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Symbol Defintion
Example : x^2 is square of x
CA 09 01. The length of arc
Definition
- Rectangular coordinate :
- y = F(x), find arc length from A(x1,y1) to B(x2,y2)
- Distance between A(x1,y1) and B(x2,y2) = Sqr((x2-x1)^2 + (y2-y1)^2)
- It is Lim[Sqr((dx)^2 + (dy)^2))] as B approach A
- It is dL = Lim[Sqr(1 + (dy/dx)^2))]dx as B approach A
- L = ∫Sqr(1 + (y')^2)dx
- Polar coordinate
- Arc length = r*A where A is angle
- dL = r*dA = arc length of angle dA
- L = ∫∫rDA
- Parametric equation
- x = F(t) and y = G(t)
- dy/dx = (dy/dt)/(dx/dt)
- L = ∫Sqr(1 + (y')^2)dx
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Study Subjects | Analytic geometry
Analytic geometry
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CA 09 02. Circumference of circle
Prove that the circumference of circle is 2*pi*r using rectangular coordinates
- Circle : x^2 + y^2 = r^2
- y = Sqr(r^2 - x^2) is semi-circle
- dy/dx = -x/y
- dL = Sqr((1 + (dy/dx)^2))dx
- dL = Sqr((1 + (-x/y)^2))dx
- dL = Sqr((x^2 + y^2)/y^2))dx
- dL = (r/y)dx
- dL = r*(1/Sqr(r - x^2)dx
- L = ∫r*(1/Sqr(1-x^2))dx for x from 0 to r
- Let x = cos(A) and dx = -sin(A)dA
- If x = 0, 0 = cos(A) and A = pi/2
- If x = 1, 1 = cos(A) and A = 0
- L = ∫r*(-1)dA = r*pi/2 if A = pi/2 to A = 0
- Hence circumstance = 4*(r*pi/2) = 2*pr
- C = ∫ rDA for angle A from 0 to 2*pi
- C = r*A = r*(2*pi - 0)
- C = 2*pi*r
- Equation of circle : x = r*cos(A) and y = r*sin(A)
- dy/dx = (dy/dt)/(dx/dt) = cos(t)/(-sin(t) = -cot(t)
- dL = Sqr(1 + (dy/dx)^2)dx
- dL = Sqr(1 + cot(t)^2)dx
- dL = Sqr(csc(x)^2)dx and dx = -(r*sin(t))dt
- dL = r*(csc(x))*(-r*sin(t))dt
- dL = r*dt
- L = ∫r*dt = 2*pi*r if t = 0 to t = 2*pi
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CA 09 03. Aera of circle is A = pi*(r^2)
Method 1 : Polar coordinate
- dA = r*dL = dr*(r*dU)
- A = ∫r*dr*du and U from 0 to 2*pi
- A = (r^2)*U/2
- A = (r^2)*(2*pi - 0)/2
- A = pi*r^2
- Equation of circle : x^2 + y^2 = r^2
- dA = y*dx
- dA = Sqr(r^2 - x^2)*dx
- Hence A = ∫Sqr(r^2 - x^2)*dx and x form -r to r
- Let x = r*cos(U) and U from pi to 0
- dx = -r*sin(u)*dU
- Hence A =∫r*Sqr(1 - cos(U)^2)*r*sin(U)*dU
- A =∫-(r^2)*Sqr(1 - cos(U)^2)*sin(U)*dU
- A =∫-(r^2)*(sin(U)^2)*dU
- A =∫-(r^2)*(1/2 - cos(2*U))*dU
- A =-(r^2)*(1/2)*U for U from pi to 0
- A = 0 - (-1)*(r^2)*(1/2)*(pi) This is for semi-circle
- Hence area = pi*(r^2)
- ∫ (sin(x)^2)*dx
- ∫ (1/2 - cos(2*x))*dx
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CA 09 04. Area of ellipse = pi*a*b
Method 2 : Rectangular method
- Equation of ellipse : (x^2)/(a^2) + (y^2)/(b^2) = 1
- dA = y*dx
- dA = Sqr(b^2 - (b^2)*(x^2)/(a^2))*dx
- Hence A = ∫Sqr(b^2 - (b^2)*(x^2)/(a^2))*dx
- Where x from -a to a
- Let x = a*cos(U)
- If x = -a, then -a = a*cos(U) and U = pi
- If x = +a, theb +a = a*cos(U) and U = 0
- dx = -a*sin(u)*dU
- Hence A =∫b*Sqr(1 - cos(U)^2)*(-a*sin(U))*dU
- A =∫-(a*b)*Sqr(1 - cos(U)^2)*sin(U)*dU
- A =∫-(a*b)*(sin(U)^2)*dU
- A =∫-(a*b)*(1/2 - cos(2*U))*dU
- A =-(a*b)*(1/2)*U for U from pi to 0
- A = 0 - (-a*b)*(1/2)*(pi) This is for semi-circle
- Hence area = pi*(a8b)
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CA 09 05. Area enclosed by y = sin(x) and x-axis for x = 0 to x = pi
- A = ∫sin(x)*dx
- A = -cos(x) for x = 0 to x = pi
- A = -cos(pi) - (-cos(0))
- A = -(-1) + 1
- A = 2
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CA 09 06. Volume of sphere
Diagram in polar form
- OF = r and GF = r*cos(U)
- Angle EOF = U
- Angle ZOE = Angle HGF = V
- dU is increment along angle U
- dV is increment along angle V
- dr is increment along radius r
Polar form
- Arc length of angle dU is r*dU
- Arc length of angle dV is r*cos(U)*dv
- dA = (dr)*(r*dU)*(r*cos(U)*dV)
- Hence A = 8*∫r^2*cos(U)*dU*dV*dr
- Hence A = 8*((r^3)/3)*(-sin(U)*V for U = 0 to pi/2 and V = 0 to pi/2
- Hence A = 4*pi*(r^3)/3
- Rotate y = Sqr(r^2 - x^2) about axis
- dA = pi*(y^2)*dx
- Hence A = ∫ pi*(r^2 - x^2)*dx for x = -r to r
- Hence A = pi*(x - x^3)/3
- Hence A = pi*(2*r^3 - 2*(r^3)/3)
- Hence A = 4*pi*(r^3)/3
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CA 09 07. Answer
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CA 09 08. Answer
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CA 09 09. Answer
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CA 09 10. Answer
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