The Relationship Between Cartesian and Polar Coordinates | 笛卡尔坐标与极坐标的关系

📚 The Relationship Between Cartesian and Polar Coordinates | 笛卡尔坐标与极坐标的关系

Two coordinate systems are commonly used to describe the position of a point in the plane: the Cartesian coordinate system and the polar coordinate system. Understanding how these two systems relate is essential for solving problems in geometry, algebra, and calculus.

两种常用的坐标系用于描述平面上点的位置：笛卡尔坐标系和极坐标系。理解这两种坐标系之间的转换关系，对于解决几何、代数和微积分问题至关重要。

1. Understanding Cartesian Coordinates | 理解笛卡尔坐标

The Cartesian coordinate system uses two perpendicular axes, usually labelled \(x\) and \(y\), which intersect at the origin. A point is represented by an ordered pair \((x, y)\), where \(x\) is the horizontal distance from the origin and \(y\) is the vertical distance.

笛卡尔坐标系使用两条互相垂直的数轴，通常标记为 \(x\) 轴和 \(y\) 轴，它们相交于原点。一个点用有序数对 \((x, y)\) 表示，其中 \(x\) 是到原点的水平距离，\(y\) 是到原点的垂直距离。

For example, the point \((3, 4)\) is 3 units to the right of the origin and 4 units above it. This system is intuitive for describing straight lines and rectangular shapes.

例如，点 \((3, 4)\) 表示在原点右侧 3 个单位、上方 4 个单位的位置。这种坐标系对于描述直线和矩形图形非常直观。

2. Understanding Polar Coordinates | 理解极坐标

In the polar coordinate system, a point is described by an ordered pair \((r, \theta)\). Here, \(r\) is the distance from the pole (the origin) to the point, and \(\theta\) is the angle measured anticlockwise from the positive \(x\)-axis, known as the initial line.

在极坐标系中，一个点用有序数对 \((r, \theta)\) 描述。其中 \(r\) 是从极点（原点）到该点的距离，\(\theta\) 是从正 \(x\) 轴（称为极轴）逆时针方向测量的角度。

The angle \(\theta\) is usually given in radians in A-Level mathematics. By convention, \(r\) can be positive or negative. If \(r < 0\), the point is located in the opposite direction to the angle \(\theta\).

在 A-Level 数学中，角度 \(\theta\) 通常使用弧度制。按照约定，\(r\) 可以为正或负。如果 \(r < 0\)，则该点位于角度 \(\theta\) 所指示的相反方向。

3. Converting Polar to Cartesian | 极坐标转换为笛卡尔坐标

Consider a point \(P\) with polar coordinates \((r, \theta)\). We can drop a perpendicular from \(P\) to the \(x\)-axis, forming a right-angled triangle with hypotenuse \(r\). The horizontal leg is \(x\) and the vertical leg is \(y\).

考虑一个具有极坐标 \((r, \theta)\) 的点 \(P\)。我们可以从 \(P\) 向 \(x\) 轴作垂线，形成一个斜边为 \(r\) 的直角三角形。水平直角边为 \(x\)，垂直直角边为 \(y\)。

Using the definitions of cosine and sine:

根据余弦和正弦的定义：

x = r cos θ, y = r sin θ

These two formulas allow us to find the Cartesian coordinates directly from the polar coordinates. Note that these formulas hold for any angle, whether positive or negative, and for any value of \(r\).

这两个公式使我们能够直接从极坐标求出笛卡尔坐标。注意，这些公式对任何角度（正或负）以及任何 \(r\) 值都成立。

4. Converting Cartesian to Polar | 笛卡尔坐标转换为极坐标

To convert from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\), we use the same right-angled triangle. The distance \(r\) is found using Pythagoras’ theorem:

要将笛卡尔坐标 \((x, y)\) 转换为极坐标 \((r, \theta)\)，我们使用同一个直角三角形。距离 \(r\) 由毕达哥拉斯定理求出：

r = √(x² + y²)

To find the angle \(\theta\), we use the tangent ratio:

求角度 \(\theta\) 时，我们使用正切比：

tan θ = y / x, so θ = arctan(y / x)

However, the arctan function only returns values in the range \((-π/2, π/2)\). Therefore, we must adjust \(\theta\) based on the quadrant in which the point lies.

