a spherical balloon filled with gas has a leak that permits the gas to escape at a rate of 1.5 cubic meters per minute. how fast is the surface area of the balloon shrinking when the radius is 4 meters?
v = 4/3 pi r^3
dv/dt = 4pi r^2 dr/dt
-1.5 = 4pi (16) dr/dt
dr/dt = -1.5/64pi
a = 4pi r^2
da/dt = 8pi r dr/dt
= 8pi (4) (-1.5/64pi)
= -3/4 m^2/min
Well, let's calculate it step by step. First, we need to find the rate at which the radius is changing over time. The volume of a sphere is given by the formula V = (4/3)πr³, where V is the volume and r is the radius.
Now, we know that the gas is escaping at a rate of 1.5 cubic meters per minute. This means that the volume of the balloon is decreasing at a rate of 1.5 cubic meters per minute.
Differentiating the volume formula with respect to time, we get dV/dt = 4πr²(dr/dt), where dV/dt is the rate at which the volume is changing over time and dr/dt is the rate at which the radius is changing over time.
Since we have dr/dt and r = 4 meters (as given in the question), we can substitute these values into the equation to find dV/dt.
1.5 = 4π(4)²(dr/dt)
1.5 = 16π(dr/dt)
(dr/dt) = 1.5/(16π)
So, the rate at which the radius is changing is 1.5/(16π) meters per minute.
To find how fast the surface area of the balloon is shrinking, we can use the formula for the surface area of a sphere, A = 4πr².
Differentiating this formula with respect to time, we get dA/dt = 8πr(dr/dt).
Substituting the values, we have:
dA/dt = 8π(4)(1.5/(16π))
dA/dt = 8π(4)(1.5)/(16π)
dA/dt = 4(1.5)
dA/dt = 6
Therefore, when the radius is 4 meters, the surface area of the balloon is shrinking at a rate of 6 square meters per minute.
And hey, here's a funny thought - there's nothing like popping a balloon to make surface area disappear in a hurry! Just make sure you're not standing too close when it happens! 🎈💥😄
To find how fast the surface area of the balloon is shrinking, we need to differentiate the surface area formula with respect to time. First, let's find the surface area formula for a sphere.
The surface area of a sphere is given by:
A = 4πr^2
Where A is the surface area and r is the radius of the sphere.
Now, let's differentiate the formula with respect to time (t):
dA/dt = 8πr(dr/dt)
Here, dA/dt represents the rate of change of the surface area, and dr/dt represents the rate of change of the radius.
Given information:
dr/dt = rate of change of the radius = 0 (since the radius does not change in this question)
To find how fast the surface area is shrinking, we can substitute the given values into the equation:
dA/dt = 8π(4)(0) = 0
Therefore, when the radius is 4 meters and the rate of gas escape is 1.5 cubic meters per minute, the surface area of the balloon is not shrinking, as the rate of change (dA/dt) is zero.
To determine how fast the surface area of the balloon is shrinking, we need to find the rate of change of the surface area with respect to time.
Let's denote the radius of the balloon as r and the surface area as A. We know that the rate at which the volume is changing, dv/dt, is given as 1.5 cubic meters per minute.
Now, let's find an equation that relates the surface area A to the radius r. The formula for the surface area of a sphere is given by A = 4πr^2.
To find the derivative dA/dt (the rate at which the surface area is changing with respect to time), we will apply the chain rule of differentiation. Since A is a function of r and r is a function of t, we have:
dA/dt = dA/dr * dr/dt
Differentiating the equation A = 4πr^2, we get:
dA/dr = 8πr
Now, we need to find the value of dr/dt (the rate at which the radius is changing with respect to time) when the radius is 4 meters. Unfortunately, the problem statement does not provide any information about the rate of change of the radius. Hence, we cannot calculate dr/dt unless an additional equation or value is given in the problem.
In summary, without knowing the rate at which the radius is changing with respect to time, we cannot determine how fast the surface area of the balloon is shrinking when the radius is 4 meters.