1 ch 11 parametric equations and polar coordinates 參數方程式與極座標

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1 CH 11 CH 11 PARAMETRIC EQUATIONS PARAMETRIC EQUATIONS AND POLAR COORDINATES AND POLAR COORDINATES 參參參參參參參參參 參參參參參參參參參

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Page 1: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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CH 11CH 11CH 11CH 11

PARAMETRIC EQUATIONS PARAMETRIC EQUATIONS AND POLAR COORDINATESAND POLAR COORDINATES參數方程式與極座標參數方程式與極座標

Page 2: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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學習內容• 11.2 11.2 Parametric Curves Parametric Curves • 11.3 Polar Coordinates11.3 Polar Coordinates

參數曲線極座標

Page 3: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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11.211.211.211.2

Parametric CurvesParametric Curves參數曲線

Page 4: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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學習重點

• 知道函數的參數式表示法• 會求參數式曲線的切線斜率

Page 5: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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函數的表示法

• 一般表示法– y = F(x)

• 參數表示法– x = f(t) and y = g(t)

• 參數表示法代入一般式– g(t) = F(f(t))

Page 6: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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導函數之參數表示法If g, F, and f are differentiable and g(t) = F(f(t)),

then the Chain Rule gives

g’(t) = F’(f(t))f’(t) = F’(x)f’(t)

'( )'( )

'( )

g tF x

f t

if f’(t) ≠ 0

if 0

dydy dxdt

dxdx dtdt

Page 7: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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第二階導函數之參數表示法

2

2

d dyd y d dy dt dx

dxdx dx dxdt

Page 8: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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•A curve C is defined by the parametric equations x = t2, y = t3 – 3t. Show that C has two tangents at the point (3, 0) and find their equations.

Example 1 (a)

At the point (3, 0) x = t2 = 3, y = t3 – 3t = 0

y = t3 – 3t = t(t2 – 3) = 0 when t = 0 or t = ± .3

故曲線在 (3, 0)這一點通過兩次

Page 9: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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2/ 3 3 3 1

/ 2 2

dy dy dt tt

dx dx dt t t

33

1-3

2

3

dx

dy

3t

3-3

1--3-

2

3

dx

dy

3-t

3-x3y

3-x3-y

x = t2, y = t3 – 3t if 0

dydy dxdt

dxdx dtdt

Page 10: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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•A curve C is defined by x = t2, y = t3 – 3t. Find the points on C where the tangent is horizontal or vertical.

Example 1 (b)

2/ 3 3 3 1

/ 2 2

dy dy dt tt

dx dx dt t t

horizontal tangent

t

1-t

2

3 2

0dx

dy t2 = 1 t = ±1 (1, -2) and (1, 2)

Page 11: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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vertical tangent

t

1-t

2

3

dx

dy 2

t = 0 (0, 0).

Page 12: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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•A curve C is defined by x = t2, y = t3 – 3t. Determine where the curve is concave upward or downward.

Example 1 (c)

22 2

2 3

3 11 3 122 4

d dytd y dt dx t

dxdx t tdt

–The curve is concave upward when t > 0.–It is concave downward when t < 0.

2

2

d dyd y d dy dt dx

dxdx dx dxdt

t

1-t

2

3 2

Page 13: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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•A curve C is defined by x = t2, y = t3 – 3t. Sketch the curve.

Example 1 (d)

Page 14: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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Find the tangent to the cycloid x = r(θ – sin θ), y = r(1 – cos θ ) at the point where θ = π/3.

Example 2 (a)

/ sin sin

/ 1 cos 1 cos

dy dy d r

dx dx d r

1

2

sin / 3 3 / 23

1 cos / 3 1

dy

dx

θ = π/3

Page 15: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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3sin

3 3 3 2

1 cos3 2

x r r

ry r

θ = π/3

33

2 3 2

or

3 23

r r ry x

x y r

Tangent line

Page 16: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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At what points is the tangent horizontal? When is it vertical?

Example 2 (b)

Horizontal tangent

cos-1

sin

dx

dy

dy/dx = 0 sinθ = 0 and 1 – cos θ ≠ 0 θ = (2n – 1)π, n an integer ((2n – 1)πr, 2r).

x = r(θ – sin θ), y = r(1 – cos θ )

Page 17: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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dy/dx = ∞ 1- cosθ = 0 θ = 2nπ, n an integer (2nπ, 0).

Vertical tangent

cos-1

sin

dx

dy x = r(θ – sin θ), y = r(1 – cos θ )

Page 18: 1 CH 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 參數方程式與極座標

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Q1 Find an equation of the tangent line to the curve x=t sin t, y=t cos t at t =11π.

(a) y = 12π + x/(11π)

(b) y = -11π + x/(11π)

(c) y = 11π + x/(11π)

(d) y = -12π + x/(11π)