Air is being pumped into a spherical balloon so that its volume increases at a rate of 90cm^3/s. How fast is the surface area of the balloon increasing when its radius is 9cm? Recall that a ball of radius r has volume V=4/3 pi r^3 and surface area S=4/3 pi r^2
Your formula for Surface Area is incorrect, anyhow ....
V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
when r = 9, and dV/dt = 90
90 = 4π (81) dr/dt
dr/dt = 90/(324π) = (5/18π)
SA = 4πr^2 <---- the correct formula
dSA/dt = 8π r dr/dt
for the given data:
dSA/dt = 8π(9)(5/18π) = 20 cm^2/s
To find the rate at which the surface area of the balloon is increasing, we need to differentiate the surface area formula with respect to time.
Given that the volume of the balloon is increasing at a rate of 90 cm^3/s, we can write:
dV/dt = 90 cm^3/s
To relate the volume and radius of the balloon, we can use the volume formula for a sphere:
V = (4/3)πr^3
Differentiating both sides of the equation with respect to time, we get:
dV/dt = 4πr^2(dr/dt)
Now we can substitute the given value for dV/dt and the given radius (r = 9 cm) to find the rate of change of the radius:
90 = 4π(9)^2(dr/dt)
Simplifying the equation:
90 = 4π(81)(dr/dt)
90 = 324π(dr/dt)
Now, we can solve for dr/dt, which represents the rate at which the radius is changing. Divide both sides of the equation by 324π:
90 / (324π) = dr/dt
Simplifying:
(15/54π) = dr/dt
So, when the radius is 9 cm, the rate at which the radius is changing is approximately (15/54π) cm/s.
Since we need to find how fast the surface area is increasing when the radius is 9 cm, we can substitute the radius (r = 9 cm) into the surface area formula:
S = (4/3)πr^2
S = (4/3)π(9)^2
S = (4/3)π(81)
S = 108π
Now, differentiate the surface area formula with respect to time:
dS/dt = d/dt (108π)
dS/dt = 0 (since the surface area does not change with time)
Therefore, when the radius is 9 cm, the surface area of the balloon is not changing, and the rate at which the surface area is increasing is 0 cm^2/s.
To find how fast the surface area of the balloon is increasing, we need to take the derivative of the surface area function with respect to time and apply the chain rule.
Given that the volume of the balloon is increasing at a rate of 90 cm^3/s, we are given that dV/dt = 90 cm^3/s.
Recall that the volume of a sphere is given by V = (4/3)πr^3, where r is the radius of the sphere.
Differentiating both sides of the equation with respect to time, we get:
dV/dt = d/dt[(4/3)πr^3]
To find the rate of change of the surface area, we need to differentiate the surface area formula with respect to time.
The surface area of the balloon is given by S = (4/3)πr^2.
Differentiating both sides of the equation with respect to time, we get:
dS/dt = d/dt[(4/3)πr^2]
Using the chain rule, the derivative of r^2 with respect to time is 2r*(dr/dt), since the radius is a function of time.
Therefore, we have:
dS/dt = (4/3)π * 2r * (dr/dt)
Now we can substitute the given values to find the rate at which the surface area is increasing when the radius is 9 cm.
Let's substitute r = 9 cm and dV/dt = 90 cm^3/s into the equation and solve for dS/dt:
dS/dt = (4/3)π * 2 * 9 * (dr/dt)
dS/dt = 24π * (dr/dt)
Now we can find dS/dt by multiplying 24π by the rate at which the radius is changing, which is represented by dr/dt.
Keep in mind, without knowing the value of dr/dt, we cannot determine the precise value of dS/dt.