然而，\(\arctan\) 函数只返回 \((-π/2, π/2)\) 范围内的值。因此，我们必须根据点所在的象限来调整 \(\theta\)。

5. Understanding the Quadrant Issue | 理解象限问题

When using \(\theta = \arctan(y/x)\), the calculator may give the correct angle for points in the first quadrant, but for points in other quadrants, we must add or subtract \(\pi\).

当使用 \(\theta = \arctan(y/x)\) 时，计算器可能只对第一象限的点给出正确角度；对于其他象限的点，我们必须加上或减去 \(\pi\)。

Quadrant	Sign of x	Sign of y	Adjustment for θ
First	+	+	None (θ = arctan(y/x))
Second	–	+	θ = arctan(y/x) + π
Third	–	–	θ = arctan(y/x) – π or + π
Fourth	+	–	θ = arctan(y/x) + 2π (or take negative value)

For example, the point \((-1, 1)\) has \(r = √2\). The arctan(–1) gives \(-π/4\), but since the point is in the second quadrant, \(\theta = 3π/4\).

例如，点 \((-1, 1)\) 的 \(r = √2\)。\(\arctan(-1) = -π/4\)，但该点在第二象限，所以 \(\theta = 3π/4\)。

6. Worked Example: Polar to Cartesian | 实例：极坐标转笛卡尔坐标

Convert the polar point \((5, π/3)\) to Cartesian coordinates.

将极坐标点 \((5, π/3)\) 转换为笛卡尔坐标。

Using \(x = r cos θ\) and \(y = r sin θ\):

使用 \(x = r cos θ\) 和 \(y = r sin θ\)：

x = 5 cos(π/3) = 5 × 0.5 = 2.5

y = 5 sin(π/3) = 5 × (√3/2) = (5√3)/2

Therefore the Cartesian coordinates are \((2.5, (5√3)/2)\).

因此笛卡尔坐标为 \((2.5, (5√3)/2)\)。

For a negative \(r\), say \((-3, π/4)\):

对于负的 \(r\)，例如 \((-3, π/4)\)：

x = –3 cos(π/4) = –(3√2)/2

y = –3 sin(π/4) = –(3√2)/2

This point lies in the third quadrant, as expected for a negative \(r\).

该点位于第三象限，这与负 \(r\) 的预期一致。

7. Worked Example: Cartesian to Polar | 实例：笛卡尔坐标转极坐标

Convert the Cartesian point \((1, –1)\) to polar coordinates.

将笛卡尔坐标点 \((1, –1)\) 转换为极坐标。

First, find \(r\):

首先求 \(r\)：

r = √(1² + (–1)²) = √2

Next, find \(\theta\). Since \(x > 0\) and \(y < 0\), the point lies in the fourth quadrant, so \(\theta = \arctan(–1) = –π/4\).

接着求 \(\theta\)。由于 \(x > 0\)，\(y < 0\)，该点位于第四象限，所以 \(\theta = \arctan(–1) = –π/4\)。

Thus the polar coordinates are \((√2, –π/4)\), or equivalently \((√2, 7π/4)\).

因此极坐标为 \((√2, –π/4)\)，等价于 \((√2, 7π/4)\)。

Another example: \((0, –3)\). Since \(x = 0\), \(r = 3\) and \(\theta = –π/2\).

另一个例子：\((0, –3)\)。由于 \(x = 0\)，\(r = 3\)，\(\theta = –π/2\)。

8. Converting Equations: Cartesian to Polar | 方程转换：笛卡尔转极坐标

To convert an equation from Cartesian to polar form, substitute \(x = r cos θ\) and \(y = r sin θ\), then simplify.

要将方程从笛卡尔形式转换为极坐标形式，代入 \(x = r cos θ\) 和 \(y = r sin θ\)，然后化简。

For example, consider the line \(y = 2x\). Substituting gives:

例如，考虑直线 \(y = 2x\)。代入得到：

r sin θ = 2r cos θ

Assuming \(r \neq 0\), we divide both sides by \(r\):

假设 \(r \neq 0\)，两边同除以 \(r\)：

sin θ = 2 cos θ → tan θ = 2

So the polar equation is simply \(\theta = \arctan 2\), a constant angle.

