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\n\\( l = - 37.60 + 3.71 t - 6.32 \\times 10 ^ { - 4 } t ^ { 3 } \\)\n\ncalculate the rate of growth at t = 15 weeks.\n\n\\( \\frac { d l } { d t } = 3.28 \\) mm/week\n(round to two decimal places as needed.)\n\ncalculate the rate of growth at t = 25 weeks.\n\n\\( \\frac { d l } { d t } = 2.52 \\) mm/week\n(round to two decimal places as needed.)

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\n\\( l = - 37.60 + 3.71 t - 6.32 \\times 10 ^ { - 4 } t ^ { 3 } \\)\n\ncalculate the rate of growth at t = 15 weeks.\n\n\\( \\frac { d l } { d t } = 3.28 \\) mm/week\n(round to two decimal places as needed.)\n\ncalculate the rate of growth at t = 25 weeks.\n\n\\( \\frac { d l } { d t } = 2.52 \\) mm/week\n(round to two decimal places as needed.)

Answer

Explanation:

Step1: Differentiate (L) with respect to (t)

Given (L=-37.60 + 3.71t-6.32\times10^{-4}t^{3}). Using the power rule (\frac{d}{dt}(t^{n})=nt^{n - 1}) and (\frac{d}{dt}(C)=0) (where (C) is a constant). (\frac{dL}{dt}=\frac{d}{dt}(-37.60)+\frac{d}{dt}(3.71t)-\frac{d}{dt}(6.32\times 10^{-4}t^{3})) (\frac{dL}{dt}=0 + 3.71-3\times6.32\times10^{-4}t^{2}) (\frac{dL}{dt}=3.71-1.896\times10^{-3}t^{2})

Step2: Calculate (\frac{dL}{dt}) at (t = 10)

Substitute (t = 10) into (\frac{dL}{dt}=3.71-1.896\times10^{-3}t^{2}) (\frac{dL}{dt}\mid_{t = 10}=3.71-1.896\times10^{-3}\times(10)^{2}) (=3.71-1.896\times0.1) (=3.71 - 0.1896) (=3.5204\approx3.52)

Step3: Calculate (\frac{dL}{dt}) at (t = 15)

Substitute (t = 15) into (\frac{dL}{dt}=3.71-1.896\times10^{-3}t^{2}) (\frac{dL}{dt}\mid_{t = 15}=3.71-1.896\times10^{-3}\times(15)^{2}) (=3.71-1.896\times0.225) (=3.71-0.4266) (=3.2834\approx3.28)

Step4: Calculate (\frac{dL}{dt}) at (t = 25)

Substitute (t = 25) into (\frac{dL}{dt}=3.71-1.896\times10^{-3}t^{2}) (\frac{dL}{dt}\mid_{t = 25}=3.71-1.896\times10^{-3}\times(25)^{2}) (=3.71-1.896\times0.625) (=3.71 - 1.185) (=2.525\approx2.53)

Answer:

When (t = 10), (\frac{dL}{dt}\approx3.52) mm/week; when (t = 15), (\frac{dL}{dt}\approx3.28) mm/week; when (t = 25), (\frac{dL}{dt}\approx2.53) mm/week. Since (3.52>3.28>2.53), the rate of growth of the fetus decreases as it ages.