A balloon is being filled at a rate of 4000 cm^3/min. How fast is the surface area of the ballon increasing when the radius is 125 cm? V=4/3pir^3, SA=4pir^2

V = (4/3)π r^3

dV/dt = 4π r^2 dr/dt
when r = 125
4000 = 4π(125^2) dr/dt
dr/dt = 4000/(62500π) =8/(125π)

SA = 4πr^2
d(SA)/dt = 8πr dr/dt
= 8π(125)(8/125π) = 64

check my arithmetic and put in the units. (64 cm^2)

Find the equation of the tangent line to the curve y=e^2x + 4x at the point (0,1)

the tangent line at any point (x,y) has slope y'

y = e^2x + 4x
y' = 2e^2x + 4

y'(0) = 6

so, now you have a point, (0,1), and a slope, 6.

using the point-slope form for a line,

(y-1)/(x-0) = 6
y-1 = 6x
y = 6x+1

To find the rate at which the surface area of the balloon is increasing, we need to differentiate the formula for surface area (SA) with respect to time (t).

The formula for surface area of a balloon in terms of its radius (r) is SA = 4πr^2.

To find dSA/dt, we need to differentiate both sides of the equation with respect to time:

dSA/dt = d(4πr^2)/dt

Since r is also changing with respect to time, we need to use the chain rule. Let's differentiate each term step by step.

dSA/dt = d(4πr^2)/dt
= 8πr * dr/dt

Now, we need to find the value of dr/dt, which represents the rate at which the radius is changing over time.

We are given that the balloon is being filled at a rate of 4000 cm^3/min. Since the volume of a balloon is given by V = (4/3)πr^3, we can differentiate this equation with respect to time (t) to find dr/dt.

dV/dt = d((4/3)πr^3)/dt
= 4πr^2 * dr/dt

Now, we can solve for dr/dt:

dr/dt = dV/dt / (4πr^2)
= 4000 cm^3/min / (4π(125 cm)^2)
= 4000 cm^3/min / (4π(15625 cm^2))
= 4000 cm^3/min / (62500π cm^2)
≈ 0.0202 cm/min

Finally, substitute the value of dr/dt into the rate of change equation for surface area:

dSA/dt = 8πr * dr/dt
= 8π(125 cm) * 0.0202 cm/min
= 251.3274 cm^2/min

Therefore, when the radius is 125 cm, the surface area of the balloon is increasing at a rate of approximately 251.3274 cm^2/min.