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Mathematics Dictionary
Dr. K. G. Shih
Derivative : Parametric equations and Polar functions
Subjects

  • CA 07 01 | - First derivative
  • CA 07 02 | - Second derivative
  • CA 07 03 | - Describe the curve of x = sec(t) and y = tan(t)
  • CA 07 04 | - Describe the curve of x = tan(t) and y = sec(t)
  • CA 07 05 | -
  • CA 07 06 | -
  • CA 07 07 | -
  • CA 07 08 | - Polar coordinates
  • CA 07 09 | - Spiral : R = e^(A)
  • CA 07 10 | - If R = cos(A), find slope of tangent at A = pi/4
    • Answers


    CA 07 01. First derivative with respect ot x

    Parametric equations
    • x = F(t)
    • y = G(t)
    first derivative
    • dy/dx = = (dy/dt)/(dx/dt)
    Example : x = cos(t) and y = sin(t), find dy/dx
    • dy/dx = (dy/dt)/(dx/dt)
    • dy/dx = cos(t)/(-sin(t))
    • dy/dx = -cot(t)
    • dy/dx = -x/y
    Example : x = cos(t) and y = sin(t), find equation of tangent at t = pi/4
    • When t = pi/4, x = Sqr(2)/2 and y = Sqr(2)/2
    • When t = pi/4, slope = s = dy/dx = -cot(t) = -1
    • Equation is y = s*x + b
    • Substitute values of x and y at t = pi/4, Sqr(2)/2 = (-1)*(Sqr(2)/2) + b
    • Hence b = Sqr(2)
    • Hence equation is y = -1*x + Sqr(2)

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    CA 07 02. Second derivative with respective to x

    Parametric equations
    • x = F(t)
    • y = G(t)
    Second derivative
    • First derivative is y' = dy/dx = = (dy/dt)/(dx/dt)
    • Second derivative is y" = dy'/dx = (dy'/dt)/(dx/dt)
    Example : x = cos(t) and y = sin(t), find y"
    • dy'/dx = (dy'/dt)/(dx/dt)
    • dy/dx = (d/dt(-cot(t))/(-sin(t))
    • dy/dx = (csc(x)^2)/(-sin(t))
    • dy/dx = -csc(t)^3 = -1/(y^3)
    Example : x = cos(t) and y = sin(t), find concavity of curve for t = 0 to pi/2
    • When t = 0 to t = pi/2, sin(t) is positive and then csc(t) is positive
    • Hence y" is positive.
    • Hence the curve is concave downward

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    CA 07 03. Describe the curve of x = sec(t) and y = tan(t)

    Diagram Progam
     First derivative
    • y' = dy/dx = (dy/dt)/(dx/dt)
    • y' = dy/dx = (sec(t)^2)/(sec(t)*tan(t))
    • y' = dy/dx = sec(t)/tan(t)
    Second derivative
    • y" = d/dx(dy/dx) = (d/dt(dy/dx))/(dx/dt)
    • y" = d/dx(dy/dx) = (d/dt(sec(t)/tan(t))/(sec(t)*tan(t))
    • y" = d/dx(dy/dx) = ((sec(t)*tan(t)*tan(t) - sec(t)*sec(t)^2)/tan(t)^2)/(sec(t)*tan(t)
    • y" = d/dx(dy/dx) = (Sec(t)*tan(t)^2 - sec(t)^3)/(sec(t)*tan(t)^3)
    • y" = d/dx(dy/dx) = (tan(t)^2 - sec(t)^2/(tan(t)^3)
    • Since tan(t)^2 + 1 = sec(t)^2
    • Hence y" = -1/tan(t)^3
    For t = 0 to t = 90
    • When t = 0 x = 1 and y = 0
    • When t = 90 x = infinite and y = infinite (1st quadrant)
    • Since x = (+) and y = (+) Hence the curve is in first quadrant
    • Since y' = +, the curve is increasing from (1,0)
    • Since y" = -, the curve is concave downward
    • Asymptote is y = x
    For t = 90 to t = 180
    • When t = 90 x = - infinite and y = - infinite (3rd quadrant)
    • When t = 180 x = -1 and y = 0
    • Since x = (-) and y = (-) Hence the curve is in 3rd quadrant
    • Since sec(t) = (-) and tan(t) = (-), hence y' = +, the curve is increasing to (-1,0)
    • Since tan(t) = (-), hence y" = (+), the curve is concave upward
    • Asymptote is y = x
    For t = 180 to t = 270
    • When t = 180 x = -1 and y = 0
    • when t = 270 x = infinite and y = infinite (2nd quadrant)
    • Since x = (-) and y = (+) Hence the curve is in 2nd quadrant
    • Since sec(t) = (-) and tan(t) = (+), Hence y' = (-), the curve is decreasing to (-1,0)
    • Since since tan(t) = (+), hence y" = (-), the curve is concave downward
    • Asymptote is y = -x
    For t = 270 to t = 360
    • When t = 270 x = +infinite and y = -infinite (4th quadrant)
    • when t = 360 x = 1 and y = 0 (2nd quadrant)
    • Since x = (+) and y = (-) Hence the curve is in 4thd quadrant
    • Since sec(t) = (+) and tan(t) = (-), Hence y' = (-), the curve is decreasing from (1,0)
    • Since since tan(t) = (-), hence y" = (+), the curve is concave upward
    • Asymptote is y = -x
    Diagram
    • See 07 01

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    CA 07 04. x = tan(t) and y = sec(t)

    Diagram Progam
     Describe the curve of x = tan(t) and y = sec(t) using y' and y"
    • Questions for various t values as below
      • Curve in which quadrant
      • Signs of y' and y"
      • Start point = ? End point = ?
      • Asymptotes = ?
      • See diagram 07 02
    • For t = 000 - 090 degree
    • For t = 090 - 180
    • For t = 180 - 270
    • For t = 270 - 360

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    CA 07 05.


