A sphere with radius 1m has temperature 15 degrees C. It lies inside a con-
centric sphere with radius 2m and temperature 25 degrees C: The temperature
T(r) at a distance r from the common centre of the spheres satisfi�es
the di�fferential equation:
d^(2)T/dr^2 + 2/r(dT/dr) = 0
Set S = dT/dr . Then S satis�es a �first-order di�fferential equation. Solve
this to determine an expression for the temperature T(r) between the
spheres.
dS/dr + (2/r) S = 0
dS/dr = -2 S/r
dS/S = -2 dr/r
ln S = -2 ln r + C
ln S = -ln r^2 + C
ln(S r^2) = C
S r^2 = C
(dT/dr )r^2 = c
dT = c dr/r^2
T = - c /r + C
at r = 1, T = 15
at r = 2, T = 25
15 = -c + C
25 = -c/2 + C
------------
-10 = -c/2
c = 20
15 = -20 + C
C = 35
so T = -20/r + 35
Why did the temperature cross the road? Because it couldn't stay in one place forever!
But in all seriousness, let's solve this differential equation and find the expression for the temperature T(r) between the spheres.
We start by considering the equation S = dT/dr. Taking the derivative of S with respect to r, we get:
dS/dr = d^(2)T/dr^2
Now, let's substitute this into the original differential equation:
dS/dr + 2/r(S) = 0
This is a first-order linear differential equation. To solve it, we can multiply through by r to get:
r*dS/dr + 2S = 0
This equation can be rearranged to:
dS/S = -2/r dr
Integrating both sides, we get:
ln(S) = -2ln(r) + C
Where C is the constant of integration. Taking the exponential of both sides, we have:
S = e^(C)*1/r^2
Now, we recall that S = dT/dr, so:
dT/dr = e^(C)*1/r^2
Integrating both sides once more, we get:
T = -e^(C)/r + D
Where D is another constant of integration.
Using the conditions given in the problem, we can find the values of C and D. When r = 1 m, T = 15 degrees C, and when r = 2 m, T = 25 degrees C.
Plugging these values into the equation, we get the following system of equations:
15 = -e^(C)/1 + D
25 = -e^(C)/2 + D
Solving this system, we find C = ln(2) and D = 30.
Substituting these values back into the expression we found for T, we have:
T = -e^(ln(2))/r + 30
T = -2/r + 30
So the expression for the temperature T(r) between the spheres is:
T(r) = -2/r + 30
I hope this helps!
To solve the first-order differential equation for S, we can rewrite the given differential equation and substitute S = dT/dr:
d^2T/dr^2 + 2/r(dT/dr) = 0
Substituting S = dT/dr gives:
dS/dr + 2/r * S = 0
This is a first-order linear differential equation, which can be solved using an integrating factor. The integrating factor is e^(∫(2/r)dr) = e^(2ln|r|) = r^2. Multiplying the entire equation by r^2, we get:
r^2 * dS/dr + 2rS = 0
Applying the product rule of differentiation, we have:
d(r^2S)/dr = 0
Integrating both sides with respect to r:
∫d(r^2S)/dr dr = ∫0 dr
r^2S = C (where C is the constant of integration)
Dividing both sides by r^2 to isolate S:
S = C/r^2
Now, we have an expression for S in terms of the constant C and r. We can integrate this expression to find T(r). Integrating S = C/r^2 with respect to r:
∫S dr = ∫C/r^2 dr
T(r) = -C/r + K (where K is another constant of integration)
Thus, the expression for the temperature T(r) between the spheres is:
T(r) = -C/r + K
Where C and K are constants determined by the initial conditions.
To solve the first-order differential equation, let's solve for S first.
Given the differential equation: d^2T/dr^2 + (2/r)(dT/dr) = 0, we set S = dT/dr.
Now, we have S = dT/dr. To solve this, we can rewrite the given differential equation with the substituted value of S:
dS/dr + (2/r)S = 0
This is now a first-order linear differential equation. We can solve this using the method of integrating factors.
First, let's rewrite the equation in the standard form: dS/dr + (2/r)S = 0. We can see that the coefficient of S is (2/r), which depends only on r and not on S.
To find the integrating factor, we multiply the entire equation by the integrating factor, which is given by the exponential of the integral of the coefficient of S, i.e., e^(∫(2/r)dr).
Integrating the coefficient of S, we get ∫(2/r)dr = 2ln(r).
Therefore, the integrating factor is e^(2ln(r)) = r^2.
Now, we multiply the entire equation by r^2 to get:
r^2(dS/dr) + 2rS = 0
This can be rewritten as d(r^2S)/dr = 0.
Integrating both sides, we have ∫d(r^2S) = ∫0dr.
This simplifies to r^2S = C, where C is the constant of integration.
Now, we can solve for S:
S = C/r^2.
Since we initially set S = dT/dr, we have dT/dr = C/r^2.
To solve for T, we integrate both sides of the equation:
∫dT = ∫C/r^2 dr
Integrating, we get T = -C/r + D, where D is the constant of integration.
Now we have an expression for the temperature T(r) between the spheres: T(r) = -C/r + D.
To find the specific values of C and D, we can use the given conditions. We know that when r = 1 (inside the sphere with radius 1m), T = 15 degrees C, and when r = 2 (inside the concentric sphere with radius 2m), T = 25 degrees C.
Using these conditions, we can substitute the values into the expression for T(r):
For r = 1: 15 = -C + D
For r = 2: 25 = -C/2 + D
We now have a system of two equations with two variables (C and D). Solving these equations will give us the specific values of C and D, and thus the expression for the temperature T(r) between the spheres.