4.10-12: The size of an unborn fetus of a certain species depends on its age. Data for Head circumference (H) as a function of age (t) in weeks were fitted using the formula H= - 29.03 + 1.763t - 0.0379t log t.dH(a) Calculate the rate of fetal growthdt(b) IsdHlarger early in development (say at t= 8 weeks) or late (say at t= 36 weeks)?dt1 dH(c) Repeat part (b) but for fractional rate of growthH dt

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Description: 4.10-12: The size of an unborn fetus of a certain species depends on its age. Data for Head circumference (H) as a function of age (t) in weeks were fitted using the formula H= - 29.03 + 1.763t - 0.0379t log t.dH(a) Calculate the rate of fetal growth

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Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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4.10-12:

The size of an unborn fetus of a certain species depends on its age. Data for Head circumference (H) as a function of age (t) in weeks were fitted using the formula H= - 29.03 + 1.763t - 0.0379t log t.
dH
(a) Calculate the rate of fetal growth
dt
(b) Is
dH
larger early in development (say at t= 8 weeks) or late (say at t= 36 weeks)?
dt
1 dH
(c) Repeat part (b) but for fractional rate of growth
H dt

Expert Solution

Step 1

Given

$H = - 29.03 + 1.763 t^{2} - 0.0379 t^{2} \log t$

Step 2

Solve (a)

$T h e r a t e o f f e t a l g r o w t h \frac{d H}{d t} = \frac{d}{d t} (- 29.03 + 1.763 t^{2} - 0.0379 t^{2} \log t) \frac{d H}{d t} = (0 + 1.763 (2 t) - 0.0379 (2 t \log t + t^{2} \times \frac{1}{t})) \frac{d H}{d t} = (3.526 t - 0.0379 (2 t \log t + t)) \frac{d H}{d t} = (3.526 t - 0.0758 t \log t - 0.0379 t) \frac{d H}{d t} = (3.4881 t - 0.0758 t \log t)$

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