Question

researchers fit data from over 6000 fetal ultrasounds. they measured the femur length, l, (in millimeters) as a function of the fetus age, t, (in weeks) and found the formula below. calculate the rate of growth, \\( \frac { d l } { d t } \\), at t = 10, 15, and 25 weeks. does the rate of growth of the fetus increase or decrease as it ages?
\\( l = - 37.60 + 3.71 t - 6.32 \times 10 ^ { - 4 } t ^ { 3 } \\)
write a function for \\( \frac { d l } { d t } \\).
\\( \frac { d l } { d t } = \square \\)
(use integers or decimals for any numbers in the expression.)

Explanation:

Step1: Differentiate the constant term

The derivative of a constant \(C\) is \(0\). For the term \(-37.60\), its derivative is \(0\).

Step2: Differentiate the linear term

Using the power rule \(\frac{d}{dt}(at)=a\) (where \(a = 3.71\)), the derivative of \(3.71t\) is \(3.71\).

Step3: Differentiate the cubic term

Using the power rule \(\frac{d}{dt}(bt^{n})=nbt^{n - 1}\), for the term \(-6.32\times10^{-4}t^{3}\), we have \(n = 3\) and \(b=-6.32\times 10^{-4}\). So its derivative is \(3\times(-6.32\times 10^{-4})t^{2}=- 0.001896t^{2}\).

Combining these results, \(\frac{dL}{dt}=3.71-0.001896t^{2}\)

Answer:

\(\frac{dL}{dt}=3.71 - 0.001896t^{2}\)