Objectives of the lesson: to verify the existence of a buoyant force, to understand the reasons for its occurrence and to derive rules for its calculation, to promote the formation of a worldview idea of the knowability of phenomena and properties of the surrounding world.
Lesson objectives: Work on developing the skills to analyze properties and phenomena based on knowledge, highlight the main reason influencing the result. Develop communication skills. At the stage of generating hypotheses, develop oral speech. To check the level of independent thinking of the student in terms of the students’ application of knowledge in various situations.
Archimedes is an outstanding scientist of Ancient Greece, born in 287 BC. in the port and shipbuilding city of Syracuse on the island of Sicily. Archimedes received an excellent education from his father, the astronomer and mathematician Phidias, a relative of the Syracuse tyrant Hiero, who patronized Archimedes. In his youth, he spent several years in the largest cultural center in Alexandria, where he developed friendly relations with the astronomer Conon and the geographer-mathematician Eratosthenes. This was the impetus for the development of his outstanding abilities. He returned to Sicily as a mature scientist. He became famous for his numerous scientific works, mainly in the fields of physics and geometry.
The last years of his life, Archimedes was in Syracuse, besieged by the Roman fleet and army. The 2nd Punic War was underway. And the great scientist, sparing no effort, organizes the engineering defense of his hometown. He built many amazing combat vehicles that sank enemy ships, smashed them to pieces, and destroyed soldiers. However, the army of the city’s defenders was too small compared to the huge Roman army. And in 212 BC. Syracuse was taken.
The genius of Archimedes was admired by the Romans and the Roman commander Marcellus ordered his life to be spared. But the soldier, who did not know Archimedes by sight, killed him.
One of his most important discoveries was the law, later called Archimedes' law. There is a legend that the idea of this law came to Archimedes while he was taking a bath, with the exclamation “Eureka!” he jumped out of the bath and ran naked to write down the scientific truth that had come to him. The essence of this truth remains to be clarified; we need to verify the existence of a buoyant force, understand the reasons for its occurrence and derive rules for calculating it.
The pressure in a liquid or gas depends on the depth of the body's immersion and leads to the appearance of a buoyant force acting on the body and directed vertically upward.
If a body is lowered into a liquid or gas, then under the action of a buoyant force it will float up from deeper layers to shallower ones. Let us derive a formula for determining the Archimedes force for a rectangular parallelepiped.
The fluid pressure on the upper face is equal to
where: h1 is the height of the liquid column above the top edge.
Pressure force on the top the edge is equal
F1= p1*S = w*g*h1*S,
Where: S – area of the top face.
The fluid pressure on the lower face is equal to
where: h2 is the height of the liquid column above the bottom edge.
The pressure force on the lower edge is equal to
F2= p2*S = w*g*h2*S,
Where: S is the area of the bottom face of the cube.
Since h2 > h1, then р2 > р1 and F2 > F1.
The difference between the forces F2 and F1 is equal to:
F2 – F1 = f*g*h2*S – f*g*h1*S = f*g*S* (h2 – h1).
Since h2 – h1 = V is the volume of a body or part of a body immersed in a liquid or gas, then F2 – F1 = w*g*S*H = g* w*V
The product of density and volume is the mass of the liquid or gas. Therefore, the difference in forces is equal to the weight of the fluid displaced by the body:
F2 – F1= mf*g = Pzh = Fout.
The buoyancy force is the Archimedes force, which defines Archimedes' law
The resultant of the forces acting on the side faces is zero, therefore it is not involved in the calculations.
Thus, a body immersed in a liquid or gas experiences a buoyant force equal to the weight of the liquid or gas displaced by it.
Archimedes' Law was first mentioned by Archimedes in his treatise On Floating Bodies. Archimedes wrote: “bodies heavier than the liquid, immersed in this liquid, will sink until they reach the very bottom, and in the liquid they will become lighter by the weight of the liquid in a volume equal to the volume of the immersed body.”
Let's consider how the Archimedes force depends and whether it depends on the weight of the body, the volume of the body, the density of the body and the density of the liquid.
Based on the Archimedes force formula, it depends on the density of the liquid in which the body is immersed and on the volume of this body. But it does not depend, for example, on the density of the substance of the body immersed in the liquid, since this quantity is not included in the resulting formula.
