Problem 85
Question
Ultrasound is often used to make images of developing fetuses. In particular, by measuring the size of a fetus ultrasound technicians can estimate its age and then predict its birthdate. To do this requires formulas for fetus size as a function of age. Verburg et al. (2008) fit data from over 6000 fetal ultrasounds. They measured the femur length, \(L\), (in \(\mathrm{mm}\) ) as a function of the fetus age, \(t\), (in weeks) and found the following formula: $$ L=-37.50+3.71 t-6.33 \times 10^{-4} t^{3} $$ Calculate the rate of growth, \(d L / d t\), at \(t=15,20\), and 30 weeks. Does the rate of growth of the fetus increase or decrease as it ages?
Step-by-Step Solution
Verified Answer
The rate of fetal growth decreases as the fetus ages.
1Step 1: Understand the problem
We need to calculate the rate at which the femur length, \(L\), changes with respect to time \(t\), in other words, \(\frac{dL}{dt}\). This requires us to differentiate the given formula for \(L(t)\).
2Step 2: Differentiate the formula
The formula given is \(L = -37.50 + 3.71t - 6.33 \times 10^{-4} t^{3}\). We need to find its derivative, \(\frac{dL}{dt}\).Differentiate each term separately:- The derivative of \(-37.50\) is 0.- The derivative of \(3.71t\) is \(3.71\).- The derivative of \(-6.33 \times 10^{-4} t^3\) is \(-3 \times 6.33 \times 10^{-4} t^2 = -1.899 \times 10^{-3} t^2\).Thus,\[ \frac{dL}{dt} = 3.71 - 1.899 \times 10^{-3} t^2. \]
3Step 3: Calculate rate of growth at t=15 weeks
Substitute \(t = 15\) into the derived rate of growth formula:\[ \frac{dL}{dt}(15) = 3.71 - 1.899 \times 10^{-3} \times (15)^2. \]Perform the calculation:\(\frac{dL}{dt}(15) = 3.71 - 1.899 \times 10^{-3} \times 225 = 3.71 - 0.427275 = 3.282725\, \text{mm/week}.\)
4Step 4: Calculate rate of growth at t=20 weeks
Substitute \(t = 20\) into the rate of growth formula:\[ \frac{dL}{dt}(20) = 3.71 - 1.899 \times 10^{-3} \times (20)^2. \]Perform the calculation:\(\frac{dL}{dt}(20) = 3.71 - 1.899 \times 10^{-3} \times 400 = 3.71 - 0.7596 = 2.9504\, \text{mm/week}.\)
5Step 5: Calculate rate of growth at t=30 weeks
Substitute \(t = 30\) into the rate of growth formula:\[ \frac{dL}{dt}(30) = 3.71 - 1.899 \times 10^{-3} \times (30)^2. \]Perform the calculation:\(\frac{dL}{dt}(30) = 3.71 - 1.899 \times 10^{-3} \times 900 = 3.71 - 1.7091 = 2.0009\, \text{mm/week}.\)
6Step 6: Analyze the results
From the calculated rates of growth, we have:- At \(t=15\) weeks, \(\frac{dL}{dt}\) is \(3.282725\, \text{mm/week}\).- At \(t=20\) weeks, \(\frac{dL}{dt}\) is \(2.9504\, \text{mm/week}\).- At \(t=30\) weeks, \(\frac{dL}{dt}\) is \(2.0009\, \text{mm/week}\).The rate of growth decreases as the fetus ages.
Key Concepts
Rate of ChangeFetal GrowthUltrasound Measurements
Rate of Change
When examining fetal growth, understanding the rate of change is crucial. The rate of change tells us how quickly a fetus's body part, like the femur, is growing over time. By using differentiation in calculus, we can determine the rate at which the femur length, denoted as \( L \), changes with respect to time \( t \). This is expressed as \( \frac{dL}{dt} \).
Differentiation allows us to find the instantaneous rate of change at a specific time point. For the exercise given, the femur length equation was:
By substituting the weeks, such as 15, 20, and 30 into the derivative formula, we discovered that the growth rate decreases as the fetus ages. Understanding these rates of growth is crucial for monitoring normal fetal development.
Differentiation allows us to find the instantaneous rate of change at a specific time point. For the exercise given, the femur length equation was:
- \( L = -37.50 + 3.71t - 6.33 \times 10^{-4} t^3 \)
By substituting the weeks, such as 15, 20, and 30 into the derivative formula, we discovered that the growth rate decreases as the fetus ages. Understanding these rates of growth is crucial for monitoring normal fetal development.
Fetal Growth
Fetal growth is a key indicator of a developing baby's health. During pregnancy, doctors and technicians measure different parts of the fetus to ensure it is growing at a healthy rate. One such measurement is the femur length, which is used because it correlates well with gestational age.
The growth rate of the fetus is not constant. In the initial weeks, rapid growth is typical. However, as the fetus matures, the rate typically decreases, which was evident in our example exercise. At 15 weeks, the growth rate was observed to be faster at 3.282725 mm/week and slowed to 2.0009 mm/week by 30 weeks.
The growth rate of the fetus is not constant. In the initial weeks, rapid growth is typical. However, as the fetus matures, the rate typically decreases, which was evident in our example exercise. At 15 weeks, the growth rate was observed to be faster at 3.282725 mm/week and slowed to 2.0009 mm/week by 30 weeks.
- This decline in growth rates can indicate normal maturation processes.
- It is important to track these changes to ensure the fetus is developing as expected.
Ultrasound Measurements
Ultrasound is a safe and non-invasive method used to monitor fetal development. Ultrasounds use sound waves to create images of the fetus inside the womb. This method is pivotal for obtaining key fetal measurements, such as the femur length, head circumference, and abdominal circumference.
Ultrasound measurements help healthcare professionals estimate the fetus's age and predict the due date. This estimation is more reliable than using the mother's last menstrual period alone, as every pregnancy is unique. Additionally, by having accurate sizes and rates of growth from ultrasounds:
Ultrasound measurements help healthcare professionals estimate the fetus's age and predict the due date. This estimation is more reliable than using the mother's last menstrual period alone, as every pregnancy is unique. Additionally, by having accurate sizes and rates of growth from ultrasounds:
- Doctors can assess the health and development of the fetus more comprehensively.
- Potential issues like growth restrictions can be identified and addressed early.
Other exercises in this chapter
Problem 84
Investigated how maximal heart rate depends on age. They found that if age \(x\) is given in years, then the maximum heart rate of a healthy adult can be predic
View solution Problem 85
Assume that \(f(x)\) is differentiable. Find an expression for the derivative of \(y\) at \(x=2\), assuming that \(f(2)=-1\) and \(f^{\prime}(2)=1\) \(y=\frac{x
View solution Problem 86
Assume that \(f(x)\) is differentiable. Find an expression for the derivative of \(y\) at \(x=2\), assuming that \(f(2)=-1\) and \(f^{\prime}(2)=1\) \(y=[f(x)]^
View solution Problem 86
To measure the physical properties of cells, a piezoelectric probe is used. The force applied by the probe is compared against how much the cell deforms. If \(F
View solution