Should the radius of curvature of a cardioid be transformed into a parametric equation in Cartesian coordinates, or can we just directly find the derivative of R with respect to θ?

It's going to be Cartesian

In Cartesian coordinate system, the parameter equation of circle C is x = 2cos θ y = 2 + 2Sin θ (θ is a parameter), then the ordinary equation of circle C is The polar coordinate system is established with the origin o as the pole and the positive half axis of X axis as the polar axis, then the polar coordinate of the center of circle C is ．

(1) ∵ curve C: x = 2cos θ y = 2 + 2Sin θ (θ is a parameter), ∵ 2cos θ = x, 2Sin θ = Y-2, the sum of the squares of the two formulas is: x2 + (Y-2) 2 = 4, that is, the curve C is transformed into an ordinary equation. (2) by using ρ cos θ = x, ρ sin θ = y, ρ 2 = x2 + Y2, the substitution is: ρ 2-4, ρ sin θ = 0, that is

The cycloid parameter equation is explained in detail thank

Cycloid is one of the most attractive curves in mathematics. It is defined as follows: if a circle rolls slowly along a straight line, the track of a fixed point on the circle is called cycloid
x＝a（φ－sinφ）,y＝a（1－cosφ）
Let the initial coordinates of the point be (0,0) and the center coordinates be (0, a)
When the circle rotates, the coordinates of the center of the circle are (a, φ, a)
The coordinates of the point relative to the center of the circle are (- asin φ, - ACOS φ)
So the coordinates of this point are (a (φ - sin φ), a (1-cos φ))
That is, x = a (φ - sin φ), y = a (1-cos φ)
I'll give you a parameter equation of hypocycloid
Hypocycloid
When a moving circle rolls along a fixed line without sliding, it is the locus of a certain point outside or inside the moving circle. As shown in the figure, the Cartesian coordinate system is established. If the radius of the moving circle is a, and the distance from the center of the circle to the outer (inner) point m is B, then the parameter equation of the hypocycloid is x = a φ - bsin φ, y = a-bcos φ. When b > A, it is a long amplitude cycloid, when B < A, it is a short amplitude cycloid, and when B = a, it is a cycloid

SW how to make equation line, parameter equation
y=sin(x)

What about y ^ 2 = 2 * a * x + b * x ^ 2 (a model of aspheric lens)

In the V-T image, if the area surrounded by the graph line and the time axis is both below and above, which is the displacement and which is the distance

If the area above the time axis is positive and the area below is negative, the sum is the displacement. If both are positive, the sum is the distance

How to estimate the displacement of uniformly accelerated image on V-T diagram

For example, for the displacement from T1 to T2, make the vertical line of X axis through T1 and T2 respectively. The area enclosed by the two vertical lines and the image and the axis is the displacement. If there is an image under the X axis, subtract the area under the X axis from the area on the X axis. Note that the displacement is a loss, and the positive and negative directions are expressed

The displacement in uniform linear motion can be expressed by the rectangular area between the velocity graph line and the time axis, because s = vt. but why can the displacement in uniform linear motion be expressed by the area between the velocity graph line and the time axis?

In fact, no matter what motion, its displacement can be expressed by the area between the velocity graph line and the time axis. The displacement is equal to the integral of VT between two moments. The displacement in uniform linear motion s = VT is a special form of integration

If the area between the image of the straight line y = x + B and the figure enclosed by the two coordinate axes is equal to 4, then B = (urgent)

2 root 2

Find the area of the figure enclosed by the coordinate axis of the line y = 2x + 6

When x = 0
y=6
The point of intersection with y axis is (0,6)
When y = 0
0=2x+6
∴x=-3
The point of intersection with x-axis is (- 3,0)
∴S=6×3/2=9

First draw the image of the line y = - 2x + 5, and then calculate the area of the triangle formed by the line and the two coordinate axes?
I don't need the figure. How can I find the area?

The intersection point is: x = 0 = > y = 5
y=0=》x=5/2
That is, the length of the line segment intersecting the two coordinate axes is 5 and 5 / 2
So the area is: 1 / 2 * 5 * 5 / 2 = 25 / 4

Should the radius of curvature of a cardioid be transformed into a parametric equation in Cartesian coordinates, or can we just directly find the derivative of R with respect to θ?