x, y, and z, for a change in distance of x,y,and z

    For the purposes of this example I am going to say that the heat at any point equals 1 divided by the square of the distance from q.

    In the x direction
    [itex] t = \frac{1}{(xq-x)^{2}}[/itex]

    In the y direction
    [itex] t = \frac{1}{(yq-y)^{2}}[/itex]

    In the z direction
    [itex] t = \frac{1}{(zq-z)^{2}}[/itex]

    Where (xq,yq,zq)=(0,2.5,2.5) And (x,y,z) is the point you are at

    So, [itex] \frac{dt}{dx} = 2(xq-x)^{-1}, \frac{dt}{dy} = 2(yq-y)^{-1}, \frac{dt}{dz} = 2(zq-z)^{-1}[/itex]

    And the final answer is:

    [itex] dt =(2(xq-x)^{-1}\hat{x} + 2(yq-y)^{-1}\hat{y} + 2(zq-z)^{-1}\hat{z}) . dr[/itex]

    Is this correct?

    Thank you
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 9, 2013 #2
    The gradient points in the direction of greatest increase. Since temperature is decreasing, you will want to take the negative gradient of your function. You also differentiated incorrectly. The derivative of x^(-2) is not 2^(-1). It's -2^(-3). Remember: d/dx(x^n) = nx^(n-1).
  4. Feb 9, 2013 #3
    Ah yes I can see I have differentiated incorrectly, although I think it would be 2(xq-x)^-3 and not -2(xq-x)^-3 because of the chain rule:
    derivative of (xq-x) with repect to x = -1, so the whole thing will be multiplied by -1.

    And can you explain exactly what I did wrong apart from this, I do not really understand why I should have to take the negative gradient, or what this actually even means.

    When you say:
    I've always thought that this just meant that if you plot the gradient vector it would point in the direction of greatest increase, but that was just a property of the gradient function, and the true reason you would use the gradient function is to see how much change one variable would change (dt) as three other variables are changing (dr)
    [itex]dt = \nabla .dr[/itex]
  5. Feb 9, 2013 #4
    I totally missed that -x part. Thanks!

    The fact that the gradient points in the direction of greatest increase is the reasons for the negative sign in your case. The temperature is decreasing as you go away from the source. Without this negative sign, the resulting vector field would be pointing away from the source which implies that temperature is increasing.

    By multiplying the gradient by -1, the resulting vector field would be pointing towards the source which means that temperature increases towards the source. Gradients are not only used for seeing *how much* a variable changes, but we are also interested in *which direction* the greatest increase/decrease occurs (for other directions we use directional derivatives). In your case, you specified that temperature drops away from the source.

    It's a subtle point, but it's important.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Is this use of the grad function correct?
  1. Is this correct (Replies: 1)