A spherical balloon is being inflated at a rate of 10 cubic centimeters per second.
A. Find an expression for dr/dt, the rate at which the radius of the balloon is increasing.
B. How fast is the radius of the balloon increasing when the diameter is 40 cm?
C. How fast is the surface area of the balloon increasing when the radius is 5 cm?
(A) v = 4/3 πr^3
dv/dt = 4πr^2 dr/dt
dr/dt = 10/(4πr^2)
now B is easy, and
(C) A = 4πr^2
dA/dt = 8πr dr/dt
A. To find an expression for dr/dt, we can use the relationship between the volume of a sphere and its radius. The volume of a sphere is given by V = (4/3)πr^3.
Taking the derivative of both sides with respect to time t, we have dV/dt = (4/3)(3πr^2)(dr/dt).
Given that dV/dt = 10 cubic centimeters per second, we can rewrite the expression as 10 = 4πr^2(dr/dt).
Simplifying, we have dr/dt = 10 / (4πr^2).
B. We are given that the diameter of the balloon is 40 cm. Since diameter = 2 * radius, we have r = 40 / 2 = 20 cm.
Plugging this value into the expression we found in part A, we have dr/dt = 10 / (4π(20)^2).
Simplifying, we find dr/dt = 10 / (4π(400)) = 10 / (1600π) = 0.00198 cm/s (rounded to 5 decimal places).
Therefore, the radius of the balloon is increasing at a rate of 0.00198 cm/s when the diameter is 40 cm.
C. We are given that the radius of the balloon is 5 cm.
To find how fast the surface area is increasing, we need to find dA/dt, the derivative of the surface area of a sphere with respect to time.
The surface area of a sphere is given by A = 4πr^2.
Taking the derivative of both sides with respect to time t, we have dA/dt = 4π(2r)(dr/dt).
Given that r = 5 cm and dr/dt is the rate at which the radius is increasing, we can plug in these values to find dA/dt.
dA/dt = 4π(2(5))(dr/dt) = 40π(dr/dt).
Since we know dr/dt = 10 / (4πr^2), we can substitute this into the equation.
dA/dt = 40π(10 / (4πr^2)) = (400 / r^2) square centimeters per second.
Plugging in r = 5 cm, we have dA/dt = 400 / (5^2) = 400 / 25 = 16 square centimeters per second.
Therefore, the surface area of the balloon is increasing at a rate of 16 square centimeters per second when the radius is 5 cm.
A. The volume of a sphere is given by the formula V = (4/3)πr^3, where V is the volume and r is the radius. We are given that the volume is increasing at a rate of 10 cubic centimeters per second. Therefore, we can differentiate the volume formula with respect to time to find the rate of change of the volume with respect to time:
dV/dt = d/dt [(4/3)πr^3]
Using the chain rule, we get:
dV/dt = (4/3)π * 3r^2 * (dr/dt)
Simplifying further:
dV/dt = 4πr^2 * (dr/dt)
Therefore, the expression for dr/dt, the rate at which the radius of the balloon is increasing, is:
dr/dt = (dV/dt) / (4πr^2)
B. We are given that the diameter is 40 cm. The radius of the sphere is half the diameter: r = 40/2 = 20 cm. We need to find how fast the radius is increasing, so we substitute the given values into the expression we found in part A:
dr/dt = (dV/dt) / (4πr^2)
Substituting dV/dt = 10 cubic centimeters per second and r = 20 cm:
dr/dt = 10 / (4π(20)^2)
Simplifying:
dr/dt = 10 / (4π(400))
dr/dt ≈ 10 / (4π(400))
dr/dt ≈ 0.0126 cm/s
Therefore, the radius of the balloon is increasing at a rate of approximately 0.0126 cm/s when the diameter is 40 cm.
C. We are given that the radius is 5 cm. We need to find how fast the surface area is increasing, so we need to differentiate the surface area formula with respect to time. The surface area of a sphere is given by the formula A = 4πr^2. Differentiating this formula with respect to time, we get:
dA/dt = d/dt (4πr^2)
Using the chain rule, we get:
dA/dt = 4π * 2r * (dr/dt)
Substituting the given values of r = 5 cm and dr/dt (which we calculated in part A):
dA/dt = 4π * 2(5) * (dr/dt)
Simplifying:
dA/dt = 40π * (dr/dt)
Substituting dr/dt ≈ 0.0126 cm/s from part B:
dA/dt = 40π * 0.0126
dA/dt ≈ 1.588 cm^2/s
Therefore, the surface area of the balloon is increasing at a rate of approximately 1.588 cm^2/s when the radius is 5 cm.
To solve this problem, we can use the formulas for the volume and surface area of a sphere.
A. The volume of a sphere is given by V = (4/3)πr^3, where r is the radius. We are given that the volume is increasing at a rate of 10 cubic centimeters per second, so we can differentiate the formula with respect to time, t, to find the rate at which the radius is increasing:
dV/dt = (4/3)π * d(r^3)/dt
We can simplify this expression by applying the chain rule:
10 = (4/3)π * 3r^2 * dr/dt
Simplifying further, we get:
dr/dt = 10 / ((4/3)π * 3r^2)
B. To find how fast the radius is increasing when the diameter is 40 cm, we first need to find the radius. The diameter is twice the radius, so if the diameter is 40 cm, the radius is 20 cm.
Now we can substitute the value of the radius into the expression we found in part A:
dr/dt = 10 / ((4/3)π * 3(20^2))
Simplifying further, we get:
dr/dt = 10 / (320π)
So the rate at which the radius is increasing when the diameter is 40 cm is 10 / (320π) cubic centimeters per second.
C. To find how fast the surface area of the balloon is increasing when the radius is 5 cm, we need to find the formula for the surface area of a sphere. The surface area is given by A = 4πr^2.
Differentiating the formula with respect to time, we get:
dA/dt = 4π * d(r^2)/dt
Again, applying the chain rule, we get:
dA/dt = 4π * 2r * dr/dt
Substituting the value of the radius, which is 5 cm, we get:
dA/dt = 4π * 2(5) * dr/dt
Simplifying further, we get:
dA/dt = 40π * dr/dt
Since we already know from part A that dr/dt = 10 / ((4/3)π * 3r^2), we can substitute this value into the equation:
dA/dt = 40π * (10 / ((4/3)π * 3(5^2)))
Simplifying further, we get:
dA/dt = 400π / (100π)
Finally, we simplify to get the final answer:
dA/dt = 4 square centimeters per second.