Rate of increase of the surface area

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Homework Help Overview

The problem involves finding the rate of increase of the surface area of a spherical balloon as it is inflated, using the formula for surface area \( S = 4 \pi r^2 \). The original poster attempts to calculate this rate at specific radii (1 ft, 2 ft, and 3 ft) and expresses uncertainty about their results and the units involved.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the derivative of the surface area formula and question whether additional information, such as \( dr/dt \), is necessary for the problem. There is also a request for clarification on the original poster's notation and calculations.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's approach and questioning the clarity of their expressions. Some guidance has been offered regarding the interpretation of the rate of change, but no consensus has been reached on the correctness of the original calculations.

Contextual Notes

There appears to be confusion regarding the notation used by the original poster, particularly the phrase "i put 4 2R," which some participants find unclear. Additionally, the discussion touches on the importance of units in the context of the problem.

afcwestwarrior
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a spherical balloon is being inflated. find the rate of increase of the surface area (S=4 pie r squared) with respect to radius r when r is (A) 1 ft, (B) 2 ft, (C) 3ft.

Here's what i did

i found the derivative of s and i put 4 2R and then i plugged in the numbers in R
and i got 8 ft sq/ft 16 ft sq/ft and 24 ft sq/ft

except I'm not sure if I am doing it right, although those are the right answers i didnt get the units along with them
 
Last edited:
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is there simple algebra involved in this calculus problem
 
are you not given dr/dt or dv/dt?
 
afcwestwarrior said:
a spherical balloon is being inflated. find the rate of increase of the surface area (S=4 pie r squared) with respect to radius r when r is (A) 1 ft, (B) 2 ft, (C) 3ft.

Here's what i did

i found the derivative of s and i put 4 2R and then i plugged in the numbers in R
and i got 8 ft sq/ft 16 ft sq/ft and 24 ft sq/ft

except I'm not sure if I am doing it right, although those are the right answers i didnt get the units along with them

You've got a graph there, simply find the gradient of the slope, that's the rate of increase.

(1,8)
(2,16)
(3,24)
 
Last edited:
afcwestwarrior said:
a spherical balloon is being inflated. find the rate of increase of the surface area (S=4 pie r squared) with respect to radius r when r is (A) 1 ft, (B) 2 ft, (C) 3ft.

Here's what i did

i found the derivative of s and i put 4 2R and then i plugged in the numbers in R
and i got 8 ft sq/ft 16 ft sq/ft and 24 ft sq/ft

except I'm not sure if I am doing it right, although those are the right answers i didnt get the units along with them

Give more detail about what you did (I, for one, don't know what "i put 4 2R" means!). What is the formula for surface area (I see you give that)? what is its derivative?

Plastic Photon, since the problem asks for the rate of increas eof surface area with respect to r you don't need to know dr/dt.

If the surface area is in square feet and the radius in feet, then the "rate of change of surface area with respect ot radius" is in (square feet)/feet or just feet!
 
Last edited by a moderator:
ok so ur saying i don't need to find the derivative of it, ok check this mate,
the i put 4 2 R means that's the derivative of the surface area
 
No, "i put 4 2 R" doesn't mean anything!
 

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