Let us now determine the weight of a body immersed in a liquid (or gas). Since the two forces acting on the body in this case are directed in opposite directions (the force of gravity is downward, and the Archimedean force is upward), then the weight of the body in the liquid will be less weight bodies in vacuum by the Archimedean force:
P A = m t g – m f g = g (m t – m f)
Thus, if a body is immersed in a liquid (or gas), then it loses as much weight as the liquid (or gas) it displaced weighs.
Hence:
The Archimedes force depends on the density of the liquid and the volume of the body or its immersed part and does not depend on the density of the body, its weight and the volume of the liquid.
Determination of Archimedes' force by laboratory method.
Equipment: glass with clean water, glass of salt water, cylinder, dynamometer.
Work progress:
- determine the weight of the body in the air;
- determine the weight of the body in the liquid;
- find the difference between the weight of a body in air and the weight of a body in liquid.
4. Measurement results:
Conclude how the Archimedes force depends on the density of the liquid.
The buoyancy force acts on bodies of any geometric shape. In technology, the most common bodies are cylindrical and spherical shapes, bodies with a developed surface, hollow bodies in the shape of a ball, a rectangular parallelepiped, or a cylinder.
The gravitational force is applied to the center of mass of a body immersed in a liquid and is directed perpendicular to the surface of the liquid.
The lifting force acts on the body from the side of the liquid, is directed vertically upward, and is applied to the center of gravity of the displaced volume of liquid. The body moves in a direction perpendicular to the surface of the liquid.
Let's find out the conditions for floating bodies, which are based on Archimedes' law.
The behavior of a body located in a liquid or gas depends on the relationship between the modules of gravity F t and the Archimedes force F A , which act on this body. The following three cases are possible:
- F t > F A - the body drowns;
- F t = F A - the body floats in a liquid or gas;
- F t< F A - тело всплывает до тех пор, пока не начнет плавать.
Another formulation (where P t is the density of the body, P s is the density of the medium in which it is immersed):
- P t > P s - the body sinks;
- P t = P s - the body floats in a liquid or gas;
- P t< P s - тело всплывает до тех пор, пока не начнет плавать.
The density of organisms living in water is almost the same as the density of water, so they don’t need strong skeletons! Fish regulate their diving depth by changing the average density of their body. To do this, they only need to change the volume of the swim bladder by contracting or relaxing the muscles.
If a body lies at the bottom in a liquid or gas, then the Archimedes force is zero.
Archimedes' principle is used in shipbuilding and aeronautics.
Floating body diagram:
The line of action of the force of gravity of the body G passes through the center of gravity K (center of displacement) of the displaced volume of fluid. In the normal position of a floating body, the center of gravity of the body T and the center of displacement K are located along the same vertical, called the axis of swimming.
When rolling, the center of displacement K moves to point K1, and the force of gravity of the body and the Archimedean force FA form a pair of forces that tends to either return the body to its original position or increase the roll.
In the first case, the floating body has static stability, in the second case there is no stability. The stability of the body depends on relative position the center of gravity of the body T and the metacenter M (the intersection point of the line of action of the Archimedean force during a roll with the axis of navigation).
In 1783, the MONTGOLFIER brothers made a huge paper ball, under which they placed a cup of burning alcohol. The balloon filled with hot air and began to rise, reaching a height of 2000 meters.
The reason for the emergence of Archimedean force is the difference in pressure of the medium at different depths. Therefore, Archimedes' force occurs only in the presence of gravity. On the Moon it will be six times, and on Mars it will be 2.5 times less than on Earth.
In weightlessness there is no Archimedean force. If we imagine that the force of gravity on Earth suddenly disappeared, then all the ships in the seas, oceans and rivers will go to any depth at the slightest push. But something independent of gravity will not allow them to rise up. surface tension water, so they won’t be able to take off, they’ll all drown.
How does the power of Archimedes manifest itself?
The magnitude of the Archimedean force depends on the volume of the immersed body and the density of the medium in which it is located. Its exact definition in modern terms is as follows: a body immersed in a liquid or gaseous medium in the field of gravity is acted upon by a buoyant force exactly equal to the weight of the medium displaced by the body, that is, F = ρgV, where F is the Archimedes force; ρ – density of the medium; g – free fall acceleration; V is the volume of liquid (gas) displaced by the body or an immersed part of it.
If in fresh water For every liter of volume of a submerged body, a buoyant force of 1 kg (9.81 N) acts, then in sea water, the density of which is 1.025 kg*cub. dm, the Archimedes force of 1 kg 25 g will act on the same liter of volume. For a person of average build, the difference in the support force of sea and fresh water will be almost 1.9 kg. Therefore, swimming in the sea is easier: imagine that you need to swim across at least a pond without a current with a two-kilogram dumbbell in your belt.
