Finding Arc Lengths of Curves Given by Parametric Equations · 求由参数方程给出的曲线的弧长
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| arc length/ɑːk leŋθ/ | 弧长 | hú zhǎng |
| speed/spiːd/ | 速率 | sù lǜ |
How far a parametric point travels
- A parametric point $\big(x(t),y(t)\big)$ moves through the plane. How long is the path it traces?
- That's the arc length 弧长 of the parametric curve, from $t=a$ to $t=b$.
- Each tiny time step moves the point by $dx=\tfrac{dx}{dt}\,dt$ across and $dy=\tfrac{dy}{dt}\,dt$ up.
- Add up those tiny hypotenuses to get the total length.
参数点走了多远
- 参数点 $\big(x(t),y(t)\big)$ 在平面上移动。它描出的路径有多长?
- 那就是参数曲线从 $t=a$ 到 $t=b$ 的弧长。
- 每一微小时间步把点横向移 $dx=\tfrac{dx}{dt}\,dt$、纵向移 $dy=\tfrac{dy}{dt}\,dt$。
- 把这些微小斜边加起来得到总长度。
The formula
- The little step length is $\sqrt{dx^2+dy^2}=\sqrt{\big(\tfrac{dx}{dt}\big)^2+\big(\tfrac{dy}{dt}\big)^2}\,dt$.
- Integrating over the time interval:
-
$$L=\int_a^b \sqrt{\Big(\tfrac{dx}{dt}\Big)^2+\Big(\tfrac{dy}{dt}\Big)^2}\,dt$$
- Both component derivatives are squared, added, rooted, and integrated in $t$.
公式
- 小步长度是 $\sqrt{dx^2+dy^2}=\sqrt{\big(\tfrac{dx}{dt}\big)^2+\big(\tfrac{dy}{dt}\big)^2}\,dt$。
- 在时间区间上积分:
-
$$L=\int_a^b \sqrt{\Big(\tfrac{dx}{dt}\Big)^2+\Big(\tfrac{dy}{dt}\Big)^2}\,dt$$
- 两个分量导数都平方、相加、取根,关于 $t$ 积分。
The arc length of a parametric curve is $\int_a^b(\ ?\ )\,dt$ where the integrand is... · 参数曲线的弧长为 $\int_a^b(\ ?\ )\,dt$，其中被积函数为...
Both derivatives squared, added, under a root. · 两个导数平方相加，再开根号。
For · 支持 $x=t^2$, $y=t^3$: $(dx/dt)^2+(dy/dt)^2=$ · 对于 $x=t^2$, $y=t^3$: $(dx/dt)^2+(dy/dt)^2=$
$(2t)^2+(3t^2)^2=4t^2+9t^4$.
The integrand is the speed
- $\sqrt{\big(\tfrac{dx}{dt}\big)^2+\big(\tfrac{dy}{dt}\big)^2}$ is exactly the speed 速率 of the moving point.
- Speed is distance per unit time; integrating speed over time gives total distance.
- So arc length = the integral of speed — the parametric version of "distance = ∫ speed dt."
- This connects directly to motion problems (lesson 9.6).
被积函数就是速率
- $\sqrt{\big(\tfrac{dx}{dt}\big)^2+\big(\tfrac{dy}{dt}\big)^2}$ 恰好是动点的速率。
- 速率是单位时间的距离;对速率关于时间积分得到总距离。
- 所以弧长 = 速率的积分——"距离 = ∫ 速率 dt"的参数版本。
- 这直接联系到运动问题(9.6 课)。
Length is the integral of speed · 长度是速率的积分
y = ax^{1.5}
The parametric arc-length integrand is the point's speed — integrating speed over time gives the distance traveled. · 参数弧长被积函数是点的速率 —— 对速率关于时间积分得到行进距离。
The parametric arc-length integrand equals the point's ____. · 参数弧长被积函数等于点的 ____。
Integral of speed = distance. · 速率的积分 = 距离。
The parametric arc length is the same integral used for a particle's... · 参数弧长与粒子...使用的积分相同
Integral of speed = distance traveled. · 速率的积分 = 行进距离。
Setting it up in practice
- Compute $\tfrac{dx}{dt}$ and $\tfrac{dy}{dt}$, square and add, take the root.
