The Arc Length of a Smooth, Planar Curve and Distance Traveled · 光滑平面曲线的弧长与行进距离
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Arc length/ɑːk leŋθ/ | 弧长 | hú zhǎng |
Measuring the length of a curve
- Area and volume were integrals; so is the length of a curve itself.
- Arc length 弧长 adds up the lengths of tiny straight pieces along the curve.
- On each little piece, the curve is nearly a straight line — a hypotenuse of a small right triangle.
- Sum those hypotenuses (integrate) to get the total length.
度量曲线的长度
- 面积和体积都是积分;曲线的长度本身也是。
- 弧长把曲线上微小直线段的长度加起来。
- 在每一小段上,曲线几乎是一条直线——一个小直角三角形的斜边。
- 把这些斜边求和(积分)得到总长度。
From a small triangle to the formula
- A tiny piece has horizontal run $dx$ and vertical rise $dy=f'(x)\,dx$.
- Its length is $\sqrt{dx^2+dy^2}=\sqrt{1+\big(f'(x)\big)^2}\,dx$ (Pythagoras).
- Add them from $a$ to $b$:
-
$$L=\int_a^b \sqrt{1+\big(f'(x)\big)^2}\,dx$$
从小三角形到公式
- 一小段有水平变化 $dx$ 和竖直变化 $dy=f'(x)\,dx$。
- 它的长度是 $\sqrt{dx^2+dy^2}=\sqrt{1+\big(f'(x)\big)^2}\,dx$(勾股定理)。
- 从 $a$ 到 $b$ 相加:
-
$$L=\int_a^b \sqrt{1+\big(f'(x)\big)^2}\,dx$$
A curve is longer than its chord · 曲线比其弦长
y = a·x^{1.5}
Arc length sums tiny hypotenuses $\sqrt{1+(f')^2}\,dx$ — always at least the horizontal distance $b-a$. · 弧长累加微小的斜边 $\sqrt{1+(f')^2}\,dx$ —— 始终至少等于水平距离 $b-a$。
The arc length of $y=f(x)$ on $[a,b]$ is... · $y=f(x)$ 在 $[a,b]$ 上的弧长为...
Square the derivative, add 1, take the root. · 对导数平方，加1，再开方。
Using the formula
- Compute $f'(x)$, square it, add $1$, take the square root, and integrate over $[a,b]$.
- The integrand is always $\ge1$, so arc length is at least the horizontal distance $b-a$ (a curve is longer than the straight line).
- Many arc-length integrals are hard by hand; on the BC exam you often set them up and evaluate numerically.
- The key skill is building the integrand correctly from $f'$.
使用公式
- 算出 $f'(x)$,平方,加 $1$,取平方根,在 $[a,b]$ 上积分。
- 被积函数总 $\ge1$,所以弧长至少是水平距离 $b-a$(曲线比直线长)。
- 许多弧长积分手算很难;BC 考试中你常建立它并用数值求值。
- 关键技能是从 $f'$ 正确地构造被积函数。
For · 支持 $y=\tfrac23 x^{3/2}$, $f'(x)=x^{1/2}$. The arc-length integrand is... · 对于 $y=\tfrac23 x^{3/2}$, $f'(x)=x^{1/2}$。弧长被积函数为...
$(f')^2=x$, so $\sqrt{1+x}$. · $(f')^2=x$，所以 $\sqrt{1+x}$。
Evaluate · 评价 $\int_0^3 \sqrt{1+x}\,dx$ (a decimal). · 计算 $\int_0^3 \sqrt{1+x}\,dx$（小数形式）。
$\tfrac23(8-1)=\tfrac{14}{3}\approx4.667$.
Because the integrand is $\ge 1$, the arc length is always at least... · 因为被积函数是 $\ge 1$，所以弧长始终至少为...
A curve is at least as long as the straight run. · 曲线长度至少等于直线运行距离。
Arc length as distance traveled
- If a particle moves along $y=f(x)$, the arc length is the distance it travels along the path.
- (For motion given by $x(t),y(t)$, the parametric version comes in lesson 9.3.)
- So this same integral measures both a curve's geometric length and a path's traveled distance.
- Setup is identical; only the interpretation changes.
弧长即行进距离
- 若一个质点沿 $y=f(x)$ 运动,弧长就是它沿路径行进的距离。
- (对由 $x(t),y(t)$ 给出的运动,参数版本在 9.3 课。)
- 所以同一个积分既度量曲线的几何长度,也度量路径的行进距离。
- 建立方式相同;只是解读不同。
The arc-length integrand uses the derivative $f'(x)$, not $f(x)$. · 弧长被积函数使用导数 $f'(x)$，而非 $f(x)$。
It is · 它是 $\sqrt{1+(f'(x))^2}$. · 它是 $\sqrt{1+(f'(x))^2}$。
Along a path $y=f(x)$, the arc length also measures the ____ traveled. · 沿路径 $y=f(x)$，弧长也衡量 ____ 行进距离。
Arc length = distance traveled along the path. · 弧长 = 沿路径行进的距离。
The integrand is $\sqrt{1+(f'(x))^2}$ — the derivative is squared, and there's a $\boldsymbol{+1}$ inside the root, not outside. Don't write $\sqrt{1+f(x)^2}$ (using $f$ instead of $f'$) or forget the $+1$. And the whole thing is under one square root before integrating.
被积函数是 $\sqrt{1+(f'(x))^2}$——导数被平方,根号里有个 $\boldsymbol{+1}$,不在外面。别写成 $\sqrt{1+f(x)^2}$(用 $f$ 而非 $f'$)或忘掉 $+1$。而且整个东西在积分前都在一个平方根下。
Set up the arc length of $y=\tfrac23 x^{3/2}$ from $x=0$ to $x=3$.
- $f'(x)=x^{1/2}$, so $\big(f'(x)\big)^2=x$.
- $L=\displaystyle\int_0^3 \sqrt{1+x}\,dx=\Big[\tfrac23(1+x)^{3/2}\Big]_0^3=\tfrac23(8-1)=\tfrac{14}{3}$.
建立 $y=\tfrac23 x^{3/2}$ 从 $x=0$ 到 $x=3$ 的弧长。
- $f'(x)=x^{1/2}$,所以 $\big(f'(x)\big)^2=x$。
- $L=\displaystyle\int_0^3 \sqrt{1+x}\,dx=\Big[\tfrac23(1+x)^{3/2}\Big]_0^3=\tfrac23(8-1)=\tfrac{14}{3}$。
The arc length of $y=f(x)$ on $[a,b]$ is $L=\int_a^b\sqrt{1+\big(f'(x)\big)^2}\,dx$ — summing tiny hypotenuses $\sqrt{1+(f')^2}\,dx$. Square the derivative, add $1$ inside the root, integrate. It also gives the distance traveled along the path.
$y=f(x)$ 在 $[a,b]$ 上的弧长是 $L=\int_a^b\sqrt{1+\big(f'(x)\big)^2}\,dx$——把微小斜边 $\sqrt{1+(f')^2}\,dx$ 相加。把导数平方,根号内加 $1$,积分。它也给出沿路径的行进距离。