因此极坐标方程简化为 \(\theta = \arctan 2\)，即一个恒定角度。

Now consider the circle \(x^2 + y^2 = 16\). Using \(r^2 = x^2 + y^2\):

再考虑圆 \(x^2 + y^2 = 16\)。利用 \(r^2 = x^2 + y^2\)：

r² = 16 → r = 4

This shows that circles centred at the origin have very simple polar equations.

这表明圆心在原点的圆具有非常简洁的极坐标方程。

9. Converting Equations: Polar to Cartesian | 方程转换：极坐标转笛卡尔坐标

To convert from polar to Cartesian, use the relationships \(x = r cos θ\), \(y = r sin θ\), and \(r^2 = x^2 + y^2\). It is often useful to multiply both sides by \(r\) to obtain \(r^2\) terms.

要将极坐标转换为笛卡尔坐标，利用关系式 \(x = r cos θ\)，\(y = r sin θ\) 和 \(r^2 = x^2 + y^2\)。通常将两边同乘以 \(r\) 以获得 \(r^2\) 项会很有帮助。

Consider the polar equation \(r = 2a cos θ\). Multiply both sides by \(r\):

考虑极坐标方程 \(r = 2a cos θ\)。两边同乘以 \(r\)：

r² = 2a r cos θ

Substitute \(r^2 = x^2 + y^2\) and \(r cos θ = x\):

代入 \(r^2 = x^2 + y^2\) 和 \(r cos θ = x\)：

x² + y² = 2ax

Completing the square gives:

配方得到：

(x – a)² + y² = a²

This is the equation of a circle with centre \((a, 0)\) and radius \(a\).

这是一个圆心在 \((a, 0)\)、半径为 \(a\) 的圆的方程。

10. Common Pitfalls and Exam Tips | 常见陷阱与考试技巧

The conversion between coordinate systems often leads to several common errors. Being aware of these can save valuable marks in the exam.

坐标系之间的转换常常导致一些常见错误。意识到这些错误可以在考试中为你节省宝贵的分数。

Always check whether the angle is required in degrees or radians. In A-Level polar work, radians are standard.
不要忘记检查角度是使用角度制还是弧度制。在 A-Level 极坐标部分，弧度制是标准。
Remember to adjust the angle for the correct quadrant when using \(\arctan\).
使用 \(\arctan\) 时，务必根据象限调整角度。
A negative \(r\) is allowed but changes the direction of the point; always interpret negative \(r\) carefully.
负的 \(r\) 是允许的，但会改变点的方向；务必仔细解读负的 \(r\)。
When converting equations, avoid dividing by \(r\) if \(r = 0\) is a possible solution. Consider whether the origin is included in the curve.
转换方程时，如果 \(r = 0\) 可能是解，则要避免除以 \(r\)。同时考虑原点是否包含在曲线中。
Always sketch a quick diagram to verify that the converted point lies in the expected quadrant.
总是快速画一个草图，以验证转换后的点位于预期象限中。

11. Applications and Further Study | 应用与深入学习

The relationship between Cartesian and polar coordinates plays a crucial role in many areas of mathematics. In complex numbers, a complex number \(z = x + iy\) can be written in polar form as \(z = r(cos θ + i sin θ)\), which leads to Euler’s formula \(e^{iθ} = cos θ + i sin θ\).

笛卡尔坐标与极坐标之间的关系在数学的许多领域中都起着关键作用。在复数中，复数 \(z = x + iy\) 可以写成极坐标形式 \(z = r(cos θ + i sin θ)\)，进而引出欧拉公式 \(e^{iθ} = cos θ + i sin θ\)。

In physics, circular motion and wave functions are often expressed using polar coordinates. In calculus, polar coordinates simplify the integration of regions with circular symmetry, such as areas inside curves like the cardioid or rose.

在物理学中，圆周运动和波动函数通常使用极坐标表示。在微积分中，极坐标简化了具有圆形对称性的区域积分，例如心形线或玫瑰线内部的面积。

Mastering the conversion between these two systems not only boosts exam performance but also provides a deeper geometric insight into the structure of mathematical objects.

掌握这两种坐标系之间的转换，不仅能提升考试成绩，还能让你更深入地理解数学对象的结构与几何本质。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

The Relationship Between Cartesian and Polar Coordinates | 笛卡尔坐标与极坐标的关系