As shown in the figure below, point a on the circumference moves in a circular motion at a constant speed in a counter clockwise direction. It is known that point a turns θ (0

As shown in the figure below, point a on the circumference moves in a circular motion at a constant speed in a counter clockwise direction. It is known that point a turns θ (0

14. Point a turns 2 θ in 2 minutes, and π < 2 θ < π
After 14 minutes, it returned to its original position, ∧ 14 θ = 2K π,
Theta =, and ＜ theta ＜ π,
Ψ θ = π or π

A point m on the positive half axis of the x-axis moves in a circular motion at a constant speed in a counter clockwise direction. If point m turns a in one minute, it is called (0

3π/4>a>π/2
14a=2kπ
k=(1,2,3,4.)
So I need your answer

As shown in the figure, point a on the circle moves in a circular motion at a constant speed in a counter clockwise direction, and the known point a turns the angle of θ (0

Let's go back to the origin after n turns
So 14 θ = 2pai * n
Pai< 2θ

An indefinite integral: ∫ √ (1 + cost ^ 2) DT
Cost ^ 2 is the square of the cosine,

Is cost ^ 2 the cosine of square t or the square of cosine t?
Well, it's like ∫ √ (1-ksint ^ 2) DT (0

How to find the indefinite integral of (1 + tant) under the root of 1 / (cost's Square) *
Urgent, to answer the process in detail!

∫1/[(cos²t)√(1+tant)]dt=∫1/√(1+tant)dtant
Do you know how to ask
This is a simple differential 1 / [(COS & # 178; t) DT = dtant

How to calculate ∫ 1 / Sint ^ 2 DT

Negative cot

Finding (1 + cost) DT under the definite integral ∫ (- π / 2,0) cost / root sign

∫(-π/2,0) cost/√(1+cost)dt
=∫(-π/2,0)(2cos²2t-1)/(√2cos2t)dt
=∫(-π/2,0)(2cos²2t-1)/(√2cos2t)dt
=∫(-π/2,0)(2cos²2t-1)/(√2cos2t)dt
2t=x
t=-π/2 x=-π
=1/2∫(-π,0)(2cos²x-1)/(√2cosx)dx
=√2/2∫(-π,0)cosxdx-1/2∫(-π,0)1/(√2cosx)dx
The integral of cosx in (- π, 0) interval is 0
=0

How to find the arc length: for the curve X = e ^ t * Sint, y = e ^ t * cost, what is the length of an arc from t = 0 to t = 1
Just give me the final result of this question. I'll check it with myself

DX = (e ^ t * Sint + e ^ t * cost) dtdy = (e ^ t * cost-e ^ t * Sint) DT (DX) ^ 2 + (Dy) ^ 2 = 2 ((e ^ t * Sint) ^ 2 + (e ^ t * cost) ^ 2) (DT) ^ 2 arc length is sqrt (2 ((e ^ t * Sint) ^ 2 + (e ^ t * cost) ^ 2)) integral result of T from 0-1 is sqrt (2) * (E-1)

Find the maximum value of function f (T) = (1 + Sint) / (2 + cost)

Find the maximum value of function f (T) = (1 + Sint) / (2 + cost)
Let k = (1 + Sint) / (2 + cost), then K can be regarded as the slope of a line passing through points a (cost, Sint) and B (- 2, - 1) in the coordinate plane xoy
Because a moves on the circle x ^ 2 + y ^ 2 = 1, it can be seen that when the line Ba is the tangent of the circle, the slope k reaches the extreme value
Let the tangent equation passing through point B be y + 1 = K (x + 2) y = KX + 2k-1
｜2k-1｜=√(1+k^2) .(2k-1)^2=1+k^2.
The solution is: K1 = 0, K2 = 4 / 3
So the minimum value of F (T) is 0 and the maximum value of F (T) is 4 / 3
Let Tan (A / 2) = x, then Sina = 2x / (1 + x ^), cosa = (1-x ^ 2) / (1 + x ^ 2)
By substituting f (a), we get: let y = f (a)
y=(x^2+2x+1)/(x^2+3)
[y-1]x^2+2x+3y-1=0
Because f (a) is a real number
4-4(y-1)*(3y-1)≥0
3y^2-4y≤0
The solution of this inequality is: 0 ≤ y ≤ 4 / 3
So the maximum value of F (a) = (sina-1) / (cosa-2) is 4 / 3, and the minimum value is 0

The second problem of solving the convolution of the sin t * cost function is to use the convolution theorem to find the inverse Laplace transform f (s) = s ^ 2 / (s)^
Convolution of sin t * cost function
The second problem uses convolution theorem to find the inverse Laplace transform
F (s) = s ^ 2 / (s ^ 2 + 1) ^ 2 and f (s) = S + 2 / (s ^ 2 + 4S + 5) ^ 2