Verified answer

Both the volume and radius of the sphere are functions of time so

V(t) = (4/3)π[r(t)]³

Differentiate implicitly to get V' then solve for r'

V' = 4πr²r'

r' = V'/(4πr²)

From the problem statement, the rate of change of volume: V' = 3 in³/s

So when r = 5 cm,

r' = (3 in³/s)/(4π*(5 in)²)

r' = 0.010 in/s

or

r' = 9.55 x 10⁻³ in/s

since the balloon is being inflated at a rate of 3 cubic inches per second that means that the rate of change or the derivative of the volume of the sphere is changing at that rate so what you do is take the derivative of the formula for the volume of a sphere with respect to time and just plug in 5 inches into the formula and you should have your answer

V = 4/3πR³

dV/dt = 4πR²dR/dt

dR/dt = (dV/dt)/4πR²

Given

R = 5 inches

dV/dt = 3 cubic inches/sec

dR/dt = 3/4π5² = .010 in/sec ( to 3 decimal places) = 0.00955 (to 3 significant digits)

For a sphere,

V=(4/3)pi*R^3

dV=(4/3)pi*3R^2 dR

dV=4*pi*R^2*dR

dV=3 cu.in/s

R=5

dR=dV/(4*pi*R^2)=3/(4*pi*25)=0.00955 in/s

dx/dt = one million feet/sec x^2 + y^2 = 4 hundred 2x(dx/dt) + 2y(dy/dt) = 0 x(dx/dt) + y(dy/dt) = 0 dy/dt = -(x/y)(dx/dt) =-(3/10)(one million) = -0.3 feet/sec is how briskly your face will hit the floor