Solution 这是重力作用下抛体的运动方程. 对方程积分一次得

$$\frac{d\mathbf{r}}{dt} = -g\mathbf{t}\mathbf{k} + \mathbf{c}$$

Since $\dot{\mathbf{r}}(0) = \mathbf{V}$ , 我们有

$$c = V \quad \text{and} \quad \frac{d\mathbf{r}}{dt} = -g\mathbf{t}\mathbf{k} + \mathbf{V}$$

再积分一次，得

$$\mathbf{r}(t) = \mathbf{Vt - \frac{1}{2}gt^2\mathbf{k} + \mathbf{a}$$

Since $\mathbf{r}(0) = 0$ , 我们有 $\mathbf{a} = 0$ , giving

$$r = Vt - \frac{1}{2}gt^2k$$

Now $\mathbf{r} = (x, y, z)$ , 使得当 $\mathbf{V} = (u, 0, v)$ , 我们有

$$\begin{aligned} (x, y, z) &= (u, 0, v)t + (0, 0, -\frac{1}{2}gt^2) \ &= (ut, 0, vt) + (0, 0, -\frac{1}{2}gt^2) \ &= (ut, 0, vt - \frac{1}{2}gt^2) \end{aligned}$$

Thus

$$x = ut, \quad y = 0 \quad \text{and} \quad z = vt - \frac{1}{2}gt^2$$

Substituting $t = x/u$ 代入……的方程中 $z$ gives, 经过适当的整理后,

$$z = \frac{1}{2} \frac{v^2}{g} - \frac{g}{2u^2} \left( x - \frac{uv}{g} \right)^2$$

这是一条顶点在的抛物线 $(uv/g, 0, v^2/2g)$ .

9.5.2 Exercises

32 If $\mathbf{r} = (t, t^2, t^3)$ , find $\dot{\mathbf{r}}(t)$ and $\ddot{\mathbf{r}}(t)$ .

33 给定向量

$$\mathbf{r} = (1 + t)\mathbf{i} + t^2\mathbf{j} + \frac{2}{3}t^3\mathbf{k}$$

evaluate $d\mathbf{r}/dt$ 并将其写成如下形式

$$\frac{d\mathbf{r}}{dt} = f(t) \hat{\mathbf{T}}(t)$$

where $\hat{T}$ 是单位切向量方向. Calculate $d\hat{T}/dt$ 以其最简形式表示，并证明它与……垂直 $\hat{T}$ .

$$\hat{r} = \cos\theta i + \sin\theta j$$

$$\hat{\boldsymbol{\theta}} = -\sin\theta \mathbf{i} + \cos\theta \mathbf{j}$$

因此，由定义可得 $\mathbf{r} = \mathbf{r}\hat{\mathbf{r}}$ 证明

$$\frac{d\mathbf{r}}{dt} = \frac{d\mathbf{r}}{dt} \hat{\mathbf{r}} + r\omega \hat{\boldsymbol{\theta}} \quad \text{where} \quad \omega = \frac{d\theta}{dt}$$

推导出

$$\frac{d\hat{r}}{dt} = \omega\hat{\theta} \quad \text{and} \quad \frac{d\hat{\theta}}{dt} = -\omega\hat{r}$$

and

$$\frac{d^2 \mathbf{r}}{dt^2} = \left( \frac{d^2 r}{dt^2} - r\omega^2 \right) \hat{\mathbf{r}} + \left( 2\omega \frac{dr}{dt} + r \frac{d\omega}{dt} \right) \hat{\boldsymbol{\theta}}$$