The necessary and sufficient condition for the curve of equation x ^ 2 + y ^ 2-3x-2y + k = 0 to pass through the origin is K=

This is just substituting x = 0 and y = 0 into the equation, so k = 0

A necessary and sufficient condition for the circle (x-a) 2 + (y-b) 2 = R2 to pass through the origin is______ ．

Circle (x-a) 2 + (y-b) 2 = R2 passes through the origin, which is equivalent to the coordinates of the origin. It is suitable for circle (x-a) 2 + (y-b) 2 = R2. That is, A2 + B2 = R2, so the answer is A2 + B2 = R2

Under what condition does the curve of equation (x-a) Λ2 + (y-b) Λ2 = R Λ 2 pass through the origin?

a²+b²=r²

A necessary and sufficient condition for the curve of equation y = ax & sup2 + BX + C to pass through the origin

C = 0 or C = a = O

Why is x still equal to a (1-cost) and y = cot (T / 2) when seeking the second derivative

'x is still equal to a (1-cost) and y = cot (T / 2)'
It's impossible. You must be wrong
X = a (1-cost) y = a (t-sint) should be y = a (1-cost), x = a (t-sint) this is a cycloid equation. Your X, y are reversed

Given the parameter equation x = cost, y = cost, how much is d ^ 2Y of DX ^ 2

A:
x=cost,y=cost
So: x = y
So: y '= dy / DX = 1
So: y '' = 0
That is: D & # 178; Y / DX & # 178; = 0

X = Sint, y = TSINT + cost, find d ^ 2Y / DX ^ 2It = π / 4
dx/dt=cost
Dy / dt = Sint + tcost Sint = tcost (the question is this! Why not dy / dt = tcost Sint? How did the front Sint come from
y'=dy/dx=(dy/dt)/(dx/dt)=t
y"=d^2y/dx^2=d(y')/dx=d(y')/dt/(dx/dt)=1/cost
When t = π / 4, Y "= 1 / cos (π / 4) = √ 2

Because TSINT is a compound function, we need to use the multiplication derivation rule to find the derivative. First, we use the derivative of t to multiply Sint, and then add the derivative of Sint to multiply t

∫ y DS, where l is the curve integral of cycloid arch x = a (t-sint) y = a (1-cost)
32a^2 / 3

t：0→2π
ds=√[(dx/dt)²+(dy/dt)²] dt=√[a²(1-cost)²+a²sin²t] dt=a√(2-2cost)dt=a√[4sin²(t/2)]dt=2asin(t/2)dt
∫ y ds
=∫[0→2π] 2a²(1-cost)sin(t/2) dt
=4a²∫[0→2π] sin³(t/2) dt
=8a²∫[0→2π] sin³(t/2) d(t/2)
=-8a²∫[0→2π] sin²(t/2) d[cos(t/2)]
=-8a²∫[0→2π] [1-cos²(t/2)] d[cos(t/2)]
=-8a²[cos(t/2) - (1/3)cos³(t/2)] |[0→2π]
=-8a²(-1 - 1/3 - 1 - 1/3)
=32a²/3
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Y = (1 + Sint) / cost, find dy

（1+sint）/（cost)^2*dt

The necessary and sufficient condition for the curve of equation x ^ 2 + y ^ 2-3x-2y + k = 0 to pass through the origin is K=