Rate of change (surface area/radius)

In summary, the conversation involves solving a related rates problem involving the surface area of a spherical balloon as it is being inflated. The first part (a) is correctly solved using the equation dA/dr = 8πr. The second part (b) involves solving for the radius using the equation 4πr^2 = 200, and then using the derivative equation to find the rate of change of the surface area. The correct answer is 100.27 cm2/cm.
  • #1
321study
3
0
Homework Statement
I am 99% positive I did a) correctly, as the answer is correct according to the answer in the back of my textbook. However, I cannot do b).
The question: A spherical balloon is being inflated. Find the rate of change of the surface area with respect to the radius when:

a) The radius is 5cm
b) The surface area is 200cm2

The attempt at a solution

a) 8[itex]\pi[/itex]*5cm
=125.66 cm2/cm (or 40[itex]\pi[/itex] cm2/cm)

b) I am unsure of what to do for this. Is it something like:

[itex]\sqrt{}200/4\pi[/itex]

=3.9894

And then do something else, which I don't know what that is.

The answer to this question, according to the book, is 100.27cm2/cm

Sorry if the equations don't look very clear. This is my first post and I will learn how to write it all up neatly when I am not so tired.
 
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  • #2
Welcome to Physics Forums.

Since you have solved the first part, I assume that you have worked our that

[tex]\frac{dS}{dr} = 8\pi r\;.[/tex]

Now, what you need to ask yourself if: "How is the radius related to the surface area?". :wink:
 
  • #3
There's an excellent video here:
http://press.princeton.edu/video/banner/

which, I think, should give you what you need to know. It's 50 minutes into video 5, and you shouldn't need to watch any of the others in order to follow the explanation (although they are all a good intuitive primer to calculus). Professor Banner covers related rates problems for about 20 minutes, and it should address your question.
 
  • #4
I presume you know that the surface area of a sphere of radius r is [itex]A= 4\pi r^2[/itex]. That gives, of course, [itex]dA/dr= 8\pi r[/itex] which is where you got "8π*5cm" for (a).

For (b), instead of being given r, you are given A= 200. Solve [itex]4\pi r^2= 200[/itex] for r, then do it the same as (a).
 
  • #5
Unfortunately I won't be able to view the video as it will take a while to download and I've got to get going to school for an excursion. Thank you though.

So if I try:

200*4pi

= 2513.27

Get the square root of that makes 50.1325

Multiple by 2 for some reason? Makes 100.265. Rounded off gives 100.27. I am pretty sure I tried this method before, however I got 100.29 (rounding error) and I forgot how I did it all. Nevertheless, I don't really understand why I must multiple my answer by 2.
Thanks for all your help so far.
 
  • #6
321study said:
So if I try:
200*4pi
= 2513.27
Get the square root of that makes 50.1325

Try to solve for r again. Look at the equation [itex]4\pi r^2= 200[/itex], make sure you're correctly solving for r. Then, once you have r, you can use your derivative equation to get the answer.
 
  • #7
Ok I did it.

Here is my working:

4[itex]\pi[/itex]r2 = 200
r2 = 200/4[itex]\pi[/itex]
r = Square root of 50/[itex]\pi[/itex]

dS/dr = 8[itex]\pi[/itex]r
= 8[itex]\pi[/itex]Square root of 50/[itex]\pi[/itex]
= 100.27 cm2/cm

Thanks everyone for all your help :smile:
 

1. What is the rate of change of surface area with respect to radius?

The rate of change of surface area with respect to radius is a measure of how the surface area of an object changes as the radius of the object changes. It is calculated using the derivative of the surface area formula with respect to the radius.

2. How is the rate of change of surface area related to the rate of change of volume?

The rate of change of surface area and the rate of change of volume are related through the derivative of the volume formula, which involves both the surface area and the radius. The rate of change of volume can be calculated by multiplying the rate of change of surface area by the height of the object.

3. Can the rate of change of surface area be negative?

Yes, the rate of change of surface area can be negative. This occurs when the radius is decreasing, resulting in a decrease in surface area. For objects with a constant volume, the rate of change of surface area and radius are inversely proportional, so a decrease in radius will result in an increase in the rate of change of surface area.

4. How does the rate of change of surface area affect the strength of a structure?

The rate of change of surface area can affect the strength of a structure, as it determines the amount of stress and strain placed on the material. A larger surface area will result in a larger force being applied to the material, which can lead to structural failure if the material is not strong enough to withstand it.

5. Can the rate of change of surface area be used to optimize the design of an object?

Yes, the rate of change of surface area can be used to optimize the design of an object. By understanding how changes in the radius affect the surface area, engineers can adjust the dimensions of an object to achieve the desired surface area and optimize its strength and performance.

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