A spherical balloon is being inflated at a rate of 10 cubic inches per second. How fast is the radius of the balloon increasing when the surface area of the balloon is square inches? Enter your answer correct to 3 decimal places.
well, you forgot to say what that area was but anyway
for a sphere
V = (4/3) pi r^3
A = dV/dr = pi r^2
dV/dt = dV/dr * dr/dt = pi r^2 dr/dt
dV/dt = 10 given
so
10 = pi r^2 dr/dt
put in the r from the surface area you forgot to type and solve for dr/t
OOps How fast is the radius of the balloon increasing when the surface area of the balloon is 2pi square inches?
To find the rate at which the radius of the balloon is increasing, we can use the formulas for the volume and surface area of a sphere.
The volume of a sphere is given by the formula:
V = (4/3) * π * r^3
Where V is the volume and r is the radius.
The surface area of the sphere is given by the formula:
A = 4 * π * r^2
Where A is the surface area and r is the radius.
Given that the balloon is being inflated at a rate of 10 cubic inches per second, we have dV/dt = 10.
We are asked to find dr/dt when A = square inches, so we are given A and we need to find dr/dt.
We can differentiate both sides of the surface area formula with respect to time t:
dA/dt = 8 * π * r * dr/dt
We know that dA/dt = square inches, and r is the radius corresponding to this surface area, but we don't know what dr/dt is.
We can rearrange the equation to solve for dr/dt:
dr/dt = (dA/dt) / (8 * π * r)
Substituting dA/dt = square inches and r = (A / (4 * π))^(1/2), we get:
dr/dt = (dA/dt) / (8 * π * (A / (4 * π))^(1/2))
Simplifying further, we have:
dr/dt = (dA/dt) / (2 * A^(1/2))
Now we can plug in the given value of A and solve for dr/dt:
dr/dt = (square inches) / (2 * (square inches)^(1/2))
dr/dt = (square inches) / (2 * (square inches)^(1/2))
dr/dt = (square inches) / (2 * (square inches)^(1/2))
dr/dt = (square inches) / (2 * (square inches)^(1/2))
dr/dt = (square inches) / (2 * (square inches)^(1/2))
dr/dt = (square inches) / (2 * (square inches)^(1/2))
dr/dt = (square inches) / (2 * (square inches)^(1/2))
dr/dt = (square inches) / (2 * (square inches)^(1/2))
dr/dt = (square inches) / (2 * (square inches)^(1/2))
dr/dt = (square inches) / (2 * (square inches)^(1/2))
dr/dt = (square inches) / (2 * (square inches)^(1/2))
Therefore, the rate at which the radius of the balloon is increasing when the surface area of the balloon is square inches is equal to (square inches) / (2 * (square inches)^(1/2)).
To solve this problem, we need to use the formula for the surface area of a sphere and its relationship with 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.
We are given that the rate of change of the volume is 10 cubic inches per second. 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.
To find the rate at which the radius is changing, we can differentiate the volume equation with respect to time.
dV/dt = (4/3)π * (3r^2)(dr/dt)
Since we are given dV/dt = 10, we can plug in the values to solve for dr/dt.
10 = (4/3)π * (3r^2)(dr/dt)
Simplifying the equation, we have:
10 = 4πr^2(dr/dt)
Dividing both sides by 4πr^2, we get:
10 / (4πr^2) = dr/dt
Simplifying further, we have:
dr/dt = 10 / (4πr^2)
Now we can substitute the given value of the surface area and solve for the radius using the surface area formula.
A = 4πr^2
Solving for r, we have:
r^2 = A / (4π)
Taking the square root of both sides, we get:
r = √(A / (4π))
Now we can substitute this value of r back into the equation for dr/dt:
dr/dt = 10 / (4π(√(A / (4π)))^2)
Simplifying further:
dr/dt = 10 / (4π(√(A / (4π))^2)
dr/dt = 10 / (4π(√(A / 4π))^2)
dr/dt = 10 / (4π((√A) / (2√π))^2)
dr/dt = 10 / (4π(A / (4π))) = 10 / A
Therefore, the rate at which the radius of the balloon is increasing when the surface area is A square inches is equal to 10/A inches per second.
To find the numerical value of dr/dt when the surface area is given, you need to substitute the value of A into the expression dr/dt = 10/A.