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Math Help - temperature at (x,y,z)

  1. #1
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    temperature at (x,y,z)

    Suppose the temperature are point (x,y,z) in space is T(x,y,z) = x^2+y^2+z^2

    Let a particle follow the right circular helix \sigma(t) = (cos (t), sin (t), t) and let T(t) be its temperature at time t

    a) what is T'(t)

    \frac{dT}{dt} = \frac{\partial x}{\partial t}\frac{dx}{dt} + \frac{\partial y}{\partial t}\frac{dy}{dt} + \frac{\partial z}{\partial t}\frac{dz}{dt} ??

    =-2cos(t)sin(t)+2sin(t)cos(t)+2t

    theres a part (b) but I'll wait till I get this part
    Last edited by Jonroberts74; June 28th 2014 at 02:07 PM.
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  2. #2
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    Re: temperature at (x,y,z)

    So \frac{dT}{dt}= 2t. Now, what is your question?
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    Re: temperature at (x,y,z)

    Quote Originally Posted by HallsofIvy View Post
    So \frac{dT}{dt}= 2t. Now, what is your question?

    b) find an approximate value for the temperature at t=\frac{\pi}{2}+0.01

    so just put (x,y,z) in terms of t and solve?
    Last edited by Jonroberts74; June 28th 2014 at 04:08 PM.
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    Re: temperature at (x,y,z)

    I think you have the wrong formula for the derivative of T with respect to t. If x, y, and z were measured in metres and t in seconds then dT/dt in your formula would have units of metres squared per second squared

    To my knowledge the total derivative is
    \frac{dT}{dt}= \frac{dT}{dt}\frac{\partial t}{\partial t}+\frac{dT}{dx}\frac{\partial x}{\partial t}+\frac{dT}{dy}\frac{\partial y}{\partial t}+\frac{dT}{dz}\frac{\partial z}{\partial t}
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    Re: temperature at (x,y,z)

    Quote Originally Posted by Shakarri View Post
    I think you have the wrong formula for the derivative of T with respect to t. If x, y, and z were measured in metres and t in seconds then dT/dt in your formula would have units of metres squared per second squared

    To my knowledge the total derivative is
    \frac{dT}{dt}= \frac{dT}{dt}\frac{\partial t}{\partial t}+\frac{dT}{dx}\frac{\partial x}{\partial t}+\frac{dT}{dy}\frac{\partial y}{\partial t}+\frac{dT}{dz}\frac{\partial z}{\partial t}
    yeah I can see how I typed it out wrong but how do you have \frac{dT}{dt} on both sides of the equation ??

    \frac{dT}{dt} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt} + \frac{\partial T}{\partial z}\frac{dz}{dt} wouldn't it be this
    Last edited by Jonroberts74; June 28th 2014 at 04:46 PM.
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    Re: temperature at (x,y,z)

    If it is given that T is a function of x, y, and z only, while x, y, and z are functions of t, then
    \frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}

    If T is a function of x, y, z, and t, then
    \frac{dT}{dt}= \frac{\partial T}{\partial t}+ \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}
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    Re: temperature at (x,y,z)

    Quote Originally Posted by HallsofIvy View Post
    If it is given that T is a function of x, y, and z only, while x, y, and z are functions of t, then
    \frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}

    If T is a function of x, y, z, and t, then
    \frac{dT}{dt}= \frac{\partial T}{\partial t}+ \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}
    wait so which one is the correct one in this question?

    and what would \frac{\partial T}{\partial t} be?
    Last edited by Jonroberts74; June 28th 2014 at 08:57 PM.
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    Re: temperature at (x,y,z)

    for part (b)

    not even sure if everything is correct yet

    t=\frac{\pi}{2}+0.01

    T(t)=cos^2(\frac{\pi}{2} + 0.01) + sin^2 (\frac{\pi}{2} + 0.01) + (\frac{\pi}{2} + 0.01)^2 \approx 3.5
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    Re: temperature at (x,y,z)

    Quote Originally Posted by Jonroberts74 View Post
    Suppose the temperature are point (x,y,z) in space is T(x,y,z) = x^2+y^2+z^2

    Let a particle follow the right circular helix \sigma(t) = (cos (t), sin (t), t) and let T(t) be its temperature at time t

    a) what is T'(t)

    \frac{dT}{dt} = \frac{\partial x}{\partial t}\frac{dx}{dt} + \frac{\partial y}{\partial t}\frac{dy}{dt} + \frac{\partial z}{\partial t}\frac{dz}{dt} ??

    =-2cos(t)sin(t)+2sin(t)cos(t)+2t

    theres a part (b) but I'll wait till I get this part
    Did you not notice that -2 cos(t)sin(t)+ 2 sin(t)cos(t)= 0? \frac{dT}{dt}= 2t. Another way to see that is to write
    T= x^2+ y^2+ z^2= cos^2(t)+ sin^2(t)+ t^2= 1+ t^2 and then differentiate that.
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