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    Bit confused on how to answer this question, and help would be appreciated!

    The temperature T(x,y) at points of the x-y plane is given by T(x,y)=x^2+3y.

    In which direction should an ant at (2,-1) move if it wishes to cool off as quickly as possible?
    - For this I got the direction as (\frac{4}{5},\frac{3}{5})

    If the ant moves in that direction with speed s, as what rate does it experience the decrease of temperature?
    - Am I right in thinking this should be solved as \frac{dT}{ds}=\frac{dT}{dx}.\fra  c{dx}{ds}+\frac{dT}{dy}.\frac{dy  }{ds}?
    I can't really work out how to get rid of the ds terms

    Along what curve through (2,-1) should the ant move in order to continue to experience the maximum rate of cooling?
    - Not sure how to do this one at all either

    Any pointers would be much appreciated!
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    For the second part of the question, surely dx/ds and dy/ds are just the components of the ant's speed (strictly speaking, its velocity) in the x and y directions? So dx/ds = s.cos(theta) = s*(4/5) and dy/ds = s.sin(theta) = s*(3/5) where theta is the angle between the x-axis and the ant's direction of movement (which from your answer to the first part looks like a 3,4,5 triangle with the x and y axes, making s easy to resolve in x and y).

    Using these in the place of dx/ds and dy/ds, then differentiating T to find dT/dx and dT/dy, I think you'll end up with an equation in terms of s, x and y. But then, using the above equations for dx/ds and dy/ds you ought to be able to get this just in terms of s(this is just off the top of my head, I haven't checked this at all). I got this from the fact that your ant will be moving at constant speed in a given straight line, so it seems reasonable to think that the rate of temperature decrease it experiences will depend solely on its speed s (i.e. a derivative of T just based on s). I think I'm right in believing this will just be a curve in the plane between the temperature axis and the direction of the ant's movement.

    EDIT: I asked Wolfram Alpha about your function for T, it gave me this surface. I find it helps me if I can visualise what I am working on (even though from the function it's fairly obvious its a parabola that slopes upwards with y):



    EDIT 2: From looking at that surface, I don't think your answer to part 1 is right. If the ant is at (2,-1) and moves in the direction (4/5,3,5), then it looks like the temperature will be increasing...
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    Have you met the directional derivative? That's what you should be using here. It says that the rate of change in the direction of a unit vector u is given by

    u.\frac{f_x}{f_y} = af_x + bf_y where u = (a,b)

    (the LHS is the dot product of u and the vector derivative)
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    (Original post by Swayum)
    Have you met the directional derivative? That's what you should be using here. Google the formula if you don't know it.
    Out of interest, did any of what I said make any sense? I wasn't aware of the directional derivative so I just said how I would approach the question.
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    (Original post by TheNack)
    Out of interest, did any of what I said make any sense? I wasn't aware of the directional derivative so I just said how I would approach the question.
    To be honest, I'm not entirely sure, but you say that

    "For the second part of the question, surely dx/ds and dy/ds are just the components of the ant's speed (strictly speaking, its velocity) in the x and y directions? So dx/ds = s.cos(theta) = s*(4/5) and dy/ds = s.sin(theta) = s*(3/5)"

    This sounds wrong to me because dx/ds isn't just the component of the ant's velocity, it's the change in it. I think you've worked out that "dx = s.cos(theta)", not dx/ds.
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    (Original post by Swayum)
    To be honest, I'm not entirely sure, but you say that

    "For the second part of the question, surely dx/ds and dy/ds are just the components of the ant's speed (strictly speaking, its velocity) in the x and y directions? So dx/ds = s.cos(theta) = s*(4/5) and dy/ds = s.sin(theta) = s*(3/5)"

    This sounds wrong to me because dx/ds isn't just the component of the ant's velocity, it's the change in it. I think you've worked out that "dx = s.cos(theta)", not dx/ds.
    Well my thinking was that because x and y are measures of distance, then a derivative of them would represent a speed i.e. a rate of change of distance. I should have stated that more clearly in my previous post.
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    (Original post by TheNack)
    Well my thinking was that because x and y are measures of distance, then a derivative of them would represent a speed i.e. a rate of change of distance. I should have stated that more clearly in my previous post.
    Ok yeah, but even then, scos(theta) doesn't measure the velcoity; it measure the distance x.
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    (Original post by Swayum)
    Ok yeah, but even then, scos(theta) doesn't measure the velcoity; it measure the distance x.
    But s is given as a speed, so wouldn't it measure the speed in the x-direction?
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    Bit confused on how to answer this question, and help would be appreciated!

    The temperature T(x,y) at points of the x-y plane is given by T(x,y)=x^2+3y.

    In which direction should an ant at (2,-1) move if it wishes to cool off as quickly as possible?
    - For this I got the direction as (\frac{4}{5},\frac{3}{5})

    If the ant moves in that direction with speed s, as what rate does it experience the decrease of temperature?
    - Am I right in thinking this should be solved as \frac{dT}{ds}=\frac{dT}{dx}.\fra c{dx}{ds}+\frac{dT }{dy}.\frac{dy}{ds}?
    I can't really work out how to get rid of the ds terms

    Along what curve through (2,-1) should the ant move in order to continue to experience the maximum rate of cooling?
    - Not sure how to do this one at all either
    I think you actually want dT/dt, i.e. the rate of change of temperature w.r.t time. This can be found using the material derivative like this

     \frac{dT}{dt} = \frac{\partial T}{\partial t} + \mathbf{v} \bullet \nabla T

    In this case the direction of the velocity is given by the first part of the question, i.e. is the direction that the scalar field T decreases in most rapidly. This is given by

     -\nabla T

    where

     \nabla T = \frac{\partial T}{dx} \mathbf{i} +  \frac{\partial T}{dy} \mathbf{j}

    where i and j are unit vectors in the x and y direction respectively.

    This means that

     \mathbf{v} = -s \frac{\nabla T}{|\nabla T|}

    i.e. the velocity is of magnitude s and in the direction specified by a unit vector in the direction of decreasing temperature.

    Put this into the earlier equation and use that T(x,y) only (i.e. the partial derivative w.r.t. time = 0) to get

     \frac{dT}{dt} = -s \frac{\nabla T}{|\nabla T|} \bullet -\nabla T

    This is calcuable from the information in the question and I think should give you the answer...
 
 
 
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