Let a (2,0) be a certain point on the plane, and the graph of the moving point P (sin (2t-60 °), cos (2t-60 °) be c Let a (2,0) be a certain point on the plane, and the graph of moving point P (sin (2t-60 °), cos (2t-60 °) be c. when t changes from 30 ° to 45 °, the area of line AP sweeping graph C is calculated

Let a (2,0) be a certain point on the plane, and the graph of the moving point P (sin (2t-60 °), cos (2t-60 °) be c Let a (2,0) be a certain point on the plane, and the graph of moving point P (sin (2t-60 °), cos (2t-60 °) be c. when t changes from 30 ° to 45 °, the area of line AP sweeping graph C is calculated

Draw a picture. The trajectory of P point is the unit circle, t from 30 '~ 45', P point from (0,1) to (1 / 2, 3 / 2 under the root sign). Use double integral

Let (2,0) be a certain point on the plane, and the graph of the moving point P (sin (2t - π / 3), cos (2t - π / 3)) be c
When π / 3 becomes π / 4, the area of figure C swept by the line AP

When t = π / 3, P (sin π / 3, cos π / 3), namely P1 (√ 3 / 2,1 / 2)
When t = π / 4, P (sin π / 6, cos π / 6), namely P2 (1 / 2, √ 3 / 2)
When t changes from π / 3 to π / 4, the sector area is π * (2 * π / 3-2 * π / 4) / 2 π = π / 12
The area of op1a is S1 = 1 / 2 * 2 * 1 / 2 = 1 / 2
The area of op2a is S2 = 1 / 2 * 2 * √ 3 / 2 = √ 3 / 2
So the area of graph C scanned by AP is s = √ 3 / 2-1 / 2 - π / 12

Sin image and COS image
In mathematics, there is the problem of moving sin or cos image. How to enlarge the x-coordinate first and how to enlarge the x-coordinate first
And sin (x + π / 4) first shift π / 3, then enlarge X by two times

If the abscissa is expanded to 2 times, then the coefficient of X is multiplied by 1 / 2Sin (x / 2 + π / 4). Suppose that the abscissa is shifted to the left by π / 3, then x is changed into x + π / 3sin [(x + π / 3) / 2 + π / 4] = sin (x / 2 + 5 π / 12). If the abscissa is shifted to the right, then x is changed into X - π / 3. If the abscissa is shifted to the left by π / 3, then x is changed into x + π / 3 = sin [(x + π / 3) + / 4] = sin (x + 7 π / 12)

Let a (2.0) be a certain point on the plane, and the moving point P (sin (2t-60 °), cos (2t-60 °). When t changes from 20 to 40, the curve track of P moving clockwise from P1 to P2 and the graph area of AP1 and ap2 are ()

Let a (2.0) be a certain point on the plane, and the moving point P (sin (2t-60 °), cos (2t-60 °). When t changes from 20 to 40, the curve track of P moving clockwise from P1 to P2 and the graph area of AP1 and ap2 are the same

What is the formula of polar coordinates for area, D θ?
The mathematics foundation is not good, the teacher speaks time does not listen carefully
The book says, if theta increases D theta, then Da = 1 / 2 * R ^ 2 * D theta
Why D θ? What is d θ?

D θ is the increment of polar angle θ in polar coordinates
The area s is approximately equal to the area of the sector = 1 / 2 * R ^ 2D θ (where R is the polar longitude and D θ is the center angle of the circle)
Please take your time

Why does 1 / 2 appear in polar integral area formula

S＝（1／2）θR².
1 / 2 is available. This is the polar coordinate area formula and the sector area formula

How to express 2x-3y-1 = 0, x2-y2 = 16 as polar coordinate equation, P = 2sinb as general coordinate equation
emergency

Because x = ρ cos θ, y = ρ sin θ
Substituting 2x-3y-1 = 0, x ^ 2-y ^ 2 = 16
The results show that 2 ρ cos θ - 3 ρ sin θ - 1 = 0, (ρ cos θ) ^ 2 - (ρ sin θ) ^ 2 = ρ ^ 2 * cos 2 θ = 16
Because y = ρ sin θ, x = ρ cos θ, so x ^ 2 + y ^ 2 = ρ ^ 2, sin θ = Y / ρ
ρ = 2Sin θ, both sides multiply by ρ at the same time, that is, ρ ^ 2 = 2, ρ sin θ = 2Y
By substituting x ^ 2 + y ^ 2 = ρ ^ 2 into the above formula, we can get that ρ = 2Sin θ is reduced to the general coordinate equation 2Y = x ^ 2 + y ^ 2, and x ^ 2 + (Y-1) ^ 2 = 1

On polar equation
The ray from pole o intersects the straight line ρ cos = 4 at point M. take a point P on om such that om &; op = 12, and find the trajectory equation of point P
ρ=3cosθ

Let the intersection of ρ cos θ = 4 and the polar axis be point a
OAM is a right triangle with OA = 4
At point P, make Pb perpendicular to the intersection axis of OM and at point B
So triangle OPB is similar to OAM
So ob / op = OM / OA
therefore OB.OA=12 So ob = 3
So OP = 3cos θ
So the trajectory of P is ρ = 3cos θ

How to solve the polar equation of a circle?

Then, C (a, t) = C (ACOS (T), asin (T)). Therefore, the equation of circle is: (ρ cos (θ) - ACOS (T)) + (ρ sin (θ) - asin (T)) = R

What is the polar equation

The curve equation described by polar coordinate system is called polar coordinate equation, which is usually expressed as the function of ρ as the independent variable