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QuestionAdrián, un niño de 10 años, necesita espirometría. ¿Cuál es el volumen que maximiza su velocidad de espiración? Redondea a 2 decimales.

Studdy Solution

STEP 1

Assumptions1. Adrian's exhaled air volume V in liters as a function of expiration time t in seconds is modeled by V(t)=−3.5−t+1+3.5V(t)=-3.5^{-t+1}+3.5. . Adrian's expiration speed F in liters per second as a function of the volume of exhaled air x in liters is modeled by (x)={−13x(x−1.5)0≤x<1.25−169x+5691.25≤x≤3.5(x)=\left\{\begin{array}{cc}-13 x(x-1.5) &0 \leq x<1.25 \\ -\frac{16}{9} x+\frac{56}{9} &1.25 \leq x \leq3.5\end{array}\right.3. The parameters in spirometry are used to diagnose when the air comes out slower than it should, if the child presents the relationship V(1)V<70%\frac{V(1)}{V_{}}<70 \%, we diagnose obstruction. In addition, restriction is diagnosed when the total exhaled air volume VV_{} is below80% of the reference value. If it does not present these values, it is normal.
4. The expiration values of another patient of the same age as Adrian, the speed-volume graph, for the second section, is given by a part of the line y=−3x+103y=-\frac{}{3} x+\frac{10}{3}.

STEP 2

To find the volume that makes Adrian's expiration speed reach its maximum value, we need to find the maximum point of the function (x)(x).

STEP 3

The function (x)(x) is a piecewise function with two parts. The maximum point could be in either of these parts, or at the point where they meet.

STEP 4

First, let's consider the first part of the function −13x(x−1.)-13x(x-1.), which is defined for 0≤x<1.250 \leq x <1.25. This is a quadratic function, and its maximum or minimum point (if it exists) is given by the formula −b2a-\frac{b}{2a}, where aa and bb are the coefficients of x2x^2 and xx, respectively.

STEP 5

In this case, a=−13a=-13 and b=13∗1.5=19.5b=13*1.5=19.5. So, the maximum or minimum point for this part of the function is at x=−b2a=−19.52∗(−13)x=-\frac{b}{2a}=-\frac{19.5}{2*(-13)}.

STEP 6

Calculate the x-coordinate of the maximum or minimum point for the first part of the function.
x=−19.52∗(−13)=0.75x=-\frac{19.5}{2*(-13)}=0.75

STEP 7

This x-coordinate is within the domain of the first part of the function (0≤x<1.250 \leq x <1.25), so it could be the maximum point. We need to calculate the corresponding y-coordinate by substituting x=0.75x=0.75 into the first part of the function.

STEP 8

Calculate the y-coordinate of the maximum or minimum point for the first part of the function.
(0.75)=−13∗0.75∗(0.75−1.5)(0.75)=-13*0.75*(0.75-1.5)

STEP 9

Calculate the y-coordinate.
(.75)=−13∗.75∗(−.75)=7.3125(.75)=-13*.75*(-.75)=7.3125

STEP 10

So, the maximum or minimum point for the first part of the function is (0.75,7.3125)(0.75,7.3125).

STEP 11

Now, let's consider the second part of the function −169x+569-\frac{16}{9}x+\frac{56}{9}, which is defined for .25≤x≤3.5.25 \leq x \leq3.5. This is a linear function, and it doesn't have a maximum or minimum point within its domain. However, the maximum point could be at one of the endpoints of the domain.

STEP 12

Calculate the y-coordinate at the left endpoint of the domain (x=.25x=.25) by substituting x=.25x=.25 into the second part of the function.

STEP 13

Calculate the y-coordinate at the left endpoint.
(.25)=−169∗.25+569(.25)=-\frac{16}{9}*.25+\frac{56}{9}

STEP 14

Calculate the y-coordinate.
(.25)=4.4444(.25)=4.4444

STEP 15

So, the point at the left endpoint of the domain of the second part of the function is (.25,4.4444)(.25,4.4444).

STEP 16

Calculate the y-coordinate at the right endpoint of the domain (x=3.5x=3.5) by substituting x=3.5x=3.5 into the second part of the function.

STEP 17

Calculate the y-coordinate at the right endpoint.
(3.5)=−169∗3.5+569(3.5)=-\frac{16}{9}*3.5+\frac{56}{9}

STEP 18

Calculate the y-coordinate.
(3.5)=0(3.5)=0

STEP 19

So, the point at the right endpoint of the domain of the second part of the function is (3.5,)(3.5,).

STEP 20

Now, we have three potential maximum points (0.75,7.3125)(0.75,7.3125) from the first part of the function, (.25,4.4444)(.25,4.4444) and (3.5,0)(3.5,0) from the second part of the function. The maximum point of the function (x)(x) is the one with the highest y-coordinate.

STEP 21

Compare the y-coordinates of the three points to find the maximum point.

STEP 22

The point with the highest y-coordinate is (0.75,7.3125)(0.75,7.3125).
So, the volume that makes Adrian's expiration speed reach its maximum value is0.75 liters, and the maximum value is7.31 liters per second (rounded to two decimal places).

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