Asked by Dan
When you cough, your windpipe contacts. The speed v, with which air comes depends on the radius,r, of your windpipe. If R is the rest radius of your windpipe, then for r </= R the speed is given by:
v=2(R-r)r^2
i got the derivative:
v'=4rR-6r
what do i do next??
v=2(R-r)r^2
i got the derivative:
v'=4rR-6r
what do i do next??
Answers
Answered by
Damon
I do not know the question or why you took what you call the derivative but it is not the derivative.
If
v = 2 R r^2 - 2 r^3
then
dv/dr = 4 R r - 6 r^2
However I do not understand where that equation came from.
In general the speed times the area must be constant if the density does not change much.
Therefore
Vo (pi R^2) = v (pi r^2)
v = Vo (R/r)^2
Where Vo is the speed in the open windpipe.
If
v = 2 R r^2 - 2 r^3
then
dv/dr = 4 R r - 6 r^2
However I do not understand where that equation came from.
In general the speed times the area must be constant if the density does not change much.
Therefore
Vo (pi R^2) = v (pi r^2)
v = Vo (R/r)^2
Where Vo is the speed in the open windpipe.
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