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Thread: temperature at (x,y,z)

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

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

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

    a) what is $\displaystyle T'(t)$

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

    $\displaystyle =-2cos(t)sin(t)+2sin(t)cos(t)+2t$

    theres a part (b) but I'll wait till I get this part
    Last edited by Jonroberts74; Jun 28th 2014 at 01:07 PM.
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    Re: temperature at (x,y,z)

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

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

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

    so just put (x,y,z) in terms of t and solve?
    Last edited by Jonroberts74; Jun 28th 2014 at 03:08 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
    $\displaystyle \frac{dT}{dt}= \frac{dT}{dt}\frac{\partial t}{\partial t}+\frac{dT}{dx}\frac{\partial x}{\partial t}+\frac{dT}{dy}\frac{\partial y}{\partial t}+\frac{dT}{dz}\frac{\partial z}{\partial t}$
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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
    $\displaystyle \frac{dT}{dt}= \frac{dT}{dt}\frac{\partial t}{\partial t}+\frac{dT}{dx}\frac{\partial x}{\partial t}+\frac{dT}{dy}\frac{\partial y}{\partial t}+\frac{dT}{dz}\frac{\partial z}{\partial t}$
    yeah I can see how I typed it out wrong but how do you have $\displaystyle \frac{dT}{dt}$ on both sides of the equation ??

    $\displaystyle \frac{dT}{dt} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt} + \frac{\partial T}{\partial z}\frac{dz}{dt}$ wouldn't it be this
    Last edited by Jonroberts74; Jun 28th 2014 at 03:46 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
    $\displaystyle \frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}$

    If T is a function of x, y, z, and t, then
    $\displaystyle \frac{dT}{dt}= \frac{\partial T}{\partial t}+ \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}$
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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
    $\displaystyle \frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}$

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

    and what would $\displaystyle \frac{\partial T}{\partial t}$ be?
    Last edited by Jonroberts74; Jun 28th 2014 at 07:57 PM.
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    Re: temperature at (x,y,z)

    for part (b)

    not even sure if everything is correct yet

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

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

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

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

    a) what is $\displaystyle T'(t)$

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

    $\displaystyle =-2cos(t)sin(t)+2sin(t)cos(t)+2t$

    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? $\displaystyle \frac{dT}{dt}= 2t$. Another way to see that is to write
    $\displaystyle T= x^2+ y^2+ z^2= cos^2(t)+ sin^2(t)+ t^2= 1+ t^2$ and then differentiate that.
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