Air is being pumped into a spherical balloon so that its volume increases at a rate of 30 cubic centimeters per second. How fast is the surface area of the balloon increasing when its radius is 19 cm?

V= 4/3*pi*r^3
S= 4 pi r^2

Got it down to 3.157 which is correct. Thanks!

from V= 4/3*pi*r^3

dV/dt = 4pi(r^2)dr/dt
so when dV/dt=30 and r = 19
30 = 4pi(361)dr/dt
dr/dt = 30/[4pi(361)]

now in
A = 4pir^2
dA/dt = 8pi(r)dr/dt
= 8pi(19)*30/[4pi(361)]
= you finish it.

To find the rate of change of the surface area of the balloon, we need to find the derivative of the surface area equation with respect to time. Let's start by expressing the surface area equation in terms of the radius, r.

S = 4πr^2

Next, we can differentiate both sides of the equation with respect to time t, using the chain rule:

dS/dt = d/dt(4πr^2)

Now, let's differentiate each term separately:

dS/dt = d/dt(4π) * d/dt(r^2)

The derivative of a constant, such as 4π, is zero, so we can simplify:

dS/dt = 0 * d/dt(r^2)

Now, let's differentiate r^2 with respect to time:

dS/dt = 0 * 2r * dr/dt

Since we know that the volume of the balloon is increasing at a rate of 30 cubic centimeters per second, we can express dr/dt (the rate of change of the radius with respect to time) in terms of the volume:

dV/dt = 30

We can also express the volume equation in terms of the radius:

V = (4/3)πr^3

Now, we can differentiate both sides of the volume equation with respect to time:

dV/dt = d/dt[(4/3)πr^3]

Let's differentiate each term separately:

dV/dt = d/dt(4/3) * d/dt(πr^3)

The derivative of the constant (4/3) with respect to time is zero, so we can simplify:

dV/dt = 0 * d/dt(πr^3)

Now, let's differentiate πr^3 with respect to time:

dV/dt = 0 * 3πr^2 * dr/dt

Simplifying further:

dV/dt = 0

So, we have:

0 = 3πr^2 * dr/dt

Now, let's solve for dr/dt in terms of r:

dr/dt = 0 / (3πr^2)

dr/dt = 0

Therefore, when the radius is 19 cm, the rate at which the surface area of the balloon is increasing is 0 cm^2/s.

To find the rate at which the surface area of the balloon is increasing, we need to find the derivative of the surface area formula with respect to time.

First, let's express the surface area of the balloon as a function of the radius. From the given formula, the surface area of a sphere is S = 4πr^2.

Now, we want to find dS/dt, the rate at which the surface area is changing with respect to time. To do this, we need to differentiate the formula with respect to time using the chain rule.

Since r is a function of time, we can rewrite the surface area formula as S(t) = 4πr(t)^2, where r(t) is the radius of the balloon at time t.

Now let's differentiate both sides of the equation with respect to time:

dS/dt = d/dt (4πr(t)^2)

To apply the chain rule, we need to differentiate the term (r(t))^2 separately.

d(r(t))^2/dt = 2*r(t)*(dr(t)/dt)

Recall that the rate at which the volume is increasing is given by dV/dt = 30 cc/s.

We also know the formula for the volume of a sphere, V = (4/3)πr(t)^3.

Differentiating both sides of this equation with respect to time, we get:

dV/dt = d/dt ((4/3)πr(t)^3)

dV/dt = 4πr(t)^2 * (dr(t)/dt)

We can solve this equation for (dr(t)/dt):

dr(t)/dt = (dV/dt) / (4πr(t)^2)

Substitute the given value for dV/dt = 30 cc/s and the given radius r(t) = 19 cm:

dr(t)/dt = 30 / (4π(19)^2)

Now we can calculate the value of dr(t)/dt which represents the rate at which the radius is changing.