The Archimedean force does not depend on the shape of the immersed body. Take an iron cylinder and measure its force from the water. Then roll out this cylinder into a sheet, immerse it flat and edge-on in water. In all three cases, the power of Archimedes will be the same.
It may seem strange at first glance, but if a sheet is immersed flat, the decrease in pressure difference for a thin sheet is compensated by an increase in its area perpendicular to the surface of the water. And when immersed with an edge, on the contrary, the small area of the edge is compensated by the larger height of the sheet.
If the water is very saturated with salts, which is why its density has become higher than the density human body, then even a person who cannot swim will not drown in it. At the Dead Sea in Israel, for example, tourists can lie on the water for hours without moving. True, it is still impossible to walk on it - the support area is small, the person falls into the water up to his neck, until the weight of the submerged part of the body is equal to the weight of the water displaced by him. However, if you have a certain amount of imagination, you can create a legend about walking on water. But in kerosene, the density of which is only 0.815 kg*cubic. dm, even a very experienced swimmer will not be able to stay on the surface.
Archimedean force in dynamics
Everyone knows that ships float thanks to the power of Archimedes. But fishermen know that Archimedean force can also be used in dynamics. If you come across a large and strong fish (taimen, for example), then there is no point in slowly pulling it to the net (fishing for it): it will break the line and leave. You need to tug lightly first when it goes away. Feeling the hook, the fish, trying to free itself from it, rushes towards the fisherman. Then you need to pull very hard and sharply so that the fishing line does not have time to break.
In water, the body of a fish weighs almost nothing, but its mass and inertia are preserved. With this method of fishing, the Archimedean force will seem to kick the fish in the tail, and the prey itself will plop down at the angler’s feet or into his boat.
Archimedes' power in the air
Archimedes' force acts not only in liquids, but also in gases. Thanks to her they fly balloons and airships (zeppelins). 1 cu. m air at normal conditions(20 degrees Celsius at sea level) weighs 1.29 kg, and 1 kg of helium weighs 0.21 kg. That is, 1 cubic meter of a filled shell is capable of lifting a load of 1.08 kg. If the shell has a diameter of 10 m, then its volume will be 523 cubic meters. m. Having completed it from the lung synthetic material, we get a lifting force of about half a ton. Aeronauts call Archimedes' force in the air fusion force.
If you pump out the air from the balloon without allowing it to shrink, then each cubic meter of it will pull up the entire 1.29 kg. An increase of more than 20% in lift is technically very tempting, but helium is expensive and hydrogen is explosive. Therefore, projects of vacuum airships appear from time to time. But materials that can withstand large amounts (about 1 kg per sq. cm) atmospheric pressure outside onto the shell, modern technology not yet able to create.
The proof and derivation of the formula for the derivative of the cosine - cos(x) is presented. Examples of calculating derivatives of cos 2x, cos 3x, cos nx, cosine squared, cubed and to the power n. Formula for the derivative of the cosine of the nth order.
ContentSee also: Sine and cosine - properties, graphs, formulas
The derivative with respect to the variable x from the cosine of x is equal to minus the sine of x:
(cos x)′ = - sin x.
Proof
To derive the formula for the derivative of cosine, we use the definition of derivative:
.
Let's transform this expression to reduce it to the known ones mathematical laws and rules. To do this we need to know four properties.
1)
Trigonometric formulas. We will need the following formula:
(1)
;
2)
Continuity property of the sine function:
(2)
;
3)
The meaning of the first remarkable limit:
(3)
;
4)
Property of the limit of the product of two functions:
If and , then
(4)
.
Let's apply these laws to our limit. First we transform the algebraic expression
.
To do this we apply the formula
(1)
;
In our case
; . Then
;
;
;
.
Let's make a substitution. At , . We use the property of continuity (2):
.
Let's make the same substitution and apply the first remarkable limit (3):
.
Since the limits calculated above exist, we apply property (4):
.
Thus, we obtained the formula for the derivative of the cosine.
Examples
Let's consider simple examples finding derivatives of functions containing cosine. Let's find derivatives of the following functions:
y = cos 2x; y = cos 3x; y = cos nx; y = cos 2 x; y = cos 3 x and y = cos n x.
Example 1
Find derivatives of cos 2x, cos 3x And cosnx.
