Deflating Balloon

[Mooch22]

A balloon is being deflated after a fun day at the carnival. The volume of the balloon is decreasing at a rate of twice the radius of the balloon. The surface area of the balloon is 400(PIE)ft^2 when t=0. Assume the balloon is spherical.

Find the following:

a.) Rate of change of the radius.

b.) Rate of change of the surface area.

c.) When the volume of the balloon is 288(PIE)ft^3, the balloon is popped. At what time was the balloon popped?

 

[Daniel_Feldman]

I'm not positive, but I believe these are standard related rates:


a.) Recall that the volume of a sphere is V=(4/3)pi(r^3)

Differentiating both sides with respect to time yields:

dV/dt=4pi(r^2)dr/dt

It's given that the volume of the balloon is decreasing at a rate of twice the radius of the balloon, do dV/dt=-2r.

-2r=4pi(r^2)dr/dt

dr/dt=-1/(2pi r)


b.) Formula for surface area of a sphere:

S=4pi(r^2)

Differentiate both sides with respect to t:

dS/dt=8pi(r)dr/dt

dr/dt=-1/(2pi r), so

dS/dt=8pi(r)(-1/2pir)

=-4


Not sure about part c. I think we need more information.

 

[tkhunny]

A balloon is being deflated after a fun day at the carnival. The volume of the balloon is decreasing at a rate of twice the radius of the balloon. The surface area of the balloon is 400(PIE)ft^2 when t=0. Assume the balloon is spherical.

Find the following:

a.) Rate of change of the radius.

b.) Rate of change of the surface area.

c.) When the volume of the balloon is 288(PIE)ft^3, the balloon is popped. At what time was the balloon popped?

First, "pie" is for eating. "pi" is the number 3.14159265358979323846264338...

Second, it is a little subtle, but that is worded badly. "The volume of the balloon is decreasing at a rate of twice the radius of the balloon." I think it means twice the RATE at which the radius is decreasing. Unfortunately, that STILL doesn't make any sense, since the volume decreases in ft³/sec and the radius decreases in ft¹/sec. SOOOOOO, we must be thinking the numerical part of the volume decrease is twice the numerical part of the radius decrease. Maybe I'm totally off and it means the "radius" not the "rate of change of the radius". That makes even less sense. Wow! The author should rethink this one, but let's go with it for now.

Volume of a sphere

V = (4/3)*pi*r³

Rate of change of Volume of a Sphere with respect to the rate of change of the radius

dV = 4*pi*r²*dr = Surface Area of a Sphere * dr = SA*dr

That last part is a very interesting relationship.

Rate of Change of Surface Area

dSA = 8*pi*r*dr

Does anyone remember the question?

Initially, we have SA = 400*pi*ft² = 4*pi*r², making r = 10 ft

Also, we have dV/(ft²) = 2*dr

That's a whole bunch of stuff. Where can you take us?