Problem 69
Question
Doctors treated a patient at an emergency room from 2:00 P.M. to 7:00 P.M. The patient's blood oxygen level \(L\) (in percent) during this time period can be modeled by $$ L=-0.270 t^{2}+3.59 t+83.1, \quad 2 \leq t \leq 7 $$ where \(t\) represents the time of day, with \(t=2\) corresponding to 2:00 P.M. Use the model to estimate the time (rounded to the nearest hour) when the patient's blood oxygen level was \(93 \%\).
Step-by-Step Solution
Verified Answer
By solving the equation with the quadratic formula, two solutions will be obtained for \(t\) when \(L=93\). Check these times to ensure they fall between 2 P.M. and 7 P.M. The rounded final answer corresponds to the correct time within the patient's hospital visit when the oxygen level was 93 percent.
1Step 1: Writing the Equality
The problem statement tells that the blood oxygen level was 93%. Therefore, first, set up an equation by substituting \(L=93\) into the model equation: \(93=-0.270 t^{2}+3.59 t+83.1\).
2Step 2: Solving for Time \(t\)
The next step is to arrange the equation for the quadratic formula. To do this, the equation can be rearranged into the standard form of a quadratic equation ax^2 + bx + c = 0 by subtracting 93 from both sides to yield \(-0.270 t^{2}+3.59 t+(83.1-93)=0\). The values of a, b, and c are then identified as -0.270, 3.59, and -9.9 respectively. The solutions can be solved using the quadratic formula \(t=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\). This gives two solutions for \(t\), and both must then be checked against the limits given in the problem (2<=t<=7), to see which one correctly answers the problem.
3Step 3: Select Correct Time
After obtaining the two possible times \(t\), determine which one makes sense within the timeframe of the visit to the hospital between 2 P.M. and 7 P.M. Discard any value that does not fit within this timeframe.
Key Concepts
Blood Oxygen Level ModelTime EstimationQuadratic Formula Application
Blood Oxygen Level Model
Doctors often use mathematical models to understand and predict changes in a patient's health metrics. In this exercise, we're modeling the patient's blood oxygen level, a crucial indicator of respiratory and cardiovascular health, to understand how it changes over time at the hospital.
A quadratic equation is used in this model, which reflects the rate and pattern of change in blood oxygen levels within a specified period. The model given is:\[L = -0.270 t^2 + 3.59 t + 83.1\]where \(L\) is the blood oxygen level percentage, and \(t\) is the time of day (where \(t=2\) corresponds to 2:00 PM).
This equation models changes over a five-hour period from 2 PM to 7 PM. Understanding and interpreting these models helps in estimating how the levels will behave at different times, providing critical insights for medical staff.
A quadratic equation is used in this model, which reflects the rate and pattern of change in blood oxygen levels within a specified period. The model given is:\[L = -0.270 t^2 + 3.59 t + 83.1\]where \(L\) is the blood oxygen level percentage, and \(t\) is the time of day (where \(t=2\) corresponds to 2:00 PM).
This equation models changes over a five-hour period from 2 PM to 7 PM. Understanding and interpreting these models helps in estimating how the levels will behave at different times, providing critical insights for medical staff.
Time Estimation
Estimating the time when the blood oxygen level reaches a specific value involves solving the quadratic equation for \(t\). Here, we want to know when the level is exactly 93%.
- We plug \(L = 93\) into the equation, forming the equation: \(93 = -0.270 t^2 + 3.59 t + 83.1\).
- Rewriting it in standard quadratic form (\(ax^2 + bx + c = 0\)), we have: \(-0.270 t^2 + 3.59 t + (83.1 - 93) = 0\).
Quadratic Formula Application
The quadratic equation derived from the blood oxygen level model requires solving using the quadratic formula. The quadratic formula is valuable because it offers an exact solution for any quadratic equation of the form: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
- Here, \(a = -0.270\), \(b = 3.59\), and \(c = -9.9\) (calculated from the equation).
- By substituting these into the quadratic formula, we solve for \(t\), finding two potential solutions for \(t\) owing to the "\(\pm\)" sign.
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