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    CA 07 06. New

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    CA 07 07. Answer

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    CA 07 08. Polar coordinates
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    CA 07 09. Spiral : Graph of R = e^A

    Question
    • Tangent to the spiral at point P has slope = tan B
    • Direction angle A and angle B have constant difference at point P
    Proof
    • The slope = dy/dx = tan(B) = (dy/dA)/(dx/DA)
    • x = R*cos(A) = (e^A)*cos(A) and y = R*sin(A) = (e^A)*sin(A)
    • dy/dA = (e^A)*sin(A) + (e^A)*cos(A)
    • dx/dA = (e^A)*cos(A) - (e^A)*sin(A)
    • Hence dy/dx = (sin(A) + cos(A))/(cos(A) - sin(A))
    • Hence tan(B) = (sin(A) + cos(A))/(cos(A) - sin(A))
    • tan(A - B) = (tan(A) - tan(B))/(1 + tan(A)*tan(B))
    • tan(A) - tan(B)
      • tan(A) - Tan(B)
      • = tan(A) - (sin(A) + cos(A))/(cos(A) - sin(A))
      • = ((tan(A)*(cos(A) -sin(A)) - (sin(A) + cos(A)))/(cos(A) - sin(A))
      • = (sin(A) - tan(A)*sin(A) - sin(A) - cos(A))/(cos(A) - sin(A))
      • = (-cos(A) - tan(A)*sin(A))/(cos(A) - sin(A)
    • 1 + tan(A)*tan(B)
      • = 1 + tan(A)*(sin(A) + cos(A))/(cos(A) - sin(A))
      • = (cos(A) - sin(A)) + tan(A)*(cos(A) + sin(A))/(cos(A) - sin(A))
      • = (cos(A) - sin(A) + sin(A) + tan(A)*sin(A))/(cos(A) - sin(A))
      • = (cos(A) + tan(A)*sin(A))/(cos(A) - sin(A))
    • tan(A - B) = -1
    • Hence angle B - A = pi/4

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    CA 07 10. If R = cos(A), find slope of tangent at A = pi/8

    Solution
    • R = cos (A) is a cricle and center on polar axis OX with radius of 1/2
    • Let point P on circle and angle A = POX = pi/8
    • The tangent at P is perpendicular to CP and C is center (-1/2, 0)
    • Hence PCX = pi/4
    • Hence tangent makes angle with OX is 3*pi/4
    • Hence slope is -1
    Use derivative
    • dy/dx = tan(B) = (dy/DA)/(dx/dA)
    • x = R*cos(A) = cos(A)^2
    • y = R*sin(A) = cos(A)*sin(A)
    • dy/DA = -sin(A)*sin(A) + cos(A)*cos(A) = cos(2*A)
    • dy/DA = -2*cos(A)*sin(A) = -sin(2*A)
    • Hence tan(B) = -cos(2*A)/sin(2*A) = -cot(2*A)
    • Now A = pi/8
    • Hence Slope = tan(B) = -cos(pi/4) = -1

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    CA 07 00. Outline

    Derivative
    • 1st derivative : dy/dx = (dy/dt)/(dx/dt)
    • 2nd derivative : dy'/dx = (dy'/dt)/(dx/dt)
    Diagram Progam
     Describe the curve of x = sec(t) and y = tan(t) using y' and y"
    • For t = 0 - 90 degree
      • It is in 1st quadrant : x = (+) and y = (+)
      • y' = (+) and y" = (-)
      • Start point (1, 0) and end point x = + infinite and y = = + infinite
      • Asymptote : y = x and y = -x
      • See diagram 07 01
    • For t = 090 - 180
      • It is in 3rd quadrant : x = (-) and y = (-)
      • y' = (+) and y" = (+)
      • Start point x = - infinite and y = = - infinite. End point (-1, 0)
      • Asymptote : y = x and y = -x
      • See diagram 07 01
    • For t = 180 - 270
      • It is in 2nd quadrant : x = (-) and y = (+)
      • y' = (-) and y" = (-)
      • Start point x = - infinite and y = = + infinite. End point (-1, 0)
      • Asymptote : y = x and y = -x
      • See diagram 07 01
    • For t = 270 - 360
      • It is in 4th quadrant : x = (+) and y = (-)
      • y' = (-) and y" = (+)
      • Start point x = + infinite and y = = - infinite. End point (1,0)
      • Asymptote : y = x and y = -x
      • See diagram 07 01
    References
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