The original functions have a similar form. Therefore, we will find the derivative of the function y = cosnx. Then, as a derivative of cosnx, substitute n = 2 and n = 3 . And, thus, we obtain formulas for the derivatives of cos 2x And cos 3x .
So, we find the derivative of the function
y = cosnx
.
Let's imagine this function of the variable x as a complex function consisting of two functions:
1)
2)
Then the original function is a complex (composite) function composed of functions and :
.
Let's find the derivative of the function with respect to the variable x:
.
Let's find the derivative of the function with respect to the variable:
.
We apply.
.
Let's substitute:
(P1) .
Now, in formula (A1) we substitute and:
;
.
;
;
.
Example 2
Find the derivatives of cosine squared, cosine cubed and cosine to the power n:
y = cos 2 x; y = cos 3 x; y = cos n x.
In this example, the functions also have a similar appearance. Therefore, we will find the derivative of the most general function - cosine to the power n:
y = cos n x.
Then we substitute n = 2 and n = 3. And, thus, we obtain formulas for the derivatives of cosine squared and cosine cubed.
So we need to find the derivative of the function
.
Let's rewrite it in a more understandable form:
.
Let's imagine this function as a complex function consisting of two functions:
1)
Functions depending on a variable: ;
2)
Functions depending on a variable: .
Then the original function is a complex function composed of two functions and :
.
Find the derivative of the function with respect to the variable x:
.
Find the derivative of the function with respect to the variable:
.
We apply the rule of differentiation of complex functions.
.
Let's substitute:
(P2) .
Now let's substitute and:
;
.
;
;
.
Higher order derivatives
Note that the derivative of cos x first order can be expressed through cosine as follows:
.
Let's find the second-order derivative using the formula for the derivative of a complex function:
.
Here .
Note that differentiation cos x causes its argument to increase by . Then the nth order derivative has the form:
(5)
.
This formula can be proven more strictly using the method of mathematical induction. Proof for nth derivative sine is described on the page “Derivative of sine”. For the nth derivative of the cosine, the proof is exactly the same. You just need to replace sin with cos in all formulas.
See also:Proof and derivation of the formulas for the derivative of the exponential (e to the x power) and the exponential function (a to the x power). Examples of calculating derivatives of e^2x, e^3x and e^nx. Formulas for derivatives of higher orders.
ContentSee also: Exponential function - properties, formulas, graph
Exponent, e to the x power - properties, formulas, graph
Basic formulas
The derivative of an exponent is equal to the exponent itself (the derivative of e to the x power is equal to e to the x power):
(1)
(e x )′ = e x.
The derivative of an exponential function with a base a is equal to the function itself multiplied by the natural logarithm of a:
(2)
.
An exponential is an exponential function whose power base is equal to the number e, which is the following limit:
.
Here it can be either a natural number or a real number. Next, we derive formula (1) for the derivative of the exponential.
Derivation of the exponential derivative formula
Consider the exponential, e to the x power:
y = e x .
This function is defined for everyone. Let's find its derivative with respect to the variable x. By definition, the derivative is the following limit:
(3)
.
Let's transform this expression to reduce it to known mathematical properties and rules. To do this we need the following facts:
A) Exponent property:
(4)
;
B) Property of logarithm:
(5)
;
IN) Continuity of the logarithm and the property of limits for a continuous function:
(6)
.
Here is a function that has a limit and this limit is positive.
G) The meaning of the second remarkable limit:
(7)
.
Let's apply these facts to our limit (3). We use property (4):
;
.
Let's make a substitution. Then ; .
Due to the continuity of the exponential,
.
Therefore, when , . As a result we get:
.
Let's make a substitution. Then . At , . And we have:
.
Let's apply the logarithm property (5):
. Then
.
Let's apply property (6). Since there is a positive limit and the logarithm is continuous, then:
.
Here we also used the second remarkable limit (7). Then
.
Thus, we obtained formula (1) for the derivative of the exponential.
Derivation of the formula for the derivative of an exponential function
Now we derive formula (2) for the derivative of the exponential function with a base of degree a. We believe that and . Then the exponential function
(8)
Defined for everyone.
Let's transform formula (8). To do this, we will use the properties of the exponential function and logarithm.
;
.
So, we transformed formula (8) to the following form:
.
Higher order derivatives of e to the x power
Now let's find derivatives of higher orders. Let's look at the exponent first:
(14)
.
(1)
.
We see that the derivative of function (14) is equal to function (14) itself. Differentiating (1), we obtain derivatives of the second and third order:
;
.