- Integrate from the starting $t$ to the ending $t$ (the $t$-interval, not $x$ or $y$).
- Many of these integrals need a calculator; the setup is the graded skill.
- Keep both derivatives — dropping one loses the diagonal motion.
实际建立
- 算出 $\tfrac{dx}{dt}$ 和 $\tfrac{dy}{dt}$,平方相加,取根。
- 从起始 $t$ 到终止 $t$ 积分($t$ 区间,不是 $x$ 或 $y$)。
- 许多这类积分需要计算器;建立是被评分的技能。
- 保留两个导数——丢一个就失去了对角运动。
The limits of integration for parametric arc length are $t$-values. · 参数弧长的积分限是 $t$ 值。
Integrate over the time interval. · 在时间区间上积分。
You can drop $\tfrac{dy}{dt}$ if $\tfrac{dx}{dt}$ is larger. · 您可以丢弃 $\tfrac{dy}{dt}$ 如果 $\tfrac{dx}{dt}$ 更大。
Both derivatives are needed for the diagonal motion. · 对角运动需要两个导数。
The limits are $t$-values (the time interval), and both $\tfrac{dx}{dt}$ and $\tfrac{dy}{dt}$ appear, each squared under the root. Don't use $x$-limits, and don't forget one of the two derivatives. This is the same "speed integral" as distance traveled — not a signed quantity, so no absolute values needed (the root is already $\ge0$).
积分限是 $t$ 值(时间区间),而且 $\tfrac{dx}{dt}$ 和 $\tfrac{dy}{dt}$ 都出现,各在根号下平方。别用 $x$ 边界,也别忘掉两个导数之一。这与行进距离是同一个"速率积分"——不是带符号的量,所以不需要绝对值(根号已经 $\ge0$)。
Find the length of $x=t^2,\ y=t^3$ for $0\le t\le 1$.
- $\tfrac{dx}{dt}=2t$, $\tfrac{dy}{dt}=3t^2$; sum of squares $=4t^2+9t^4=t^2(4+9t^2)$.
- $L=\displaystyle\int_0^1 \sqrt{t^2(4+9t^2)}\,dt=\int_0^1 t\sqrt{4+9t^2}\,dt$.
- $u$-sub $u=4+9t^2$: $=\tfrac{1}{27}\big(13^{3/2}-8\big)\approx1.44$.
求 $x=t^2,\ y=t^3$ 在 $0\le t\le 1$ 的长度。
- $\tfrac{dx}{dt}=2t$,$\tfrac{dy}{dt}=3t^2$;平方和 $=4t^2+9t^4=t^2(4+9t^2)$。
- $L=\displaystyle\int_0^1 \sqrt{t^2(4+9t^2)}\,dt=\int_0^1 t\sqrt{4+9t^2}\,dt$。
- 换元 $u=4+9t^2$:$=\tfrac{1}{27}\big(13^{3/2}-8\big)\approx1.44$。
The arc length of a parametric curve is $L=\int_a^b\sqrt{\big(\tfrac{dx}{dt}\big)^2+\big(\tfrac{dy}{dt}\big)^2}\,dt$ over the $t$-interval. The integrand is the point's speed, so this is the integral of speed = distance traveled. Square both component derivatives under one root.
参数曲线的弧长是 $L=\int_a^b\sqrt{\big(\tfrac{dx}{dt}\big)^2+\big(\tfrac{dy}{dt}\big)^2}\,dt$,在 $t$ 区间上。被积函数是点的速率,所以这是速率的积分 = 行进距离。把两个分量导数在一个根号下平方。