researchers fit data from over 6000 fetal ultrasounds. they measured the femur length, l, (in millimeters)…

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?\n\\( l = - 37.60 + 3.71 t - 6.32 \\times 10 ^ { - 4 } t ^ { 3 } \\)\nwrite a function for \\( \\frac { d l } { d t } \\).\n\\( \\frac { d l } { d t } = \\square \\)\n(use integers or decimals for any numbers in the expression.)

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?\n\\( l = - 37.60 + 3.71 t - 6.32 \\times 10 ^ { - 4 } t ^ { 3 } \\)\nwrite a function for \\( \\frac { d l } { d t } \\).\n\\( \\frac { d l } { d t } = \\square \\)\n(use integers or decimals for any numbers in the expression.)

Answer

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})