Now, according to the requirements of the state standards and existing programs of training in medical institutions, the main task of studying of discipline “Mathematics” is equipping students with mathematical knowledge and the abilities necessary for studying of special disciplines of a base level, and in requirements to professional preparation of the expert the ability to solve professional problems with use of mathematical methods is declared. This situation cannot but affect the results of mathematical training of medics. The level of professional competence of medical staff depends to a certain extent on these results. These results show that by studying mathematics, health care workers subsequently acquire certain professionally important qualities and skills, as well as apply mathematical concepts and methods in medical science and practice.
Application of applied tasks in medicine.
The professional orientation of mathematical preparation in medical educational institutions should provide the increase of the level of mathematical competence of medical students, comprehension of value of mathematics for the future professional activity, development of professionally significant qualities and methods of mental activity, mastering by students the mathematical apparatus allowing to model, analyze and solve elementary mathematical professionally significant problems, taking place in the medical science and practice, about.
Applied problems in medicine are used to solve many calculations regarding children. For example, the calculation of the growth gain of children after one year of life is done using the formula:
R = 75 + 5 ∙n, where n is the number of years.
Task 1. Calculate the average growth of a child at the age of 2 years.
Solution: R = 75 + 5 ∙ 2
R = 85 (cm)
Answer: The average height of a 2 year old child is 85 cm.
Task #2. Calculate the increase in height of a child from 4 to 6 years old.
Solution: P = 75 + 5 ∙ 4
Р = 95
Р = 75 + 5 ∙ 6
Р = 105
Р = 105 – 95
P = 10 (cm)
Answer: From 4 to 6 years, the child has added 10 cm in height.
Also with the help of applied tasks is the calculation of weight gain of children.
This calculation is done using the table of average weight gain of children for each month of the first year of life:
Task 1. Calculate the baby’s weight at 8 months of life, if it is known that the baby’s birth weight was 2kg. 800g, and every month he was gaining weight according to the table.
Solution: 2800 + 600 + 800 + 800 + 750 + 700 + 650 + 600 + 550 = 8250 (g)
Answer: At eight months, the baby weighed 8 kg 250 g.
Problem 2. How much a 1-year-old child, born with a weight of 3 kg 300g, weighs, if you know that in the last 4 months he has put on 2 kg, and in the rest months he has put on weight according to the table.
Solution: 3300 + 2000 = 5300
5300 + 700 + 650 + 600 + 550 + 500 + 450 + 400 + 350 = 9500 (г)
Answer: At one year of age the child weighed 9 kg 500 g.
Applied problems are used to calculate the child’s nutrition (volumetric method)
The daily volume of a child under 1 year of age is:
Up to 2 months ⅕ of the child’s body weight
˃ 6 ⅛
Task 1. The baby is 5 months old. At birth he weighed 3000 g. Calculate the baby’s weight according to the table and his nutrition volume.
Solution: 3000 + 600 + 800 + 800 + 750 + 700 = 6650 (g)
6650 : 7 = 950 (г)
Answer: At 5 months, the weight of the child is 6 kg 650 g, and its V nutrition 950 g.
Problem 2. How much more food a 6-month-old baby needs than a 2-month-old baby, if you know that at 6 months the baby weighed 5800 g, and at 2 months he weighed 4000 g.
Solution: 5800 ∙ 1\7 = 828, 6
4000 ∙ 1\6 = 666, 7
828, 6 – 666, 7 = 161,9 (г)
Answer: a 6-month-old baby needs 161.9 grams more food than a 2-month-old baby.
Applied problems in the basics of nursing.
Calculating the percentage concentration of solutions (in different volumes of liquid)
The three basic math problems for percentages are as follows.
Problem 1. Find the specified percentage of a given number.
The given number is divided by 100, and the result is multiplied by the number of percentages.
Example: An average of 0.5 kg of bleach is used in the department per day. During the general cleaning of the rooms, 153% of the average daily amount of bleach was used. How much chlorine lime did the ward staff use during the general cleaning of the room?
Solution: 1) 0.5 kg : 100% = 0.005
2) 0.005 ∙ 153% = 0.765 (kg)
Answer: 0. 765 kg of bleach was used up in a day during the general cleaning.
Task 2. Find the number by the given value of its percentage.
The given value is divided by the number of percentages, and the result is multiplied by 100.
Example: The weight of the chlorine lime in the solution is 10%. How much water will be needed to dilute the solution if you know that 0.2kg of lime was taken?
