fais_attention
New member
- Joined
- Sep 10, 2008
- Messages
- 9
Hey. Doing my homework again, and getting it wrong--again.
"The radius of a spherical balloon is increasing at a rate of 4 centimeters per minute. How fast is the surface area changing when the radius is 8 centimeters? (Surface Area=4pi(r^2))"
So, it took me forever to assign values. I guessed that the given was dr/dt=4 cm. I used the relationship between surface area and radius to show that
d(SA)/dt = d(4pi(r^2))/dt
I worked out the derivative of the right side, and got 8pi(dr/dt)=d(SA)/dt. I plugged in the given for r (8) and got the answer of ROC= 64pi cm/minute.
This was one of the days I was absent from class, so I copied somebody's notes and am still lost...
"The radius of a spherical balloon is increasing at a rate of 4 centimeters per minute. How fast is the surface area changing when the radius is 8 centimeters? (Surface Area=4pi(r^2))"
So, it took me forever to assign values. I guessed that the given was dr/dt=4 cm. I used the relationship between surface area and radius to show that
d(SA)/dt = d(4pi(r^2))/dt
I worked out the derivative of the right side, and got 8pi(dr/dt)=d(SA)/dt. I plugged in the given for r (8) and got the answer of ROC= 64pi cm/minute.
This was one of the days I was absent from class, so I copied somebody's notes and am still lost...