Advanced Mathematics: Sixth chapter The application of Definite integral (2) The arc length work of plane curve and the gravity of water pressure _ higher mathematics

Arc length of §6.4 plane curve

I. The case of rectangular coordinates

The function has the first order successive derivative on the interval, calculates the curve length.

If the integral variable is taken, the length of the curve arc segment corresponding to the small interval can be approximated by its arc differential.

So, the arc length element is

Arc length is

"Example 1" calculates the arc length of the curve.

Solution:

The case of parametric equation

If the curve is by parametric equation

Given that the arc length is calculated, it is only necessary to write the arc differential

The form, thus having

"Example 2" calculates the circumference length of the radius.

Solution: The parametric equation of the circle is

Three, polar coordinates case

If the curve is by polar coordinate equation

In order to derive its arc length formula, it is necessary to convert polar coordinate equation into parametric equation, and then use the ARC length calculation formula under parametric equation.

The parametric equation of the curve is

This becomes the parameter and the arc length element is

thereby having

"Example 3" calculates the arc length of the heart line.

Solution:

§6.5 work, water pressure and gravity

The work done by a variable force along a straight line

"Example 1" radius of the ball sank into the water, the ball's upper and the surface tangent, the proportion of the ball is 1, now the ball from the water out, how much work?

Solution: Establish a coordinate system as shown in the figure

The required force for the high ball to be taken out of the water is:

Which is the gravitational force of the ball, which means that after the ball is removed, the remaining part of the ball is still immersed in the water and the buoyancy is lacking.

By the lack of ball formula there

Thus

It is obvious that the gravity of the ball that is removed from the water is missing. That is, only gravity does work, and buoyancy does not work, and this is a variable force. The work done by taking the ball out of the water equals the work done by the variable force from the time it was changed.

Take the integral variable, then, for any one of the cells, the variable force from the distance to the work done.

This is the work element, and the work is

The other solution establishes the coordinate system as shown in the figure

Take the integral variable, then,

When taking up a cell, the cell corresponds to a small slice of the sphere, the volume of which is

Because the proportion of the ball is 1, the quality of the slice is about

The work done by taking the sheet out of the water should be equal to overcoming the work done by the flake gravity, and the distance must be shifted to remove the surface from the water.

So the work element is

Second, water pressure

The pressure at the depth is, here is the weight of the water.

If a flat plate is placed horizontally in the depths of the water, the water pressure on the slab side is

If the plates are not placed horizontally in the water, the pressure at different depths is not equal. At this point, the water pressure on the side of the plate must be calculated by using a definite integral.

"Example 2" is a rectangular thin plate with a long edge and is inclined to the surface of the water, and the long edge is parallel to the water. Set, the proportion of water, try to find the water pressure plate.

Solution: Because the thin plate and the surface of the water in the angle of oblique placed in the waters, it is located in the deepest water position is

Take the integral variable, then (note: Indicate depth of water)

The area of a small narrow bar on a thin plate corresponding to the cell between a cell in the

The water pressure it bears is about

So the pressure element is

The practical significance of this result is obvious

It is precisely the pressure that the sheet is placed horizontally in the depth of water;

Instead, the pressure produced by placing the sheet diagonally is equivalent to placing the sheet horizontally in depth as the water pressure on the premises.

Third, Gravity

It is known by physics that the gravitational force between two particles of mass is

For the gravitational coefficient. The direction of gravity is along the line of the two particles.

If we want to calculate the gravitational pull of a fine rod on a particle, because the distance between the points on the bar is different from that of the particle, and the gravitational direction of the point is also changing, it cannot be calculated simply by using the above formula.

"Example 3" is provided with a circular-arc fine rod with a radius, a center angle, a constant line density and a mass particle at the centre of the circle, trying to find the gravitational pull of the fine rod on the particle.

To solve such problems, in general, you should select an appropriate coordinate system.

Solution: To establish a coordinate system as shown in the figure, the particle is located at the origin of the coordinate, and the parameter equation of the arc is

To intercept a small section on an arc bar, its length is, its mass is, the distance from the origin, the angle is, its gravitational force on the particle is about

The approximate value of the force on the horizontal direction (i.e. the axis) is

and

So, we get the elements of a fine rod's force in the horizontal direction of the gravitational pull of the particle,

So

Similar to

Therefore, the gravity is the size, and the direction points to the center of the arc.

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