Respuesta :
Answer:
[tex]\frac{dF}{F}=4\frac{dR}{R}[/tex]
So a 5% relative increase in R would mean a 20% relative increase in F
Step-by-step explanation:
First we need to remind the definition of relative increase of a variable.
For a variable A its relative increase is given by [tex]\frac{dA}{A}[/tex].
Using this, the relative increase in F is [tex]\frac{dF}{F}[/tex] and similarly the relative increase in R is given by [tex]\frac{dR}{R}[/tex].
Let's then start by deriving F with respect to R:
[tex]\frac{dF}{dR}=4kR^3[/tex]
thus
[tex]dF=4kR^3dR[/tex]
[tex]\implies \frac{dF}{F}=\frac{4kR^3dR}{F}[/tex]
[tex]\implies \frac{dF}{F}=\frac{4kR^3dR}{kR^4}[/tex]
[tex]\implies \frac{dF}{F}=4\frac{dR}{R}[/tex].
If we plug the value 5% [tex]\left( \frac{5}{100}\right)[/tex] in [tex]\frac{dR}{R}[/tex] we get
[tex]\frac{dF}{F}=4\times5\%=20\%[/tex]
The 5% increase in the radius affect the flow of blood by 21.55% approx
What is directly proportional and inversely proportional relationship?
Let there are two variables p and q
Then, p and q are said to be directly proportional to each other if
[tex]p = kq[/tex]
where k is some constant number called constant of proportionality.
This directly proportional relationship between p and q is written as
[tex]p \propto q[/tex] where that middle sign is the sign of proportionality.
In a directly proportional relationship, increasing one variable will increase another.
Now let m and n are two variables.
Then m and n are said to be inversely proportional to each other if
[tex]m = \dfrac{c}{n} \\\\ \text{or} \\\\ n = \dfrac{c}{m}[/tex]
(both are equal)
where c is a constant number called constant of proportionality.
This inversely proportional relationship is denoted by
[tex]m \propto \dfrac{1}{n} \\\\ \text{or} \\\\n \propto \dfrac{1}{m}[/tex]
As visible, increasing one variable will decrease the other variable if both are inversely proportional.
For this case, we're given that:
Flux F = k.R⁴ (R being the radius of the blood vessel).
Let the increment in radius be of [tex]\delta R[/tex], and let corresponding increment in flux be of [tex]\delta F[/tex]
Then we see:
[tex]F+\delta F = k(R+\delta R)^4\\\delta F = k(4R^3\times\delta R + 6R^2 \times (\delta R)^2 + 4 R \times (\delta R)^3 + (\delta R)^4)[/tex]
This is the relative increment in F from relative increment in value of R
If 5% increment happens, that means we have: [tex]\delta R = \dfrac{R \times 5}{100} = R/20[/tex]
Thus, we get:
[tex]\delta F = k(4R^3\times\delta R + 6R^2 \times (\delta R)^2 + 4 R \times (\delta R)^3 + (\delta R)^4)\\\delta F = k(4R^3 \times (R/20) + 6R^2 \times (R/20)^2 + 4R \times (R/20)^3 + (R/20)^4) \\\\\delta F = kR^4(4/20 +6/400 + 4/8000 + 1/160000)\\\delta F = F(0.21550625)[/tex]
Its percent with respect to F is x% (say) then:
[tex]F \times x/100 = F(0.21550625)\\x \approx 21.55 \%[/tex]
Thus, the 5% increase in the radius affect the flow of blood by 21.55% approx
Learn more about percentage here:
https://brainly.com/question/11549320
