Solving a Problem with Relative Change: A 5% Increase in Radius

In summary, the conversation discusses the concept of relative change in the context of the equation F=kR^4. The goal is to show that the relative change in F is approximately four times the relative change in R, and that a 5% increase in radius will affect the flow of blood. The conversation also includes a discussion on the different ways to approach this problem, including using basic algebra and differentiation.
  • #1
powp
91
0
Hello All

I have this problem that I have no idea how to do.

F = flux or the volume of blood to flow past a point

R = radius

F=kR^4

Show that the relative change in F is about four times the relative change in R. How will a 5% increase in radius affect the flow of Blood??

How in the world do I do this?? What is Relative Change?? My textbook does not have this term

I think I need to have 2 equations F = kR^4 and R = (F/k)^(1/4). Not sure what to do.

Please help.
 
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  • #2
I think by relative change they only mean percentage change, or absolute change divided by the level: rel. change (up to time t) = [x(t) - x(0)]/x(0) = x(t)/x(0) - 1.
 
  • #3
"Relative change" is the change "relative to" the original value: i.e. the change divided by the orginal amount. If we use [tex]\Delta R[/tex] and [tex]\Delta F[/tex] to mean the changes in R and F respectively, then their "relative changes" are [tex]\frac{\Delta R}{R}[/tex] and [tex]\frac{\Delta F}{F}[/tex].
I don't know how you should do this because I don't know what level you are at and what "mechanisms" you have available to you.

Basic but harder way: Since you have F= kR4, if "dR" is the relative change in R, then [tex]dR= \frac{\Delta R}{R}[/tex] so the actual change is [tex]\Delta R= Rdr[/tex] and the new value for R (after the change) is R+ Rdr= R(1+dr). Then the new value for F is k(R(1+dr))4= kR4(1+ dr)4. Multiplying out (1+ dr)4= 1+ 4dr+ 6(dr)2+ 4(dr)3+ (dr)4 so the new value of F is kR4(1+ 4dr+ 6(dr)2+ 4(dr)3+ (dr)4). Subtracting of the old value, kR4 tells us that the actual change in F was kR4(4dr+ 6(dr)2+ 4(dr)3+ (dr)4) (we just removed that "1" inside the parentheses).
The relative change then is kR4(4dr+ 6(dr)2+ 4(dr)3+ (dr)4) divided by kR4 which is 4dr+ 6(dr)2+ 4(dr)3+ (dr)4. If "dr" is relatively small, then those powers of dr will be even smaller- the largest term will be 4dr: that is, "about four times the relative change in R."

More sophisticated and easier way. Differentiate F= kR4 with respect to time to get [tex]\frac{dF}{dt}= 4kR^3\frac{dR}{dt}[/tex]. dividing that by F= kR4, [tex]\frac{\frac{dF}{dt}}{F}= 4\frac{\frac{dR}{dt}}{R}[/tex] which says exactly that "the relative (rate of) change in F is equal to the relative (rate of) change in R".
 
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  • #4
I think this is the way we are expected to do it.

HallsofIvyMore sophisticated and easier way. Differentiate F= kR[sup said:
4[/sup] with respect to time to get [tex]\frac{dF}{dt}= 4kR^3\frac{dR}{dt}[/tex]. dividing that by F= kR4, [tex]\frac{\frac{dF}{dt}}{F}= 4\frac{\frac{dR}{dt}}{R}[/tex] which says exactly that "the relative (rate of) change in F is equal to the relative (rate of) change in R".

This makes sense except for the one thing. This may be a silly question but can you just divide a function by another function without doing the same to both sides?

It seems like you are do the following

A = 2 + B is divided by C = D + 2 and you do the following

A 2 + B
- = ------
C D + 2

Don't you need to divide both sides by the same value? Or is it since C does equal D + 2 this is allowed?

Thanks for your help
 
  • #5
Can sombody help with this part of the previous question

How will a 5% increase in radius affect the flow of Blood??

Please Please Pretty Please!
 
  • #6
Read HallsOfIvy's post C-A-R-E-F-U-L-L-Y.
 
  • #7
Thanks.

I have reread it and still have no clue. Can anybody give me a hint??

Thanks
 

FAQ: Solving a Problem with Relative Change: A 5% Increase in Radius

How do you calculate a 5% increase in radius?

In order to calculate a 5% increase in radius, you simply need to multiply the current radius by 1.05. This will give you the new, increased radius.

What is the significance of using relative change when solving a problem?

Relative change takes into account the size of the original value and allows for comparison between different data sets. It is a more accurate representation of change than absolute values.

Can you provide an example of a problem that can be solved using relative change with a 5% increase in radius?

Sure, let's say you have a circle with a radius of 10 cm. If you increase the radius by 5%, the new radius would be 10 cm x 1.05 = 10.5 cm. This would result in a larger circle with a new area and circumference.

What other factors should be considered when solving a problem with relative change?

Some other factors to consider include the starting value, the percentage of change, and the units of measurement being used. It is also important to double check your calculations and ensure they are accurate.

How can solving a problem with relative change be useful in real-life situations?

Solving problems with relative change can be useful in a variety of fields, such as finance, economics, and science. It allows for comparison between data sets and can help identify trends or patterns. For example, a business may use relative change to track sales growth, while a scientist may use it to analyze changes in environmental data over time.

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