# Average rate of change of a sphere

• thomasrules
In summary, the formula for the volume of a sphere is V(r) = \frac{4}{3}\pi(r)^3. To find the average rate of change of volume with respect to radius as the radius changes from 10 cm to 15 cm, we use the formula \frac{V(15)-V(10)}{15-10}. This gives us a result of 1989.675 cm^3 per minute. To find the instantaneous rate of change when the radius is 8 cm, we use the derivative formula and get 256\pi cm^3 per minute. The process becomes easier with the use of derivatives.
thomasrules
the Volume of a sphere is given by $$V(r) = \frac{4}{3}\pi(r)^3$$

Find the average rate of change of volume with respect to radius as the radius changes from 10 cm to 15 cm.

Ok what I tried is: $$\lim_{r\rightarrow 15} \frac{V(r)-V(a)}{r-a}$$

$$P(a,V(a))=(15,14 137)\\ Let\ r=10 \lim_{r\rightarrow 15} \frac{\frac{4}{3}\pi(10)^3-14,137}{10-15} = 1989$$

and I would do that for 10 to 15 and get the average rate but somethign is wrong here...

Last edited:
if r goes to 15 wouldn't it be wherever there is an r a 15 would go... I think you did it backwards... it looks as if you put "a" where "r" was supposed to go and "r" where "a" is supposed to go.

P(a,V(a))=(15,14 137)\\ Let\ r=10 \lim_{r\rightarrow 15} \frac{\frac{4}{3}\pi(15)^3-frac{4}{3}\pi(10)^3}{15-10} = 1989.675

I am not sure exactly about this... although I do remember doing something similar to this last semester. Is this the whole problem? Because if it wants the rate it would usually be for ex. the sphere is increasing by 1989 cm^3 per min.I don't know if the formula will show up because it is in code right now and I am not 100% sure how this forum works with formulas. Hope this helped some. If it still seems to be wrong send me a message and I'll look at it again.

well I guess the formula didn't work lol... how did you get yours up there?

The average rate of change as the radius changes from 10 cm to 15 cm is simply

$$\frac{V(15)-V(10)}{15-10}$$

No limit, no hocus pocus involved, just plain soustraction and division of numbers.

answer says $$\frac{1900}{3}\pi$$

that is the same as 1989.675, if you want to get an answer like that just keep pie separate.

second part is: Find the rate of change of volume when the radius is 8cm...

So that means the instantaneous rgight?

hmm... do you have the answer for this one as well? and do you know how to do derivatives? This dose look like it means instantaneous.

i got it its $$256\pi$$

Used the instantaneous formula thing, i.e. the long way of derivative

yeah you will learn how to get it easier just by deriving the equation later. Is this a college course or high school? Just wondering because the part you posted in was precalculus. If you get to where you need help again, I wouldn't mind helping. I need a refresher now and then to make sure I don't forget things.

## 1. What is the formula for calculating the average rate of change of a sphere?

The formula for calculating the average rate of change of a sphere is (change in volume)/(change in time).

## 2. How is the average rate of change of a sphere related to its radius?

The average rate of change of a sphere is directly proportional to its radius. This means that as the radius increases, the average rate of change also increases, and vice versa.

## 3. Can the average rate of change of a sphere be negative?

Yes, the average rate of change of a sphere can be negative. This can occur if the sphere is shrinking over time, resulting in a decrease in its volume.

## 4. How is the average rate of change of a sphere different from its instantaneous rate of change?

The average rate of change of a sphere is calculated over a period of time, while the instantaneous rate of change is calculated at a specific point in time. The average rate of change gives an overall picture of how the sphere is changing, while the instantaneous rate of change shows the rate of change at a specific moment.

## 5. Can the average rate of change of a sphere be used to predict its future volume?

No, the average rate of change of a sphere cannot be used to accurately predict its future volume. This is because it only takes into account the change in volume over a specific period of time, and does not account for any changes in other variables that may affect the sphere's volume in the future.

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