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Multivariable calculus: find the rate of change

  1. Feb 28, 2016 #1
    • Missing homework template due to originally being posted in other forum.
    The problem is:
    The temperature (in degrees Celsius) of a metal plate, located in the xy -plane, at any point (x, y ) is given by the function of two variables T(x, y ) = x sin y + y2 sin x.
    (a) Find the rate of change in temperature in the direction of the positive x-axis at the point (π, π).
    (b) Find the rate of change in temperature in the direction of the positive y -axis at the point (π, π).

    I am thinking that you would have to find the gradient vector of T(x,y), then plug the point (π, π). But the phrase "direction of the positive x-axis/y-axis at the point" is throwing me off. Does that mean you would have to find the directional derivative instead? So find the gradient vector then multiply it by a unit vector, but what is the unit vector?

    Thanks.
     
    Last edited by a moderator: Feb 28, 2016
  2. jcsd
  3. Feb 28, 2016 #2
    1. The problem statement, all variables and given/known data
    The temperature (in degrees Celsius) of a metal plate, located in the xy -plane, at any point (x, y ) is given by the function of two variables T(x, y ) = xsiny + y2sin x.
    (a) Find the rate of change in temperature in the direction of the positive x-axis at the point (π, π).
    (b) Find the rate of change in temperature in the direction of the positive y -axis at the point (π, π).

    2. Relevant equations
    ∇T = <tx, ty>

    3. The attempt at a solution
    I am thinking that you would have to find the gradient vector of T(x,y), then plug the point (π, π). But the phrase "direction of the positive x-axis/y-axis at the point" is throwing me off. Does that mean you would have to find the directional derivative instead? So find the gradient vector then multiply it by a unit vector, but what is the unit vector?

    Thanks.
     
  4. Feb 28, 2016 #3
    The gradient of the scalar T(x,y) is given by
    ∇T = <tx, ty> = <(∂T/∂x), (∂T/∂y)>
    Here (∂T/∂x) represents the directional derivative of T w.r.t x
    and (∂T/∂y) represents the directional derivative of T w.r.t y
    The Directional derivative means rate of change of a quantity in a particular direction, so you can find gradient of T and need not to multiply by any unit vector, because it won't represent the true rate of change then(as its magnitude will be 1 then)
    Hope this helps
     
  5. Feb 28, 2016 #4
    Yes, you're right, you need to find the directional derivative. You want the change in the x-direction, so what would be the unit vector in this direction?
     
  6. Feb 28, 2016 #5
    Would it be <1,0>? Do I even need a unit vector for this?
     
  7. Feb 28, 2016 #6
    So a) and b) are pretty much asking for the directional derivative of T w.r.t the variable, so a) is w.r.t x and b) is w.r.t y? It's that simple?
     
  8. Feb 28, 2016 #7
    Yes that's the unit vector you want. Here you don't actually NEED a unit vector, since the required direction being along the x-axis you can simply take the x component of the gradient. ( but can you see that you are taking the dot product of the unit vector and the gradient vector?)
     
  9. Feb 28, 2016 #8
    yes!
     
  10. Feb 28, 2016 #9
    Thanks guys! :D
     
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