A calculus problem,,Please help me,,

Note r0=2i+2j+k

r(t)-r0|^2=(cost/sqrt2+sint/sqrt3)^2+(-cost/sqrt2+sint/sqrt3)^2+(sint/sqrt3)^2

cost^2+2sint^2/3+sint^2/3=1

Firstly, it is shown that the trajectory of the particle is on a spherical surface centered on r0 and radius 1.

r(t)-r0)·(i+j-2k)=[(cost/sqrt2+sint/sqrt3)i+(-cost/sqrt2+sint/sqrt)j+(sint/sqrt3)k]·(i+j-2k)

cost/sqrt2+sint/sqrt3)+(cost/sqrt2+sint/sqrt)-2sint/sqrt3=0

r(t)-r0) is perpendicular to the vector i+j-2k, and the motion is in a plane.

The motion of the particle is on the spherical surface, and the center of the sphere r0=(2,2,1) is on the plane x+y-2z=2 (its normal vector satisfies the above conditions), so this plane is the over-centric section of the sphere, the normal vector (1,1,-2).

The motion of the particle is a circular motion centered on (2,2,1) with a radius of 1, and the circumference is on the plane x+y-2z=2.

Hello, the ** you uploaded just gives the motion trajectory equation of an object in three-dimensional space to do a unit circular motion, and explains that this trajectory is centered on (2,2,1), and the circumference is in the plane of x+y-2z=2. But it doesn't seem to give a specific question to calculate?

Obviously an infinitive of 0 0.

Lobida's theorem.

x 0, original.

e^(x²)-1]/3x²

x²/3x²

b.The weakest condition is that it cannot be launched continuously, differentially

The answer is that f(x,y) is continuous at (x0,y0), the partial derivative cannot be continuous, and the derivability of the binary function has nothing to do with continuity. c can not be launched, obviously, according to the sufficient conditions of the differentiation. d is correct because first f(x,y) is derivative, and secondly(x0,y0) is stationary, so it is an extreme point.

The projection of the surface in the x,y plane is 1 4 circles x 2 + y 2 = 1; x>=0,y>=0

Integrate with polar coordinates.

i=∫∫x^2+y^2)dxdy=∫∫r^2*rdrdθ

The double integral needs to specify a region d, which is called the double integral that the function f(x,y) does on the closed region d. Of course, your area here is a rectangular region where x is at [0,1] and y is at [0,1].

I estimate that your f(x,y) should be a probability value masking that changes with x,y (or at different points (x,y)). This double integral is to find the total probability value in the region d. Nuclear this.

See ** for the process.

This is a double integral, f(x,y)d(x)d(y).

Or first integrate x (at this time y is a constant) to obtain the value range, in the integral y, [f(x)d(x)]d(y) is the answer to the double integral in the first muffled acre integral. Hood culture.

I can't type the formula in the process.

The answer is 2

Lobida's law up and down derivation, the following is x, and the above is what you ask for, and then substitute 1, the answer is known!

With the partial integration method, it's very simple. The result is 1 9 [in(x)chengshangdejucifang-dejucifang 9].