Repetition of the basic theory and formulas, including those that allow you to calculate the volume of a cylinder, is one of the main stages in preparing for the exam. Despite the fact that this topic is considered in sufficient detail in mathematics lessons at school, many students face the need to remember the basic material and “pump” the skill of solving problems. By understanding how to calculate the volume and other unknown parameters of a cylinder, high school students will be able to get fairly high scores on the results of passing the unified state exam.

Key points to remember

It is important to remember that:

• A cylinder is a body bounded by a cylindrical surface and two circles. The cylindrical surface is lateral. And the circles represent the bases of the figure.
• The height of a cylinder is the distance between the planes of its bases.
• All its generators are parallel and equal to each other.
• The radius of a cylinder is the radius of its base.
• A figure is called a straight line if its generators are perpendicular to the bases.

How to prepare for the exam efficiently and effectively?

Studying on the eve of passing the certification test, many students are faced with the problem of finding the necessary information. Not always a school textbook is at hand when it is required. And finding formulas that will help calculate the area and other unknown parameters of a cylinder is often quite difficult even on the Internet in online mode.

Studying together with the Shkolkovo mathematical portal, graduates will be able to avoid typical mistakes and successfully pass the unified state exam. We propose to build the preparation process in a new way, moving from simple to complex. This will allow students to identify topics that are incomprehensible to themselves and eliminate gaps in knowledge.

Graduates can find all the basic material that will help in solving problems on the topic "Cylinder" in the "Theoretical Reference" section. Shkolkovo specialists presented all the necessary definitions and formulas in an accessible form.

To consolidate their knowledge, students can practice solving problems on the topic "Cylinder" and other topics, for example,. A large, constantly updated selection of tasks is presented in the Catalog section.

In order to quickly find a specific problem on the topic “Cylinder” during preparation for the exam and refresh the algorithm for solving it, graduates can first save it to “Favorites”. Not only schoolchildren from the capital, but also students from other Russian cities have the opportunity to practice their own skills on our website.

With this service, you can find the largest and smallest value of a function one variable f(x) with the design of the solution in Word. If the function f(x,y) is given, therefore, it is necessary to find the extremum of the function of two variables . You can also find the intervals of increase and decrease of the function.

Function entry rules:

A necessary condition for an extremum of a function of one variable

A sufficient condition for an extremum of a function of one variable

F" 0 (x *) = 0
f"" 0 (x *) > 0

Then the point x * is the point of the local (global) minimum of the function.

If at the point x * the condition is met:

F" 0 (x *) = 0
f"" 0 (x *)< 0

That point x * is a local (global) maximum.

Example #1. Find the largest and smallest value functions: on a segment.
Solution.

The critical point is one x 1 = 2 (f'(x)=0). This point belongs to the segment . (The point x=0 is not critical, since 0∉).
We calculate the values ​​of the function at the ends of the segment and at the critical point.
f(1)=9, f(2)= 5 / 2 , f(3)=3 8 / 81
Answer: f min = 5 / 2 for x=2; f max =9 at x=1

Example #2. Using higher order derivatives, find the extremum of the function y=x-2sin(x) .
Solution.
Find the derivative of the function: y’=1-2cos(x) . Let us find the critical points: 1-cos(x)=2, cos(x)=1, x=± π / 3 +2πk, k∈Z. We find y''=2sin(x), calculate , so x= π / 3 +2πk, k∈Z are the minimum points of the function; , so x=- π / 3 +2πk, k∈Z are the maximum points of the function.

Example #3. Investigate the extremum function in the neighborhood of the point x=0.
Solution. Here it is necessary to find the extrema of the function. If the extremum x=0 , then find out its type (minimum or maximum). If among the found points there is no x = 0, then calculate the value of the function f(x=0).
It should be noted that when the derivative on each side of a given point does not change its sign, the possible situations are not exhausted even for differentiable functions: it may happen that for an arbitrarily small neighborhood on one side of the point x 0 or on both sides, the derivative changes sign. At these points, one has to apply other methods to study functions to an extremum.

Example #4. Divide the number 49 into two terms, the product of which will be the largest.
Solution. Let x be the first term. Then (49-x) is the second term.
The product will be maximal: x (49-x) → max

Job type: 8
Theme: Cylinder

Condition

In a cylindrical vessel, the liquid level reaches 20 cm. At what height will the liquid level be if it is poured into a second cylindrical vessel, the diameter of which is twice the diameter of the first? Express your answer in centimeters.

Solution

Let R be the radius of the base of the first vessel, then 2 R is the radius of the base of the second vessel. By condition, the volume of liquid V in the first and second vessel is the same. Denote by H - the level to which the liquid has risen in the second vessel. Then

V=\pi R^2 \cdot 20, And V=\pi (2R)^2H = 4\pi R^2H. From here \pi R^2 \cdot 20 = 4\pi R^2H, 20=4H H=5

Answer

Job type: 8
Theme: Cylinder

Condition

2000 cm 3 of water was poured into a cylindrical vessel. The liquid level turned out to be 15 cm. The part was completely immersed in water. At the same time, the liquid level in the vessel rose by 9 cm. What is the volume of the part? Express your answer in cm3.

Solution

Let R be the radius of the base of the cylinder, and h be the level of the water poured into the vessel. Then the volume of poured water is equal to the volume of a cylinder with base radius R and height h. V water \u003d S main. · h = \pi R^2\cdot h. According to the condition, the equality 2000=\pi R^2\cdot15 is fulfilled. From here, \pi R^2=\frac(2000)(15)=\frac(400)(3).

Let H be the water level in the vessel after the item is immersed in it. Then the total volume of water and the part is equal to the volume of a cylinder with a base radius R and a height H. By condition H=h+9=15+9=24. So V water + details = \pi R^2\cdot H=\frac(400)(3)\cdot24=3200. Therefore, V parts = V water + parts − V water = 3200-2000=1200.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 8
Theme: Cylinder

Condition

Find the height of a cylinder if its base radius is 8 and its side surface area is 96\pi.

Solution

S=2\pi rh,

96\pi=2\pi\cdot8h,

h=\frac(96\pi)(16\pi)=6.

Answer

Source: "Mathematics. Preparation for the exam-2016. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 8
Theme: Cylinder

Condition

500 cubic meters were poured into a cylindrical vessel. see water. Determine the volume of the part completely submerged in water if, after immersion, the liquid level increased by 1.2 times. Express your answer in cube. cm.

Solution

Let V 1 denote the initial volume of liquid in the cylinder. After the part was immersed, the volume of liquid increased by 1.2 times, which means that the final volume of liquid is V 2 = 1.2 V 1. The volume of the part is equal to the difference between the volumes before and after immersion, which means V = V_2-V_1=1.2\cdot 500-500=100 cube cm.

Answer

When a liquid is overflowed, its initial volume does not change, i.e.: V 1 \u003d V 2, which means that the equality is true: \pi\left(\frac(d_1)(2)\right)^2h_1=\pi\left(\frac(3d_1)(2)\right)^2h_2

Substitute the values ​​from the condition, simplify the expression and find the desired height of the liquid of the second vessel h 2:

\pi \enspace\frac(d_1^(2))(4)\enspace 63=\pi \enspace\frac(9d_1^(2))(4)\enspace h_2

\frac(63)(4)=\frac(9)(4)h_2

h_2=\frac(63)(9)=7