The surface area of a sphere initially zero increases uniformly at the rate of 26 cm^2/seconds. Find the rate at which the radius is increasing at the end of 2 seconds.
can i get the answer
Oh my, we're getting mathy here! Let's have some fun with this problem.
To start off, we know that the surface area of a sphere (A) can be calculated using the formula A = 4πr^2, where r is the radius.
Given that the surface area is increasing uniformly at a rate of 26 cm^2/second, we can say that dA/dt = 26 cm^2/s, where dA represents the change in surface area and dt represents the change in time.
We're asked to find the rate at which the radius is increasing at the end of 2 seconds, so we need to find dr/dt when t = 2 seconds.
Now, let's differentiate the formula for surface area with respect to time, using the chain rule:
dA/dt = 8πr * dr/dt
Remember we said that dA/dt = 26 cm^2/s, so we can plug this in:
26 = 8πr * dr/dt
Now we can solve for dr/dt:
dr/dt = 26 / (8πr)
Since we're trying to find the rate of change in radius when t = 2 seconds, let's substitute t = 2 into our equation:
dr/dt = 26 / (8πr) = 26 / (8π(2)) = 26 / (16π) = 13 / (8π)
So, at the end of 2 seconds, the rate at which the radius is increasing is 13 / (8π) cm/s.
Hope that puts a smile on your face!
To find the rate at which the radius is increasing, we need to relate the rate of change of the surface area to the rate of change of the radius.
The surface area of a sphere is given by the formula:
A = 4πr^2
Where A is the surface area and r is the radius.
Given that the surface area is increasing at a rate of 26 cm^2/second, we can express this as:
dA/dt = 26 cm^2/s
To find the rate at which the radius is increasing, we differentiate the formula for surface area with respect to time:
dA/dt = d/dt(4πr^2)
The derivative of r^2 with respect to time is 2r(dr/dt), where dr/dt represents the rate at which the radius is changing over time.
Therefore, we have:
dA/dt = 8πr(dr/dt)
Now, we can substitute the given value of dA/dt as 26 cm^2/s and the time value of 2 seconds.
26 cm^2/s = 8πr(dr/dt)
To find the rate at which the radius is increasing at the end of 2 seconds, we need to solve for dr/dt.
dr/dt = (26 cm^2/s) / (8πr)
Substituting the value of r and simplifying the equation:
dr/dt = (26 cm^2/s) / (8π(0.5)^2)
dr/dt = (26 cm^2/s) / (8π(0.25))
dr/dt = (26 cm^2/s) / (2π)
dr/dt = 13 cm^2/sπ
Therefore, the rate at which the radius is increasing at the end of 2 seconds is 13 cm^2/sπ.
To find the rate at which the radius is increasing, we need to relate the surface area of the sphere to its radius.
The formula for the surface area of a sphere is given by:
A = 4πr^2
where A is the surface area and r is the radius.
Given that the surface area is increasing uniformly at a rate of 26 cm^2/second, we can differentiate the equation with respect to time to find the rate at which the surface area is changing:
dA/dt = d(4πr^2)/dt
Now, let's differentiate both sides of the equation:
dA/dt = 8πr * dr/dt
We know that dA/dt = 26 cm^2/second (given rate of increase in surface area), and we need to find dr/dt (rate of change of the radius).
Substituting the known values into the equation, we get:
26 = 8πr * dr/dt
Now, let's solve for dr/dt:
dr/dt = 26 / (8πr)
To find the rate at which the radius is increasing at the end of 2 seconds, we need to substitute the value of r at t = 2 seconds into the equation.
Since the initial surface area is zero, we can assume the initial radius is also zero. So, at t = 2 seconds, r = 0 + (2 * dr/dt), where dr/dt is the rate we are trying to find.
Substituting the values into the equation, we get:
dr/dt = 26 / (8π(2))
dr/dt = 26 / (16π)
Therefore, the rate at which the radius is increasing at the end of 2 seconds is 26 / (16π) cm/second.
a = 4πr^2
at t=2, a=52, so r=√(13/π)
da/dt = 8πr dr/dt
so, you need
8π√(13/π) dr/dt = 26
dr/dt = 1/4 √(13/π)