F (T) = asin (ω T + φ), ω > 0, f (π / 3 + T) = f (π / 3-T), G (T) = ACOS (ω T + φ), what is g (π / 3)?

F (T) = asin (ω T + φ), ω > 0, f (π / 3 + T) = f (π / 3-T), G (T) = ACOS (ω T + φ), what is g (π / 3)?

f(π/3+t)=f(π/3-t)
So x = π / 3 is the axis of symmetry of the function
The axis of symmetry of sin is where the function takes its maximum value
That is, sin is equal to 1 or - 1
That is sin (ω * π / 3 + φ) = 1 or - 1
Then cos (ω * π / 3 + φ) = 0
So g (π / 3) = ACOS (ω * π / 3 + φ) = 0

How to express x (T) in the form ACOS (T - α) and Y (T) = bsin (T - β), etc

x(t)=cos(t)+asin(t) =√（1+a^2）cos(t-α),
Where cos α = 1 / √ (1 + A ^ 2), sin α = A / √ (1 + A ^ 2)
Similarly, y (T) = sin (T) + bcos (T) = √ (1 + B ^ 2) sin (T - β),
Where cos β = 1 / √ (1 + B ^ 2), sin β = - B / √ (1 + B ^ 2)

|X|

f(x)=cos^2(x)-acos(x)
=(cosx-a/2)^2-a^2/4
|x|≤∏/4
√2

Find the curvature of the curve X = a (COS (T)) ^ 2, y = a (sin (T)) ^ 2 at t = t0
The absolute value of k = 2 / (3a (sin (2t0)))
I look like the title is wrong

Obviously X / A + Y / b = 1 is a straight line
The curvature is zero
Is the condition wrong

What is the derivative of y = LNX?

One in X

Image drawing method of finding function y = 1 / X (x is less than 0)

One branch of the inverse proportion function falls in the third quadrant. Take the point method, take (- 1, - 1) (- 2, - 1 / 2) (- 1 / 2, - 2) (- 3, - 1 / 3) (- 1 / 3, - 3), and then connect and extend it with a smooth curve. Remember that it does not intersect with the coordinate axis but is infinitely close to it

The monotone increasing interval of the function y = 1 / 2x ^ 2-lnx is? How is this image drawn?

The domain is a positive number
Let's do it with derivative, y '= X-1 / X
Let y '= 0 and solve x = 1
So (0,1) increases, (1, + infinity) decreases
According to this, I can draw a sketch

Polar curvature

[hint] the calculation formula of curve curvature in rectangular coordinate system k = |y '' | / (1 + y '^ 2) ^ (3 / 2) (*)
The equation of the curve is x = R (T) cost, y = R (T) Sint
y'=dy/dx=(r'sint+rcost)/(r'cost-rsint)
y''=dy'/dx=…… =(r^2+2r'^2-rr'')/(r'cost-rsint)^2.
Just substitute (*)

Derivation of gradient formula in polar coordinates
I want to know why there is one more R in front of the partial derivation of the angle, and I want to know the derivation process of the whole formula

Using the coordinate transformation formula to directly transform the gradient formula of rectangular coordinate system into the product coordinate system is such a form. The essential explanation of this kind of similar coefficient before this angle or other variables is that each space (different coordinate systems) has its own metric, and the metric of three-dimensional rectangular coordinate system or Cartesian space is a 3 × 3 unit matrix, The value on the diagonal corresponds to the coefficient before each variable in the gradient

Derivation of distance formula between two points in polar coordinates?
It's best to draw a picture

Let P1 (ρ 1, θ 1) P2 (ρ 2, θ 2) Δ op1p2 be defined by the cosine theorem | OP1 | ^ 2 + | op2 | ^ 2-2 | * | op2 | * cos (θ 1 - θ 2) = | p1p2 | ^ 2 (ρ 1) ^ 2 + (ρ 2) ^ 2-2 ρ 1 ρ 2cos (θ 1 - θ 2) = | p1p2 | ^ 2 | = √ [(ρ 1) ^ 2 + (ρ 2) ^ 2 ρ 1 ρ 2cos (θ 1 - θ 2)]