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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] $.
Answer
(a) $v(r)=k\left(r_{0}-r\right) r^{2}=k r_{0} r^{2}-k r^{3} \Rightarrow v^{\prime}(r)=2 k r_{0} r-3 k r^{2} . \quad v^{\prime}(r)=0 \Rightarrow k r\left(2 r_{0}-3 r\right)=0 \Rightarrow$
$r=0$ or $\frac{2}{3} r_{0}$ (but 0 is not in the interval). Evaluating $v$ at $\frac{1}{2} r_{0}, \frac{2}{3} r_{0},$ and $r_{0},$ we get $v\left(\frac{1}{2} r_{0}\right)=\frac{1}{8} k r_{0}^{3}, v\left(\frac{2}{3} r_{0}\right)=\frac{4}{27} k r_{0}^{3}$
and $v\left(r_{0}\right)=0 .$ since $\frac{4}{27}>\frac{1}{8}, v$ attains its maximum value at $r=\frac{2}{3} r_{0} .$ This supports the statement in the text.
(b) From part (a), the maximum value of $v$ is $\frac{4}{27} k r_{0}^{3}$
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Top Calculus 2 / BC Educators
Kayleah T.
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Watch More Solved Questions in Chapter 4
Video Transcript
So here we're going to maximize a function via var by finding the value at which it is a maximum and then plugging in that value to compute the actual maximum velocity. Then afterwards we're going to sketch the graph of v just to see what it looks like. So first things first to find the maximum, We're going to need to observe the critical points. And for that we're going to need to do the derivative of our function. So if we distribute this K and R and R squared inside, we can differentiate each term individually. So derivative of K are not R squared is just going to be two. K are not our because if you remember K and are not are both constants so we can treat them like normal constants. And then for the second term we're going to get three, Okay, R squared because this is the derivative of K. R cubed. And so now to find the critical points, we're going to set the sequence zero, Set this equals zero. So let's do just that. And now we want to solve for R because that will give us the values of our at which there are critical points. So first things we can do, we're going to divide by K on both sides that will cancel our case were divided by our once. So I'll cancel this are in the square and then we're gonna add 3rd either side and divided by three. So what we'll get is r is equal to 2/3 are not. And so this is our value for a critical point and we have to make sure that it's a maximum. So what we can do is a first derivative test. So we're going to set our point to be 2/3 are not. And so we want to do is choose a value less than 2/3 or not. I'm going to use 1 3rd or not And the value greater than or not or 2/3 are not someone you are not here in green. And so we're going to plug these into our formula for the derivative and for our point are critical point here to be a maximum, We should have a positive derivative for 1/3 or not and the negative derivative for are not because then if the graph looks like this with a positive derivative and then it changes to a negative derivative. This point where it changes is going to be a maximum. So let's go ahead and plug in our values here. So if we plug in 1/3 or not here we get two K are not 1 3rd or not minus three K. Excuse me 19 are not squared. So if we simplify, we will get we can factor out a K and are not and or not squared so we get K are not squared. This will be 2/3 because we factored this out and are not and this will be minus three times 1/9 is just one third and so this is positive because K and are not are just constants that are positive. So we know that this is positive. This is correct. Now, if we plug in our not for our into our derivative equation, we get to K or not are not -3 K are not squared. And then right away you can see this becomes okay are not squared two minus three, which is clearly negative Because of two ministries -1. So we know that this is negative and it is confirmed that our critical point is a max. So there we go. Now we have to compute our maximum. So what we're gonna do is just simply plug in this value for our this this value, we're going to plug into our original the original formula. So let's go ahead and write that in. So we're going to compute max, I'm going to say the max it's equal to And so all we're doing is plugging this in for our okay are not minus 2/3 are not all times two thirds are not square. And so now it's work this out On the inside here is just going to be 1/3 or not. And then when we square this, we get four nights are not squared. So let me just write that out here 1/3 or not. 4/9 are not squared. So our maximum value is four 27th. Okay, are not cubed. So here is your maximum and lastly we're going to do a quick sketch of what this graph would look like. That's enough. So we know that Critical Point is going to be a 2/3 are not. So let me we're going to say this is are not And so we'll say this is about 2/3 That's about 1/3. And so let's say RV max, it's going to be here. And so we know that our graph will hit that point. V max at our two thirds or not. And we know that derivative is positive before this point and negative afterwards. So going to draw a curve down this way towards this point. And of course we know If we plug in zero for our velocity equation, this will just be zero because we have an R squared out here turns everything to zero. So we have a zero out here and we know that it's positive derivative all the way up until we get there. And so that is a very rough sketch of what this graph might look like. And there we go.
Top Calculus 2 / BC Educators
Kayleah T.
Harvey Mudd College
Caleb E.
Baylor University
Michael J.
Idaho State University
Joseph L.
Boston College