A high 3 math problem is solved by a master 5

Over P as PR vertical AC1From the meaning of the title, it can be seen that PR is the distance from P to the plane AA1CC1, and the distance from P to CC1 is equal to RC1

Since the distance from p to the surface a1b1c1d1 is 4 3 of the distance to 0, the distance that can be converted to p straight line ac1 is 4 3 of the distance to o. Therefore, it can be seen that the trajectory of point p is a branch of the hyperbola and the eccentricity of the hyperbola is e=4 3, and the midpoint of AC1 is O1, that is, the linear line oo1 can be the x-axis, O1C1 is the y-axis to establish a coordinate system, and O(7,0) can be obtained, so the hyperbola of C=7, A=C, E=21, 4, B=A-C=343 16So the equation for the trajectory of the point p is:

16x²/441-16y²/343=1（x>0).The maximum value is required, either p is on the straight line aa1, which is equal to 14 times the root number 2 (generally impossible, the verification is to substitute y=7 times the root number 2 into the equation to find x, if x is in 0 to 14, it is vice versa), and on ac, then substitute x=14 to find y=? , then the maximum value of p to cc1 is equal to the absolute value of y plus 7 times the root number 2.

Will the solid geometry vector method be, if it will, it's very simple.

Establish a spatial Cartesian coordinate system and use point and line equations to solve it.

Let the equation be by+cz+d=0, and the coordinates of the known points are substituted to obtain.

b+2c+d=0, b+d=0, add to get 2c+2d=0, take c=1, then d = 1, b=1, so, the plane equation is y+z-1=0.

The first question is Gaussian formula! Transformed into a triple integral. When the regional function of the triple integral is estimated to be greater than or equal to zero of the integrand, the triple integral is maximized.

The second question is simple, direct Gaussian formula.

The most important thing is to know that if you want the triple integral to be maximum, you need the region function to coincide with the part of the integrand function that is greater than or equal to zero. Remember the gradient? In fact, in the case of such a vector field, the reintegration is maximum. After all, double integration can be expressed in terms of flow.

I'm afraid that if you talk too much, you won't want to watch it and you will get dizzy.

Here's the answer:

Transform the original form into the following form and make it.

Isolate the variables.

Both sides are scored at the same time.

This differential equation, which can be observed, should be seen that it is not a differential equation of separable variables, nor is it a first-order linear differential equation, it belongs to the homogeneous equation. The form of the homogeneous equation is: y'＝f(y/x)。

That is, one side of the equation is the derivative, and the other side of the equation is.

y x, this type of differential equation is a homogeneous equation. We have a fixed solution to the homogeneous equation, that is.

Let y x u, and turn the original equation into an equation containing u and x. The detailed process can be referred to the figure below.

The detailed steps are shown in the following figure:

There are 2 intersection points, the method is substitution, the formula (sina) 2+(cosa) 2=1 is used, and the specific process is shown in the following figure

Substituting the parametric equation of the curve l into the surface S

That is, 9*(2cost) 2+4*(3sint) 2+36*t 2=72

That's 36((cost) 2+(sint) 2)+36t 2=7236+36t 2=72

t2=1t can take plus or minus 1

Bring in, t=1 is x=2cos1, y=3sin1, z=1, t=-1 is x=2cos-1=2cos1, y=3sin-1=-3sin1, z=1,

What do you want to ask. Question: Find the surface equation of the rotation of the straight line (x-3) 2=(y-1) 3=z+1 around the fixed line {x=2 y=3.

Answer: Take a point b(t+1,-3t,3t,3t) on the bus x-1=y-3=z=t, and take a certain point a(2,1,-2) on x 2=y=z-2 to find the spherical equation with a as the center of the circle ab as the radiusThen find the plane that over-changes and is perpendicular to x 2=y=z -2Then the simultaneous plane equation and the spherical equation minus the parameter t are the final answer.

Question: What does the busbar mean here? The rotation of a straight line around another straight line results in a cone, why is a spherical surface used?

Today's hardships are all paving the way for a certain day.

The difference between any two solutions of a linear nonhomogeneous equation is the solution corresponding to the homogeneous equation.

Therefore y1 y2 is y'The solution of +py=0 is y=c (y1 y2) because it is of the first order.

The general solution of the non-homogeneous equation is equal to the general solution of the corresponding homogeneous equation plus a special solution of the non-homogeneous equation.

Therefore, the general solution of the original equation is y=y+y1, or y=y+y2. Pick D