Solution: 1) 0.2 : 10 = 0.02
2) 0,02 ∙ 100 = 2 (л)
Answer: He will need 2 l of water.
Problem 3. Find the expression of one number as a percentage of another number.
Multiply the first number by 100, divide the result by the second number.
Example: In one day the department has used 765 grams of bleach instead of the average daily rate of 500 grams. How many percent more lime was used?
Solution: 1) 765 – 500 = 265
2) 265 ∙ 100 = 26500
3) 26500 : 500 = 53
Answer: 53% more chlorine lime was used per day.
By definition, the concentration of pure substance in solution is the number of grams in 100 ml. Therefore, to calculate the amount of substance in 1 ml of solution, divide the available mass of pure substance in solution by 100.
Applied Problems and Pharmacology.
Problem 1. A ratio of 5 : 200 is used for the solution. How many liters of solution can be prepared from 1.5 kg of pure substance?
Solution: 5 – 200.
1500 – Х
X = 1500 ∙ 200 : 5 = 60 000 (ml) = 60 (l)
Answer: 60 liters of solution could be prepared from 1.5 kg of pure substance.
Problem 2. Decoction of oregano, melissa and Hypericum is prepared in the ratio: oregano – 30 g, melissa – 15 g, Hypericum – 15 g, water – 800 ml. How many liters of broth can be prepared from 1 kg of oregano, 0.5kg of lemon balm and 0.3 kg of St. John’s wort?
Solution: 1000 : 30 = 33.3
500 : 15 = 33,3
300 : 15 = 20
20 ∙ 800 = 16000 (ml) = 16 (l)
Answer: It is possible to prepare 16 liters of solution.
Task 3. An injection of galantamine hydrobromide 1ml – 25% solution was made. How much dry substance was contained in the injected preparation?
Solution: 1ml – 25%.
Xml – 100%.
X = 100 : 25 = 4 (mg)
Answer: 4 mg of dry matter was contained in the drug.
Mathematics and DM in Therapy.
Task 1. The need of the polyclinic for specialists is 25 people, and there are 22 people working in total. What is the percentage?
Solution: 25 – 100%.
22 – Х%
25Х = 2200
Х = 88
Answer: 88% of specialists work.
Problem 2. There is only 0.02% furacilin in solution. How many liters of disinfectant solution can be obtained from 2 grams of furacilin?
Solution: 2 : 0.02 = 100
100 ∙ 100 = 10000ml = 10 l.
Answer: You can get 10 liters of disinfectant from 2 grams of furacilin.
Mathematics in obstetrics.
Problem 1. The weight of a 4-month-old fetus is 120g, and the weight of a 7-month-old fetus is 1100g. How many percent of the weight of a 4-month-old fetus is the weight of a 7-month-old fetus?
Solution: 4-month-old is 120g.
7-month-old fetus is 1100g
120 – Х
1100Х = 12000
Х = 11
The answer is 11% of the weight of a 4-month-old fetus from the weight of a 7-month-old fetus.
Problem 2. The blood weight of a newborn baby is 15% of its body weight. Calculate the blood weight of a newborn baby weighing 4 kg 800g.
Solution: 4800 – 100%
Х – 15%
100Х = 72000
Х = 720
Answer: 720 g is the blood mass of a newborn baby.
Mathematics in anatomy.
1.The cardiovascular system.
The mass of an adult heart is 1\220 of body weight (0.425 – 0.570 kg). The mass of the heart of a newborn is on average 0.66 – 0.80% of body weight (about 20g). Parameters of an adult heart: length h – 12-15 cm, transverse section d1 – 8-10 cm, anterior – posterior dimension d2 – 5-8 cm. to calculate the volume of the heart we use the formula of cone volume:
V = 1\3 Sh = 1\3 πR²h = 1\12πd²h
Problem 1. The mass of the heart is 1\220th of the human body mass. Calculate the heart mass of a 35-year-old man if you know that at age 28 he weighed 116 kg and lost 1.5 kg in weight each year.
Solution: 35 – X
28 – 116
35 – 28 = 7
7 ∙ 1,5 = 10,5
116 – 10,5 = 105,5
105,5 : 220 = 0,477
Answer: the mass of human heart at the age of 35 is 477 g.
