Calculus - Taylor Expansion, maybe. Not sure how to simplify.

[Tsunoyukami]

I am attempting to complete a problem for a problem set and am having difficulty simplifying an expression; any help would be greatly appreciated!

The question is a physics question which attempts to derive an equation for the temperature within a planet as a function of depth assuming spherical co-ordinates. "Conside a thin spherical shell whose inner boundary has radius r and outer boundary had radius r + ∂r. The heat flux into the shell from the inner boundary is q[itex]_{r}[/itex](r) and the heat flux out of the shell at the top boundary is q[itex]_{r}[/itex](r + ∂r). The energy produced by radioactivity in the shell is [itex]\rho\rho[/itex]HV where V is the volume of the spherical shell, H is the heat generation rate per unit mass and [itex]\rho[/itex] is the density."

To determine the volume of the shell V I have used:

V = V(r + ∂r) - V(r)

Using the fact that ∂r [itex]\rightarrow[/itex] 0 I have found:

V = 4[itex]\pi[/itex]r[itex]^{2}[/itex]∂r + 4[itex]\pi[/itex]r ∂r[itex]^{2}[/itex]

Is this a reasonable time to use the fact that ∂r [itex]\rightarrow[/itex] 0? Is the second term in my expression for V negligible?


The primary question I have is whether or not there is a way to simplify q[itex]_{r}[/itex](r + ∂r); I feel like there is - and I feel like it might have to do with a Taylor expansion, but I haven't been able to figure it out - all the links I find when I do a google search are either too simple or too complex. Any help would be greatly appreciated!

(After I do this I have to write an expression that says the sum of the heat going in, the heat leaving and the heat produced inside the shell equals 0 [by assuming in is either positive or negative and out is the opposite sign of in).

Again, any help would be greatly appreciated!

 

[LCKurtz]

I will use the notation ##\Delta V = V(r+\Delta r)-V(r)= 4\pi r^2\Delta r + 4\pi r (\Delta r)^2##. This is a consequence of a general property of differentiable functions. If ##f(x)## is differentiable at ##x##, then$$
f'(x)-\frac{f(x+h)-f(x)}{h}\rightarrow 0\hbox{ if }h\rightarrow 0$$so we might use the notation$$f'(x)-\frac{f(x+h)-f(x)}{h}=o(h)$$where ##o(h)## represents a term that goes to 0 as h does. $$hf'(x) -(f(x+h)-f(x)) = ho(h)$$ $$
\Delta f = hf'(x) -ho(h)$$That last term is h times a term that goes to zero with h, so is a higher order zero than ##hf'(x)##. That is why for ##h## small the second term is neglected and you say approximately ##\Delta f = hf'(x)##. Your example is a special case where the variable is ##r## instead of ##x## and $$
f(r)=\frac 4 3 \pi r^3$$the volume of a sphere.

 

[Tsunoyukami]

Thank you very much!

As to my second question - is there any way to write q[itex]_{r}[/itex](r + ∂r) another way so that I can simplify my equation when I do an energy balance? That is, what does q[itex]_{r}[/itex](r + ∂r) equal mathematically? (I don't need any kind of really deep proof, just some kind of evidence so I can understand how you get to what it equals).

 

[Tsunoyukami]

Alright, so I've made some progress.

Using the following:
E[itex]_{in}[/itex] = q[itex]_{r}[/itex](r) x A(r) = q[itex]_{r}[/itex](r) x 4[itex]\pi[/itex]r[itex]^{2}[/itex]
E[itex]_{out}[/itex] = q[itex]_{r}[/itex](r +[itex]\delta[/itex]r) x A(r +[itex]\delta[/itex]r) = q[itex]_{r}[/itex](r +[itex]\delta[/itex]r) x 4[itex]\pi[/itex](r +[itex]\delta[/itex]r)[itex]^{2}[/itex]
E[itex]_{produced}[/itex] = [itex]\rho[/itex]HV = [itex]\rho[/itex]H4[itex]\pi[/itex]r[itex]^{2}[/itex][itex]\delta[/itex]r

I have written the following equation:

E[itex]_{in}[/itex] - E[itex]_{out}[/itex] + E[itex]_{produced}[/itex] = mC[itex]\Delta[/itex]T = [itex]\rho[/itex]VC[itex]\Delta[/itex]T = [itex]\rho[/itex]4[itex]\pi[/itex]r[itex]^{2}[/itex][itex]\delta[/itex]r C[itex]\Delta[/itex]T (this follows from discussion in class; its part of thermal physics).

I can write out the formula (it kinda sucks):

E[itex]_{in}[/itex] - E[itex]_{out}[/itex] + E[itex]_{produced}[/itex] = 4[itex]\pi[/itex]r[itex]^{2}[/itex]q[itex]_{r}[/itex](r) - 4[itex]\pi[/itex](r +[itex]\delta[/itex]r)[itex]^{2}[/itex]q[itex]_{r}[/itex](r +[itex]\delta[/itex]r) + 4[itex]\pi[/itex][itex]\rho[/itex]Hr[itex]^{2}[/itex][itex]\delta[/itex]r = [itex]\rho[/itex]4[itex]\pi[/itex]r[itex]^{2}[/itex][itex]\delta[/itex]rC[itex]\frac{(T(t + \delta t) - T(t))}{\delta t}[/itex]

