The radius of a sphere is increasing at a rate proportional to its radius. If the radius is 4 initially, and the radius is 10 after two seconds, what will the radius be after three seconds?
dR/dt = c R
Integrating this differential equation gives you
ln R/Ro = c t
R/Ro = e^(ct)
At t = 4 seconds, ln 2.5 = 2c
Therefore the constant is
c = 0.4581
D = Ro e^(0.4581 t)
When t = 3, R/Ro = 3.9523
R = 15.81
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Well, it seems like our sphere is growing and growing! I hope it doesn't end up taking over the universe, that would be quite spherical domination! Anyway, after three seconds, the radius of the sphere will be approximately 15.81. And who knows, maybe it'll use its newfound size to become the world's biggest beach ball! Just don't forget your sunscreen when you go near it!
To solve this problem, we have the differential equation:
dy/dt = c * y,
where y represents the radius of the sphere and c is a constant.
Given that the radius is initially 4 (y0 = 4), we can integrate the equation to find the relationship between radius and time:
∫(1/y) dy = ∫c dt.
This simplifies to:
ln|y| = ct + k,
where k is the constant of integration.
Using the given information that the radius is 10 after 2 seconds (t = 2, y = 10), we can substitute these values into the equation:
ln|10| = c*2 + k.
Simplifying further:
ln(10) = 2c + k.
Now, we can solve for the constant c:
2c = ln(10) - k.
Next, we need to find the constant k. When t = 0, y = 4, so:
ln|4| = c*0 + k,
ln(4) = k.
Substituting this value for k into the equation for c:
2c = ln(10) - ln(4) = ln(10/4) = ln(2.5).
Now, we know the value of c:
c = ln(2.5) / 2.
Finally, we can find the radius after 3 seconds (t = 3):
y = e^(ct).
Substituting in the values we found:
y = e^((ln(2.5)/2)*3).
Simplifying:
y = e^(ln(2.5)*3/2),
y = e^(ln(2.5^3)/2),
y = 2.5^(3/2),
y ≈ 15.81.
Therefore, the radius of the sphere after 3 seconds is approximately 15.81.
To solve this problem, we can use the differential equation given: dR/dt = cR, where dR/dt represents the rate of change of the radius with respect to time.
First, let's find the value of the constant, c. We know that the radius is 4 initially, so R = 4 when t = 0. We are also given that the radius is 10 after two seconds, so R = 10 when t = 2.
We can plug these values into the equation R/Ro = e^(ct), where Ro represents the initial radius and t represents the time in seconds.
At t = 0, we have 4/Ro = e^(0*c) = e^0 = 1. This implies that Ro = 4.
At t = 2, we have 10/4 = e^(2c).
Simplifying, we get 2.5 = e^(2c).
To solve for c, we take the natural logarithm (ln) of both sides: ln(2.5) = ln(e^(2c)).
Using the logarithmic property ln(e^x) = x, we get ln(2.5) = 2c.
Dividing both sides by 2, we have c = ln(2.5)/2.
Now that we have the value of c, we can find the radius after three seconds.
Using the equation R/Ro = e^(ct), plugging in t = 3 and Ro = 4, we get R/4 = e^(ln(2.5)/2 * 3).
Simplifying further, we have R/4 = e^(1.5 * ln(2.5)).
Using the logarithmic property e^(a * ln(b)) = b^a, we get R/4 = (2.5)^1.5.
Solving for R, we multiply both sides by 4 to get R = 4 * (2.5)^1.5.
Calculating this value, R ≈ 15.81.
Therefore, the radius of the sphere after three seconds will be approximately 15.81 units.