求极坐标方程r=θ的二阶导数（附一阶导数推导依据）

Let's break down how to compute the second derivative ( \frac{d^2y}{dx2} ) for the polar equation ( r = \theta ). We already have the first derivative sorted out from your given work, so we'll build directly on that.

Key Background & First Derivative Recap #

First, remember the polar-to-Cartesian conversions we're working with:

• ( x = r\cos\theta = \theta\cos\theta )
• ( y = r\sin\theta = \theta\sin\theta )

And we already derived the first derivative:
$$\frac{dy}{dx} = \frac{\theta\cos\theta + \sin\theta}{-\theta\sin\theta + \cos\theta}$$
To simplify notation, let's call the numerator ( N(\theta) = \theta\cos\theta + \sin\theta ) and the denominator ( D(\theta) = -\theta\sin\theta + \cos\theta ), so ( \frac{dy}{dx} = \frac{N(\theta)}{D(\theta)} ).

Step 1: Frame the Second Derivative Calculation #

The second derivative ( \frac{d^2y}{dx2} ) is the derivative of ( \frac{dy}{dx} ) with respect to ( x ). Using the chain rule, we can rewrite this in terms of ( \theta ) (since we can't directly differentiate with respect to ( x ) easily here):
$$\frac{d^2y}{dx2} = \frac{d}{d\theta}\left( \frac{dy}{dx} \right) \cdot \frac{d\theta}{dx}$$
Since ( \frac{d\theta}{dx} = \frac{1}{\frac{dx}{d\theta}} ), and ( \frac{dx}{d\theta} = D(\theta) ) (from the first derivative's denominator formula), this simplifies to:
$$\frac{d^2y}{dx2} = \frac{1}{D(\theta)} \cdot \frac{d}{d\theta}\left( \frac{N(\theta)}{D(\theta)} \right)$$

Step 2: Differentiate the First Derivative #

Use the quotient rule for differentiation: ( \frac{d}{d\theta}\left( \frac{N}{D} \right) = \frac{N'D - ND'}{D^2} )

First calculate the derivatives of ( N(\theta) ) and ( D(\theta) ):

• ( N'(\theta) = \frac{d}{d\theta}(\theta\cos\theta + \sin\theta) = \cos\theta - \theta\sin\theta + \cos\theta = 2\cos\theta - \theta\sin\theta ) (using the product rule for ( \theta\cos\theta ))
• ( D'(\theta) = \frac{d}{d\theta}(-\theta\sin\theta + \cos\theta) = -\sin\theta - \theta\cos\theta - \sin\theta = -2\sin\theta - \theta\cos\theta ) (using the product rule for ( -\theta\sin\theta ))

Now substitute these into the quotient rule numerator:
$$N'D - ND' = (2\cos\theta - \theta\sin\theta)(-\theta\sin\theta + \cos\theta) - (\theta\cos\theta + \sin\theta)(-2\sin\theta - \theta\cos\theta)$$

Step 3: Expand and Simplify the Numerator #

Let's expand each term separately to simplify:

1. First term expansion:
   $$(2\cos\theta - \theta\sin\theta)(-\theta\sin\theta + \cos\theta) = -2\theta\sin\theta\cos\theta + 2\cos^2\theta + \theta^2\sin2\theta - \theta\sin\theta\cos\theta$$
   Combine like terms:
   $$= 2\cos^2\theta + \theta^2\sin2\theta - 3\theta\sin\theta\cos\theta$$

2. Second term (flip the negative sign first):
   $$- (\theta\cos\theta + \sin\theta)(-2\sin\theta - \theta\cos\theta) = (\theta\cos\theta + \sin\theta)(2\sin\theta + \theta\cos\theta)$$
   Expand this:
   $$= 2\theta\sin\theta\cos\theta + \theta^2\cos2\theta + 2\sin^2\theta + \theta\sin\theta\cos\theta$$
   Combine like terms:
   $$= \theta^2\cos2\theta + 2\sin^2\theta + 3\theta\sin\theta\cos\theta$$

Now add the two expanded terms together:
$$(2\cos^2\theta + \theta^2\sin2\theta - 3\theta\sin\theta\cos\theta) + (\theta^2\cos2\theta + 2\sin^2\theta + 3\theta\sin\theta\cos\theta)$$
The ( -3\theta\sin\theta\cos\theta ) and ( +3\theta\sin\theta\cos\theta ) cancel out. Use the Pythagorean identity ( \sin^2\theta + \cos^2\theta = 1 ) to simplify the rest:
$$= 2(\cos^2\theta + \sin^2\theta) + \theta^2(\sin2\theta + \cos^2\theta) = 2(1) + \theta^2(1) = \theta^2 + 2$$

Step 4: Combine All Parts for the Final Result #

Putting it all together:
$$\frac{d^2y}{dx2} = \frac{1}{D(\theta)} \cdot \frac{\theta^2 + 2}{D^2(\theta)} = \frac{\theta^2 + 2}{D^3(\theta)}$$
Substitute back ( D(\theta) = \cos\theta - \theta\sin\theta ):
$$\frac{d^2y}{dx2} = \frac{\theta^2 + 2}{(\cos\theta - \theta\sin\theta)^3}$$

内容的提问来源于stack exchange，提问作者user527827