1. ## 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

2. ## Re: temperature at (x,y,z)

So $\displaystyle \frac{dT}{dt}= 2t$. Now, what is your question?

3. ## Re: temperature at (x,y,z)

Originally Posted by HallsofIvy
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?

4. ## 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
$\displaystyle \frac{dT}{dt}= \frac{dT}{dt}\frac{\partial t}{\partial t}+\frac{dT}{dx}\frac{\partial x}{\partial t}+\frac{dT}{dy}\frac{\partial y}{\partial t}+\frac{dT}{dz}\frac{\partial z}{\partial t}$

5. ## Re: temperature at (x,y,z)

Originally Posted by Shakarri
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

6. ## 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
$\displaystyle \frac{dT}{dt}= \frac{\partial T}{\partial t}+ \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}$

7. ## Re: temperature at (x,y,z)

Originally Posted by HallsofIvy
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?

8. ## Re: temperature at (x,y,z)

for part (b)

not even sure if everything is correct yet

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

$\displaystyle T(t)=cos^2(\frac{\pi}{2} + 0.01) + sin^2 (\frac{\pi}{2} + 0.01) + (\frac{\pi}{2} + 0.01)^2 \approx 3.5$

9. ## Re: temperature at (x,y,z)

Originally Posted by Jonroberts74
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.