Problem 2. To calculate the volume of the heart of an adult person if its length
h = 12 cm and the cross section d = 8 cm (V = 1\12πd²h)
Solution: V = 1\12 ∙ 3.14 ∙ 8² ∙ 12 = 200.96
Answer: Volume of an adult heart is 200.96 cm³.
2. Bone and muscle system.
The solution of problems on this topic requires knowledge of areas and volumes of shapes.
Areas of figures.
Square: S = a² = d²\2, where a is the side, d is the diagonal.
Rectangle: S = a ∙b, where a and b are sides.
Rhombus: S = d₁d₂ : 2 = a² ∙sinα, where d₁ and d₂ are diagonals, a is a side, α is one of the angles.
Parallelogram: S = a∙h = a∙h∙sinα, where a and h are sides, h is height, α is an angle.
Trapezoid: S = (a + b) : 2 ∙h = c∙h, where a and b are bases, h is height, c is midline.
Triangle: S = 1\2 ah = √p(p-a)(p-b)(p-c) where a is the base, h is the height, p is the semiperimeter.
Circle:S = π∙d²\4 ≈ 0.875d², where d is the diameter.
Volumes of figures.
Prism:V = S ∙ ∙ ℓ, where S is the perpendicular section,ℓ is the length of the side edge.
Cube: V = a³, where a is an edge of the cube.
Pyramid:V =1\3 ∙ S∙h, where S is base area, h is height.
Cylinder: V = π ∙ r² ∙ h, where r is the base radius, h is the height.
Cone: V = 1\3 ∙ S ∙ h, where S is the base area, h is the height.
Ball: V = 3\4 ∙ π ∙ r³, where r is the radius of the ball.
Problem 1. Human shin bone has length h = 40 cm, width d = 5 cm. Calculate the volume of bone (V = S ∙ h = π ∙ d ∙ h).
Solution: V = 3,14 ∙ 40 ∙ 5 = 628.
Answer: the volume of shin bone is 628 cm³.
Problem 2. A human mass is 70 kg. The muscular system makes up 40% of his body mass. The muscles of the lower extremities account for 50% of the total number of muscles. How many kilograms is that?
Solution: 70 : 100 ∙ 40 = 28
28 : 100 ∙ 50 = 14
Answer: The muscle system of the lower extremities is 14 kilograms.
3.The spinal cord and the brain.
Problem 1. The mass of a woman at the age of 35 is 72 kg. The mass of her spinal cord is 35g. Calculate how many percent of her body weight is the weight of her spinal cord.
Solution: 72 – 100%.
Х – 0,035
Х = 0,035 ∙ 100 : 72 = 0,049%
Answer: 0.049% of the weight of her spinal cord from her body weight.
Problem 2. The weight of a man is 105 kg. How much does his spinal cord weigh, if its weight is 0,05% of his body weight?
Solution: 105 : 100 ∙ 0.05 = 0.0525kg = 52.5g
Answer: the human spinal cord weighs 52.5 g.
4. Genitourinary system.
Task 1. During a day 1,500 liters of blood pass through the kidneys. All the blood will pass through the kidneys in about 5 minutes (5-6l.). How much blood will pass through human kidneys in an hour?
Solution: Let’s make up the proportion: 24h – 1500l
1h – Xl.
Where X = 1500 ∙ 1\24 = 62.5
Answer: 62.5 L of blood passes through human kidneys in one hour.
Problem 2. The capacity of the bladder of a 3-month-old baby is 100ml. How many ml of urine are in the bladder?
Solution: 100 : 100 ∙ 25 = 25
Answer: the bladder is 25ml full.
5.Anatomy in Pediatrics.
Blood in a newborn baby is 15% of body weight, in children under a year – 11% of body weight.
Task 1. Calculate the blood mass of a newborn baby weighing 3.8kg.
Solution: 3,8 : 100 ∙ 15 = 0,57
Answer: The blood mass of a newborn baby with the weight of 3.8 kg is 0.57 kg.
Problem 2. The baby was born with the weight of 2850 grams and was gaining weight according to the table. The brain mass of a new-born baby is 400 grams. Calculate, how many percent of the body weight is the mass of the brain?
Solution: 2850 – 100%.
400 – Х
Х = 400 ∙ 100 : 2850 = 13,98%
Answer: The mass of the brain from the body mass is 13.98%.
Blood in an adult makes up 6-8% of body weight. During a day blood flows through kidneys in the amount of 1500l per day. All blood flows through the system in 5 minutes (5-6 liters).
