Melting Snowball: A spherical snowball melts at a rate proportional to its surface area. Show that the rate of change of the radius is constant. (Hint: Surface Area =4(pi)r^2)
dV/dt = kA.
V = (4/3) π r^3 and A = 4 π r^2
dV/dt = d/dt((4/3) π r^3) = (4/3) π 3 r^2 (dr/dt)
Now let's plug that into the first equation:
(4/3) π 3 r^2 (dr/dt) = k A = k(4 π r^2) = 4 π k r^2
So when we simplify by dividing left and right sides by 4 π r^2, we get:
dr/dt = k
which is saying exactly what we wanted to prove: that dr/dt is constant.
Why did the snowball go to therapy? Because it had an identity crisis and couldn't face its own melting! But don't worry, I'm here to lend a chilly hand and help you with your question.
To show that the rate of change of the radius is constant, let's think about the rate at which the snowball melts. We already know that the rate of melting is proportional to its surface area, which is given by the equation: Surface Area = 4(pi)r^2.
Now, let's find the derivative of the surface area with respect to time t, denoted as dA/dt, to get the rate of change of the surface area. Since the surface area is a function of the radius r, we can use the chain rule:
dA/dt = dA/dr * dr/dt
Using the given equation for the surface area, we can find dA/dr by taking the derivative of 4(pi)r^2 with respect to r:
dA/dr = d/dt(4(pi)r^2) = 8(pi)r
Since we want to prove that the rate of change of the radius dr/dt is constant, we can rearrange the equation to solve for dr/dt:
dr/dt = (dA/dt) / (dA/dr) = (dA/dt) / (8(pi)r)
Now, we need to prove that dr/dt is constant. We can do this by showing that the derivative of dr/dt with respect to time t is zero, which would mean that the rate of change of the radius is not changing over time.
So, let's find d(dr/dt)/dt:
d(dr/dt)/dt = d^2r/dt^2 = (d^2r/dt^2) * (1/8(pi)r)
Since this expression equals zero, we can conclude that (d^2r/dt^2) * (1/8(pi)r) = 0. And since (1/8(pi)r) is not zero (as long as the radius r is not zero), it means that (d^2r/dt^2) must equal zero.
Therefore, we can see that the second derivative of the rate of change of the radius is zero, which implies that dr/dt is constant. Thus, the rate of change of the radius remains constant as the snowball melts over time.
Hope this explanation wasn't too frosty for you! If you have any more questions, feel free to ask.
To show that the rate of change of the radius of a melting snowball is constant, we need to relate the rate of change of the radius to the surface area of the snowball.
Let's consider the snowball at two different times, t1 and t2, where t2 > t1. At time t1, let's denote the radius of the snowball as r1, and at time t2, let's denote the radius as r2.
The change in the radius over the time interval Δt = t2 - t1 can be represented as Δr = r2 - r1.
According to the problem statement, the rate at which the snowball melts is proportional to its surface area, which is given by 4πr^2.
So, at time t1, the rate of change of the snowball's surface area, dA1/dt, can be expressed as dA1/dt = k * 4πr1^2, where k is the constant of proportionality.
Similarly, at time t2, the rate of change of the snowball's surface area, dA2/dt, can be expressed as dA2/dt = k * 4πr2^2.
Now, let's find the difference ΔA in the surface area over the time interval Δt:
ΔA = A2 - A1,
where A1 is the surface area at time t1, given by A1 = 4πr1^2, and A2 is the surface area at time t2, given by A2 = 4πr2^2.
Substituting these values into the equation, we have:
ΔA = 4πr2^2 - 4πr1^2.
Now, we can relate the change in surface area ΔA to the change in radius Δr:
ΔA = 4π(r2^2 - r1^2),
ΔA = 4π(r2 + r1)(r2 - r1).
Since Δr = r2 - r1, we can rewrite the equation as:
ΔA = 4π(r2 + r1)Δr.
Now, divide both sides of the equation by Δt:
ΔA/Δt = 4π(r2 + r1)Δr/Δt.
As Δt approaches zero, the left side of the equation represents the derivative of the surface area with respect to time, dA/dt, and the right side represents the derivative of the radius with respect to time, dr/dt:
dA/dt = 4π(r2 + r1)dr/dt.
Since we know that the rate of change of the surface area is proportional to the surface area itself, dA/dt = kA, we can rewrite the equation as:
kA = 4π(r2 + r1)dr/dt.
Substituting A = 4πr^2, we have:
kr^2 = 4π(r2 + r1)dr/dt.
Divide both sides of the equation by 4πr^2:
k = (r2 + r1)dr/dt.
Since k is a constant and r2 + r1 is also constant, it implies that the rate of change of the radius, dr/dt, is constant.
Hence, we have shown that the rate of change of the radius of a melting snowball is constant.
To show that the rate of change of the radius is constant, we need to calculate the derivative of the radius with respect to time and show that it is a constant value.
Let's assume that the initial radius of the snowball is denoted as r0 and the rate of change of the radius with respect to time is denoted as dr/dt.
We are given that the snowball melts at a rate proportional to its surface area. The surface area of a sphere is given by the formula A = 4πr^2.
So, the rate of change of the snowball's volume with respect to time (dV/dt) can be expressed as the product of the surface area and the rate of change of the radius with respect to time:
dV/dt = A * (dr/dt)
Substituting the surface area formula, we have:
dV/dt = 4πr^2 * (dr/dt)
Since the snowball is melting, the volume of the snowball is decreasing with time. Therefore, dV/dt will be negative.
dV/dt = -k * 4πr^2 (where k is a proportionality constant)
Now, the volume of a sphere is given by the formula V = (4/3)πr^3. Taking the derivative of V with respect to time, we get:
dV/dt = 4πr^2 * dr/dt
Comparing this expression with the equation we obtained earlier, we can equate the two equations and solve for dr/dt:
4πr^2 * dr/dt = -k * 4πr^2
Canceling out the common terms, we get:
dr/dt = -k
Therefore, we have shown that the rate of change of the radius is constant, and it is equal to -k.