首頁 自修課程 (大學) 微積分甲 (一)
  • Curves Defined by Parametric Equations
  • Calculus with Parametric Curves
  • Polar Coordinates

# Curves Defined by Parametric Equations

# Define

Parmetric equations are equation such that xx and yy are both funtions of
a third variable tt: x=f(t),y=g(t),α≤t≤βx = f(t), y = g(t) , \alpha \le t \le \betaEach tt determines a point (x,y)(x,y).
As tt varies, (x,y)=(f(t),g(t))(x,y) = (f(t),g(t)) varies and traces out a curve, which call a parametic curve.

# Formula

Cycloid:{x=r(θ−sinθ)y=r(1−cosθ)θ∈RCycloid: \left\{\begin{matrix} x = r(\theta-sin\theta)\\ y = r(1-cos\theta) \end{matrix}\right. \theta \in \mathbb{R}.

Sketch and identify curve defined by x=t2−2tx=t^2-2t and y=t+1y=t+1

∵t=y−1∴x=(y−1)2−2(y−1)=y2−4y+3\because t = y-1 \therefore x = (y-1)^2 - 2(y-1) = y^2-4y+3
⇒x=(y−2)2−1\Rightarrow x = (y-2)^2 - 1 頂點為(−1,2)(-1,2)
Let y2−4y+3=0,y=1or3y^2-4y+3 = 0, y = 1 or 3
t=0,x=0,y=1t = 0, x = 0,y = 1, t=1,x=−1,y=2t = 1, x = -1, y=2

# Calculus with Parametric Curves

# Tangent Line

Tangent Line for x=f(t),y=g(t)x=f(t), y = g(t), we need

  1. 切點座標: (x,y)=(f(t),g(t))(x,y) = (f(t),g(t))
  2. 切線斜率: dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} if dxdt≠0\frac{dx}{dt} \neq 0
    ⇒\Rightarrow The tangent line is y−g(t)=dydx∣t=t0(x−f(t0))y-g(t) = \frac{dy}{dx} |_{t = t_0} (x-f(t_0))
Remark

dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

  1. dxdt\frac{dx}{dt}、dydt\frac{dy}{dt} 不同時為 0:
    1. dydt=0\frac{dy}{dt} = 0，有水平切線。
    2. dxdt=0\frac{dx}{dt} = 0，有垂直切線。
  2. dxdt\frac{dx}{dt}、dydt\frac{dy}{dt} 同時為 0，則什麼都有可能。

Given parametric equation C:x=t3C: x = t^3,y=t3−3ty = t^3-3t

  1. Show that CC has two tangent lines at (3,0)(3,0) and find their equations.
  2. Find points on CC where tangent is horizontal or vertical.
  3. Discuss the concavity of CC.

先確認 (3,0)(3,0) 在 CC 上，⇒t=±3\Rightarrow t = \pm \sqrt{3}
∵dydx=dydtdxdt=3t2−32t=32t−32t−1\because \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2-3}{2t} = \frac{3}{2}t - \frac{3}{2}t^{-1}.
∴dydx∣t=3=3\therefore \frac{dy}{dx} |_{t = \sqrt{3}} = \sqrt{3} and dydx∣t=−3=−3\frac{dy}{dx} |_{t = -\sqrt{3}} = -\sqrt{3}
⇒\Rightarrow the tangent lines are y=3(x−3)y = \sqrt{3}(x-3) and y=−3(x−3)y = -\sqrt{3}(x-3)

dydt=0⇒3t2−3=0⇒t=±1\frac{dy}{dt} = 0 \Rightarrow 3t^2 - 3 = 0 \Rightarrow t = \pm 1
水平切線: t=1:(1,−2)t = 1 : (1,-2)、t=−1:(1,2)t = -1 : (1,2)
dxdt=0⇒2t=0⇒t=0\frac{dx}{dt} = 0 \Rightarrow 2t = 0 \Rightarrow t = 0
垂直切線: t=0:(0,0)t = 0 : (0,0)

d2ydx2=ddtdxdt=32+32t−22t=34(1t+1t3)\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}}{\frac{dx}{dt}} = \frac{\frac{3}{2}+ \frac{3}{2}t^{-2}}{2t}= \frac{3}{4}(\frac{1}{t} + \frac{1}{t^3})

d2ydx2>0\frac{d^2y}{dx^2}>0 if t>0t>0，即 t>0t>0 時凹口向上
d2ydx2<0\frac{d^2y}{dx^2}<0 if t<0t<0，即 t<0t<0 時凹口向下

# Area

Consider x=f(t),y=g(t)x = f(t),y = g(t) and α≤t≤β\alpha \le t \le \beta.
A=∫abF(x)dx=∫βαg(t)d(f(t))=∫βαg(t)f′(t)dtA = \int^b_a F(x) dx = \int^{\alpha}_{\beta}g(t)d(f(t)) = \int^{\alpha}_{\beta}g(t)f'(t)dt.

Find the area of an arch of cycloid x=r(θ−sinθ),y=r(1−cosθ)x = r(\theta - sin\theta),y = r(1-cos \theta)

A=∫02πg(θ)f′(θ)dθ=∫02πr(1−cosθ)⋅r(1−cosθ)dθ=r2∫02π(1−cosθ)2dθ=3πr2A = \int^{2\pi}_0g(\theta)f'(\theta) d\theta = \int^{2\pi}_0 r(1-cos\theta) \cdot r(1-cos\theta)d\theta = r^2 \int^{2\pi}_0 (1-cos\theta)^2 d\theta = 3\pi r^2.