Problem 1. How much has the blood weight of an adult changed, if his initial weight was 68 kg, he has gained 8 kg in 3 months, and in the last 2 months he has lost 4 kg?
Solution: (68 + 8 – 4) : 100 ∙ 7 = 5.04 kilograms
68 : 100 ∙ 7 = 4.76kg Initial blood weight.
5.04 – 4.76 = 0.28kg = 280g
Answer: the blood mass of an adult has changed by 280 g.
Task 2. The volume of circulating blood in the body is 1\13 of body weight. The system of large circulatory system contains 75-80% of blood, and the system of small circulatory system contains 20-25% of blood. How much blood circulates in the small circulatory system of a person weighing 65kg?
Solution: 65 ∙ 1\13 = 5 kg of blood
5 : 100 ∙ 75 = 3.75 kilograms of blood in the systemic circulation
5 – 3,75 = 1,25kg of blood in the small circulation loop
Answer: 1.25 kg of blood circulates in the small circle of the blood circulation.
7.Gas exchange in the lungs.
At relative rest an adult makes approximately 16 breathing movements per 1 minute. Lung vital capacity (VCL):
VEL = DO + ROV + ROS,
Where RR is respiratory volume (0.5L)
ROB – inspiratory reserve volume (1.5 L) 3 – 4 L
EVO – exhaled reserve volume (1.5 L)
Inhaled air contains
– 20,97% oxygen
– about 79% nitrogen
– about 0,03% carbon dioxide
– a small amount of water vapor and inert gases.
The percentage composition of exhaled air is different:
– 16% oxygen
– 4% of carbon dioxide.
Task 1. When breathing normally, a man makes 16 breathing movements per minute. Under physical exertion the number of breathing movements increases by 50%. How much carbon dioxide did the person exhale during physical exertion during 2 minutes, if GEL = 4000 cm³?
Solution: (16 + 16 : 100 ∙ 50) ∙ 2 = 48 breathing movements
a human being performs in 2 minutes
1500 ∙ 48 : 100 ∙ 4 = 2880cm3 (as the exhalation reserve volume is 500cm3).
Answer: a human being exhaled 2880 cm³ of carbon dioxide during physical exertion for 2 minutes.
Task 2. Every minute a man is making 16 breathing movements, and during 1 inhalation 1500 cm3 of air come into his lungs. What is the minute ventilation of the lungs?
Solution: 16 ∙ 1500 = 24000cm³.
Answer: 24000 cm³ was inhaled by a human being during one minute.
Conclusion: The role of mathematical education in the professional training of medical workers is very great.
The processes occurring nowadays in all spheres of social life impose new requirements to professional qualities of specialists. The modern stage in the development of society is characterized by qualitative changes in the activity of medical personnel, which is associated with the widespread use of mathematical modeling, statistics and other important phenomena that take place in medical practice.
At the first sight the medicine and mathematics can seem incompatible fields of human activity. Mathematics is admittedly the “queen” of all sciences, solving problems in chemistry, physics, astronomy, economics, sociology and many other sciences. Medicine, on the other hand, has long been developing “in parallel” with mathematics, and has remained an almost unformalized science, thus confirming that “medicine is an art.”
The main problem is that there are no general criteria for health, and the set of indicators for one particular patient (the conditions when he feels comfortable) may differ significantly from the same indicators for another. Often medics are faced with common problems formulated in medical terms in order to help the patient, they do not bring ready-made problems and equations to be solved.
When properly applied, a mathematical approach is not significantly different from one based simply on common sense. Mathematical methods are simply more precise and use clearer language and a broader set of concepts, but ultimately they must be compatible with conventional verbal reasoning, though they probably go further than that.
The problem-setting stage can be laborious and time-consuming, and often goes on practically until a solution is obtained. But it is the different perspectives of mathematicians and physicians, who are representatives of two different methodologies, that help to obtain the result.
All medical discoveries must rely on numerical relationships. The methods of probability theory (taking into account the statistics of morbidity depending on various factors) are a necessary thing in medicine. Without mathematics, it is impossible to make a move in medicine. Numerical relationships, for example, accounting for dosage and frequency of medication. Numerical accounting of associated factors, such as: age, physical parameters of the body, immunity, etc.
My opinion stands firmly on the fact that medics should not turn a blind eye to at least elementary mathematics, which is simply necessary for organizing fast, clear and high quality work. Every student should note for himself from the first year of training: knowledge is important and makes life much easier.