4[itex]\pi[/itex]r[itex]^{2}[/itex]q[itex]_{r}[/itex](r) - 4[itex]\pi[/itex](r[itex]^{2}[/itex] + 2r[itex]\delta[/itex]r + ([itex]\delta[/itex]r)[itex]^{2}[/itex])q[itex]_{r}[/itex](r +[itex]\delta[/itex]r) + 4[itex]\pi[/itex][itex]\rho[/itex]Hr[itex]^{2}[/itex][itex]\delta[/itex]r = [itex]\rho[/itex]4[itex]\pi[/itex]r[itex]^{2}[/itex][itex]\delta[/itex]rC[itex]\frac{(T(t + \delta t) - T(t))}{\delta t}[/itex]

4[itex]\pi[/itex]r[itex]^{2}[/itex]q[itex]_{r}[/itex](r) - (4[itex]\pi[/itex]r[itex]^{2}[/itex]q[itex]_{r}[/itex](r +[itex]\delta[/itex]r) + 8[itex]\pi[/itex]r[itex]\delta[/itex]rq[itex]_{r}[/itex](r +[itex]\delta[/itex]r) + 4[itex]\pi[/itex]([itex]\delta[/itex]r)[itex]^{2}[/itex]q[itex]_{r}[/itex](r +[itex]\delta[/itex]r)) + 4[itex]\pi[/itex][itex]\rho[/itex]Hr[itex]^{2}[/itex][itex]\delta[/itex]r = [itex]\rho[/itex]4[itex]\pi[/itex]r[itex]^{2}[/itex][itex]\delta[/itex]rC[itex]\frac{(T(t + \delta t) - T(t))}{\delta t}[/itex]4[itex]\pi[/itex]r[itex]^{2}[/itex]q[itex]_{r}[/itex](r) - 4[itex]\pi[/itex]r[itex]^{2}[/itex]q[itex]_{r}[/itex](r +[itex]\delta[/itex]r) - 8[itex]\pi[/itex]r[itex]\delta[/itex]rq[itex]_{r}[/itex](r +[itex]\delta[/itex]r) - 4[itex]\pi[/itex]([itex]\delta[/itex]r)[itex]^{2}[/itex]q[itex]_{r}[/itex](r +[itex]\delta[/itex]r) + 4[itex]\pi[/itex][itex]\rho[/itex]Hr[itex]^{2}[/itex][itex]\delta[/itex]r = [itex]\rho[/itex]4[itex]\pi[/itex]r[itex]^{2}[/itex][itex]\delta[/itex]rC[itex]\frac{(T(t + \delta t) - T(t))}{\delta t}[/itex]

r[itex]^{2}[/itex]q[itex]_{r}[/itex](r) - r[itex]^{2}[/itex]q[itex]_{r}[/itex](r +[itex]\delta[/itex]r) - 2r[itex]\delta[/itex]rq[itex]_{r}[/itex](r +[itex]\delta[/itex]r) - ([itex]\delta[/itex]r)[itex]^{2}[/itex]q[itex]_{r}[/itex](r +[itex]\delta[/itex]r) + [itex]\rho[/itex]Hr[itex]^{2}[/itex][itex]\delta[/itex]r = [itex]\rho[/itex]r[itex]^{2}[/itex][itex]\delta[/itex]rC[itex]\frac{(T(t + \delta t) - T(t))}{\delta t}[/itex]

- (r[itex]^{2}[/itex](q[itex]_{r}[/itex](r +[itex]\delta[/itex]r) -q[itex]_{r}[/itex](r)) + 2r[itex]\delta[/itex]rq[itex]_{r}[/itex](r +[itex]\delta[/itex]r) + ([itex]\delta[/itex]r)[itex]^{2}[/itex]q[itex]_{r}[/itex](r +[itex]\delta[/itex]r)) + [itex]\rho[/itex]Hr[itex]^{2}[/itex][itex]\delta[/itex]r = [itex]\rho[/itex]r[itex]^{2}[/itex][itex]\delta[/itex]rC[itex]\frac{(T(t + \delta t) - T(t))}{\delta t}[/itex]

So far I have expanded a perfect square, divided through by 4[itex]\pi[/itex] and factored to simplify somewhat. (My solution on paper looks a bit nicer, but I'm not very good at coding this so that it looks proper so I apologize if it looks really messy). Next, I divide through by what is left of the volume term, r[itex]^{2}[/itex][itex]\delta[/itex]r. It's hard for me to get fractions to appear nicely on here, so I'm going to skip a step (the step in which I would show everything in both the numerator and denominator) and then just show you what I'm left with:

[itex] -(\frac{r^{2}(q_{r}(r + \delta r) -q_{r}(r)}{r^{2} \delta r} + \frac{2r\delta rq_{r}(r + \delta r)}{r^{2} \delta r} + \frac{(\delta r)^{2}q_{r}(r +\delta r)}{r^{2} \delta r}) + \frac{\rho Hr^{2}\delta r}{r^{2} \delta r} = \frac {\rho r^{2}\delta rC \frac{(T(t + \delta t) - T(t))}{\delta t}}{r^{2} \delta r}[/itex]

[itex] -(\frac{(q_{r}(r + \delta r) -q_{r}(r)}{\delta r} + \frac{2q_{r}(r + \delta r)}{r} + \frac{\delta rq_{r}(r +\delta r)}{r^{2}}) + \rho H = \rho C \frac{(T(t + \delta t) - T(t))}{\delta t}[/itex]

Next we take the limit as [itex]\delta r[/itex] and [itex]\delta t[/itex] approach 0 - then I get the right answer (as I was typing this I found I had missed a 2 going from one step to the next which was throwing me off, but I was so close to finishing typing this up that I decided I might as well continue). Anyways, disregard my previous questions; if you find anything wrong with this please let me know!