# Arc Length

Consider x=f(t),y=g(t)x = f(t),y=g(t) and α≤t≤β\alpha \le t \le \beta
If f′f' and g′g' are continuous on [α,β][\alpha,\beta], then L=∫ab1+(F′(x))2dx=∫αβ1+(dydtdxdt)2dxdt⋅dtL = \int^b_a \sqrt{1+(F'(x))^2}dx = \int^{\beta}_{\alpha}\sqrt{1 + (\frac{\frac{dy}{dt}}{\frac{dx}{dt}})^2} \frac{dx}{dt} \cdot dt
⇒L=∫αβ(dxdt)2+(dydt)2dt\Rightarrow L = \int^{\beta}_\alpha \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt

Find the arch length of an arch of cycloid x=r(θ−sinθ),y=r(1−cosθ)x = r(\theta - sin\theta),y = r(1-cos \theta)

L=∫αβ(dxdθ)2+(dydθ)2dθ=∫02πr2(1−cosθ)2+r2sin2θdθ=8rL = \int^{\beta}_\alpha \sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} d\theta = \int^{2\pi}_0 \sqrt{r^2(1-cos\theta)^2 + r^2sin^2\theta} d\theta = 8r

# Surface Area

Consider C:x=f(t),y=g(t)C:x = f(t),y=g(t) and α≤t≤β\alpha \le t \le \beta 對 x 軸旋轉得到的旋轉體
f′f'、g′g' are continuous on [α,β][\alpha,\beta] and g(t)≥0g(t) \ge 0 on [α,β][\alpha,\beta]
⇒\Rightarrow the surface area is ∫αβ2πy(dxdt)2+(dydt)2dt\int^{\beta}_{\alpha}2\pi y \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt

C:x=rcost,t=rsint,0≤t≤πC: x = rcost , t = rsint , 0 \le t \le \pi 繞 x 軸旋轉得到球面，求此球面的表面積。

A=∫0π2π(rsint)(−rsint)2+(rcost)2dt=4πr2A = \int^{\pi}_0 2\pi (rsint)\sqrt{(-rsint)^2 + (rcost)^2}dt = 4\pi r^2

# Polar Coordinates

P[r,θ],r,θ∈RP[r,\theta], r, \theta \in \mathbb{R}.

  1. r=0r = 0: 圖形為一點 (極點)
  2. r>0r > 0: (rcosθ,rsinθ)(rcos\theta,rsin\theta)
  3. r<0r < 0: [r,θ]=[∣r∣,θ+π][r,\theta] = [|r|,\theta + \pi]

# Relationship between Polar and Cartesian Coordinates

x=rcosθ,y=rsinθx = rcos\theta, y = rsin\theta, r,θ∈Rr,\theta \in \mathbb{R}.
⇒r2=x2+y2\Rightarrow r^2 = x^2+y^2, tanθ=yxtan\theta = \frac{y}{x}.

# Calculus with Parametric Curves

# Tangent Line

  1. 切點: (x,y)=(rcosθ0,rsinθ0)=(f(θ0)cosθ0,f(θ0)sinθ0)(x,y) = (rcos\theta_0, rsin\theta_0) = (f(\theta_0)cos\theta_0, f(\theta_0)sin\theta_0)
  2. 斜率: dydx=dydθdxdθ\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} if dxdθ≠0\frac{dx}{d\theta} \neq 0
    Tangent Line is y−f(θ0)sinθ0=dydx∣θ=θ0(x−f(θ0)cosθ0)y - f(\theta_0)sin\theta_0 = \frac{dy}{dx}|_{\theta = \theta_0}(x-f(\theta_0)cos\theta_0)

# Area

  1. Area bounded by f(θ)f(\theta)
    Let RR be the region bounded by r=f(θ)r = f(\theta), θ=a\theta = a, θ=b\theta = b, where f(θ)≥0f(\theta) \ge 0 on [a,b][a,b] and 0<b−a≤2π0 < b-a \le 2\pi.
    Then the area AA of RR is A=∫ab12((f(θ))2)dθA = \int^b_a \frac{1}{2} ((f(\theta))^2)d\theta(or A=∫ab12r2dθA = \int^b_a \frac{1}{2}r^2 d\theta)

  2. Area bounded by f(θ)f(\theta) and g(θ)g(\theta)
    f(θ),g(θ)≥0f(\theta),g(\theta) \ge 0 on [a,b][a,b] and 0<b−a≤2π0 < b-a \le 2\pi
    ⇒A=12∫ab((f(θ))2)dθ−12∫ab((g(θ))2)dθ=12∫ab((f(θ))2−(g(θ))2)dθ\Rightarrow A = \frac{1}{2} \int^b_a ((f(\theta))^2)d\theta - \frac{1}{2} \int^b_a ((g(\theta))^2)d\theta =\frac{1}{2} \int^b_a ((f(\theta))^2 - (g(\theta))^2)d\theta

# Arc Length

The arc length of L=∫ab(dxdθ)2+(dydθ)2dθ=∫abr2+(drdθ)2dθL = \int^b_a \sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} d\theta = \int^b_a \sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta

# Reference

  • 蘇承芳老師 - 微積分甲（一）109 學年度 - Calculus (I) Academic Year 109
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