This shows that the nth order derivative is also equal to the original function:
.
Derivatives of higher orders of the exponential function
Now consider an exponential function with a base of degree a:
.
We found its first-order derivative:
(15)
.
Differentiating (15), we obtain derivatives of the second and third order:
;
.
We see that each differentiation leads to the multiplication of the original function by . Therefore, the nth order derivative has the following form:
.
The basis of the proof is the determination of the limit of the function. You can use another method using trigonometric formulas reductions for cosine and sine angles. Express one function through another - cosine through sine, and differentiate the sine with a complex argument.
Let's consider the first example of derivation of the formula (Cos(x))"
We give a negligibly small increment Δx to the argument x of the function y = Cos(x). With a new value of the argument x+Δx, we obtain a new value of the function Cos(x+Δx). Then the increment of the function Δу will be equal to Cos(x+Δx)-Cos(x).
The ratio of the function increment to Δx will be as follows: (Cos(x+Δx)-Cos(x))/Δx. Let's carry out identity transformations in the numerator of the resulting fraction. Let us recall the formula for the difference between the cosines of angles, the result will be the product -2Sin(Δx/2) multiplied by Sin(x+Δx/2). We find the limit of the partial lim of this product on Δx as Δx tends to zero. It is known that the first (it is called remarkable) limit lim(Sin(Δx/2)/(Δx/2)) is equal to 1, and the limit -Sin(x+Δx/2) is equal to -Sin(x) with Δx tending to zero.
Let's write down the result: the derivative (Cos(x))" is equal to - Sin(x).
Some people prefer the second way to derive the same formula
From the trigonometry course we know: Cos(x) is equal to Sin(0.5·∏-x), similarly Sin(x) is equal to Cos(0.5·∏-x). Then we differentiate the complex function - the sine of the additional angle (instead of the cosine x).
We obtain the product Cos(0.5·∏-х)·(0.5·∏-х)", because the derivative of sine x is equal to cosine x. We turn to the second formula Sin(x) = Cos(0.5·∏- x) replacing cosine with sine, we take into account that (0.5·∏-x)" = -1. Now we get -Sin(x).
So, the derivative of the cosine has been found, y" = -Sin(x) for the function y = Cos(x).
A frequently used example where the derivative of the cosine is used. The function y = Cos 2 (x) is complex. We first find the differential of a power function with exponent 2, it will be 2 Cos(x), then multiply it by the derivative (Cos(x))", which is equal to -Sin(x). We get y" = -2 Cos(x) Sin(x). When we apply the formula Sin(2 x), the sine of the double angle, we obtain the final simplified
answer y" = -Sin(2 x)
Hyperbolic functions
They are used in the study of many technical disciplines: in mathematics, for example, they facilitate the calculation of integrals and solutions. They are expressed through trigonometric functions with an imaginary argument, for example, hyperbolic cosine ch(x) = Cos(i x), where i is an imaginary unit, hyperbolic sine sh(x) = Sin(i x).
The derivative of the hyperbolic cosine is calculated quite simply.
Consider the function y = (e x +e -x)/2, this is the hyperbolic cosine ch(x). We use the rule for finding the derivative of the sum of two expressions, the rule for moving the constant factor (Const) beyond the sign of the derivative. The second term 0.5 e -x is a complex function (its derivative is equal to -0.5 e -x), 0.5 e x is the first term. (ch(x)) "=((e x +e - x)/2)" can be written differently: (0.5 e x +0.5 e - x)" = 0.5 e x -0.5 e - x, because the derivative (e - x)" is equal to -1 multiplied by e - x. The result is a difference, and this is the hyperbolic sine sh(x).
Conclusion: (ch(x))" = sh(x).
Let's look at an example of how to calculate the derivative of the function y = ch(x 3 +1).
According to the hyperbolic cosine with a complex argument, y" = sh(x 3 +1)·(x 3 +1)", where (x 3 +1)" = 3·x 2 +0.
Answer: the derivative of this function is 3 x 2 sh(x 3 +1).
The derivatives of the considered functions y = ch(x) and y = Cos(x) are tabulated
When solving examples, there is no need to differentiate them each time according to the proposed scheme; it is enough to use the derivation.
Example. Differentiate the function y = Cos(x)+Cos 2 (-x)-Ch(5 x).
It’s easy to calculate (we use tabular data), y" = -Sin(x)+Sin(2 x)-5 Sh(5 x).
