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Problem 75 Hard Difficulty

When a foreign object lodged in the trachea (wind pipe) forces a person to cough, the diaphragm thrusts upward causing an increase in pressure in the lungs. This is accompanied by a contraction in the trachea, making a narrower channel for the expelled air to flow through. For a given amount of air to escape in a fixed time, it must move faster through the narrower channel than the wider one. The greater the velocity of the air-stream, the greater the force on the foreign object. X rays show that the radius of the circular tracheal tube contracts to about two-thirds of its normal radius during a cough. According to a mathematical model of coughing, the velocity $v$ of the air-stream is related to the radius $r$ of the trachea by the equation
$$ v(r) = k(r_o - r) r^2 $$ $$ \frac{1}{2}r_o \leqslant r \leqslant r_o $$
where $k$ is constant and $r_o$ is the normal radius of the trachea.
The restriction of $r$ is due to the fact that the tracheal wall stiffens under pressure and a contraction greater than $ \frac{1}{2}r_o $ is prevented (otherwise the person would suffocate.
(a) Determine the value of $r$ in the interval $ [\frac{1}{2}r_o, r_o] $ at which $v$ has an absolute maximum. How does this compare with experimental evidence?
(b) What is the absolute maximum value of $v$ on the interval?
(c) Sketch the graph of $v$ on the interval $ [0, r_o] $.

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Video Transcript

Alright, here's a fun problem. We have coughing scenario where the velocity of air stream is a function of the radius of the trachea and we are going to analyze this so pretty cool. The first thing we want to find is the max um velocity of airstream. So I'm gonna rewrite VFR I'm going to distribute out um the are squares just to make it a little bit easier when I do power role and uh okay, so there's my velocity as a function of R. I just distributed the K. And the R squared to both parts in the parentheses. The derivative then um is equal to three K. R zero R -3 K. R. Squared. And that's my power role. So we noticed that we have a uh let's see, I think I actually put a wrong value here. This should be a two power role would be giving me to ours. That should be a two. I thought something was funny. Okay, so we pull out the cave and the r. That's in comment And that leaves us to our 0 -3 are. And to find our max velocity, we want to first find our critical points. We're going to set Are the prime of our equal to zero. We noticed that is true and are equal zero or R equals Well we can solve for it up here, we want to R zero minus three. R. Two equals zero. Remember either this equals zero or this equals zero to get net zero for the total derivative. So if I solve for R. I will get -3. R 0/-3. So basically I just get oh I keep putting a keep wanting to put a three there. Okay, say that again? Uh It's two. R zero minus three are so are then will be minus two. R zero because I subtract from both sides of the divide by minus three, so our is two thirds are zero. We want to show that it really is a max. We need our, what we want to show is that f prime um of our goes from plus to minus at R equals R0, which is our critical point. So if we do a little sign chart here with zero and we're just going to put up our critical points here, zero and two thirds R zero. And what we're gonna do is we just want to see if the sign is plus or minus in each interval. So if I make are super big, we can see what gets a minus here and this will still be a plus. So for a very big R I get minus derivative. If R is in between the two and um this becomes plus, so this is plus and if r is negative which makes no sense, negative radius so don't worry about that case. So this is therefore our max because indeed we have the prime of our going from plus to minus. So we know that V max occurs at Um 2/3 of our zero And we can plug that in via 2/3 are zero equals k. Times are zero minus two thirds are zero counts are zero books. I need to plug in another two thirds. Let's do that. Um Let's go back Times. are zero, squished that in there squared. Ok, so that gives me um K Times 1 3rd of our zero times four nights of R zero squared. So we can clean that all up to be four K over 27. R zero square nope cubed because we have three of them. Okay, so our fastest speed is four K over 27. R zero Q. Cool. So that's our formula. And um if we want to kind of grab this um to kind of get a sense of what this looks like. We know we have, if we want to graph VFR versus our We we know at 2/3 2/3 are zero, then we have our max. We know at zero ft plug in zero, we'll get zero And we know we also get zero at R0. Right? Because if I if I look at my original, the zeros are when R equals zero or in articles are zero and if we grab this will end up with something like this. It turns out if I take the next derivative, I'll see a point of inflection here and we actually get something that looks like this shape. So um anyway, hopefully that helped to get a rough idea of um the graph. Um We do have maybe just enough time to find out that where that point of inflection is and to find that we're looking at B double prime, let's make some room here. We want V double prime of our to change sign. So plus to minus or vice versa, plus two minuses. Think that a minus. Um Okay, so if we take the next derivative then B double prime of our um equals will use um this one and take the derivative that will give me two K. Are zero minus six K. So we set that equal to zero to find a break point where we can change sign. Um We get our equals um let's see our zero will be six K over two K. R zero. Six K. Over. Oh um I think I did that backwards, so let's fix that. Let's do that one again. Okay so we are oh I've forgotten our that's what it was. Equal zero. Okay, so we're solving for R so R will be minus two K. Are zero Over -6 K. Or R 0/3. So that's our point of inflection. That's actually at one third of our zero. So that's a little squishy in there. I'm going to just one more time. Draw it nice and neat for you. So you can have a nice final Whoa, Nice final graph. Let's go backwards. Make a nice straight line. Just so you've got that final graph where it wasn't so squished in there. Okay so just really quickly we have our cold breakpoints. We have R 0/3, We have 2 R0 over three and we have our zero. So we have our max two hours 0/3. And that value for V. Um The of the Vfr whoops that should be an art B. F. R. Versus our Vfr that high value was four K. Are zero cuc over 27. And then what happens is we have a point of inflection here and our graph looks something like this at the mat whoops, hard to draw it on my little board here. Anyway I'll try one more time. We're gonna have concave up and then concave down all the way to here, whoops that last bit undo again and try to make it actually hit pretty close. Okay and just one more time since I did it so squishy before. Um If I look at we actually are and the double prime of R. R. Zero was at R 0/3. And our formula was um two K. R. Zero minus six K. R. And so what happens is if I make are really big I get a negative for concrete down. This was a zero. If I make are really small I get a positive. So you can see I'm concave and then I go can keep down and it all works so we're able to graph of our function. So pretty cool. Um with how the trachea, you know, decreases in size so that the air goes through faster. Very cool. Okay. Hopefully that helped to have a wonderful day.

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James Stewart

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Joseph Lentino

James StewartCalculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Joseph Lentino

James Stewart

Calculus: Early Transcendentals