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### **Position Vector**
[
r(t) = t^2 \mathbf{i} + t \mathbf{j}
]
### **Point**
[
(4,, 2)
]
---
### **(a)** Find the velocity vector (v(t)), speed (s(t)), and acceleration vector (a(t)) of the object.
[
v(t) = \langle 2t,\ 1 \rangle
]
[
s(t) = \sqrt{4t^2 + 1}
]
[
a(t) = \langle 2,\ 0 \rangle
]
---
### **(b)** Evaluate the velocity vector and acceleration vector of the object at the given point.
[
v(2) = \langle 4,\ 1 \rangle
]
[
a(2) = \langle 2,\ 0 \rangle
]
1.讲解题目是什意思? 2.讲解解题过程](https://manimvideo.explanation.fun/video/cover/578766165717835777.png)
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## The position vector **r** describes the path of an object moving in the xy-plane. ### **Position Vector** [ r(t) = t^2 \mathbf{i} + t \mathbf{j} ] ### **Point** [ (4,, 2) ] --- ### **(a)** Find the velocity vector (v(t)), speed (s(t)), and acceleration vector (a(t)) of the object. [ v(t) = \langle 2t,\ 1 \rangle ] [ s(t) = \sqrt{4t^2 + 1} ] [ a(t) = \langle 2,\ 0 \rangle ] --- ### **(b)** Evaluate the velocity vector and acceleration vector of the object at the given point. [ v(2) = \langle 4,\ 1 \rangle ] [ a(2) = \langle 2,\ 0 \rangle ] 1.讲解题目是什意思? 2.讲解解题过程
![Find the gradient of the function at the given point. \( f(x, y) = 3x + 4y^2 + 1,\quad (1, 3) \) \[ \nabla f(1, 3) = \boxed{\ } \]
解释一下什么是梯度gradient,在现实世界中gradient表示什么?](https://manimvideo.explanation.fun/video/cover/578763141711753217.png)
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Find the gradient of the function at the given point. \( f(x, y) = 3x + 4y^2 + 1,\quad (1, 3) \) \[ \nabla f(1, 3) = \boxed{\ } \] 解释一下什么是梯度gradient,在现实世界中gradient表示什么?
![Find the gradient of the function at the given point.
\( f(x, y) = 3x + 4y^2 + 1,\quad (1, 3) \)
\[
\nabla f(1, 3) = \boxed{\ }
\]
注意使用 MathTex 显示公式](https://manimvideo.explanation.fun/video/cover/578760753267503105.png)
▶
Find the gradient of the function at the given point. \( f(x, y) = 3x + 4y^2 + 1,\quad (1, 3) \) \[ \nabla f(1, 3) = \boxed{\ } \] 注意使用 MathTex 显示公式
![**Discuss the continuity of the function**
[
f(x, y, z) = \frac{61}{\sqrt{x^2 + y^2 + z^2}}
]
**Options:**
* ○ continuous everywhere
* ○ continuous except where (x^2 + y^2 + z^2 < 1)
* ○ continuous except where (x^2 + y^2 + z^2 > 1)
* ○ continuous except where (x^2 + y^2 + z^2 > 0)
* ○ continuous except at ((0, 0, 0))](https://manimvideo.explanation.fun/video/cover/578588523731120129.png)
▶
**Discuss the continuity of the function** [ f(x, y, z) = \frac{61}{\sqrt{x^2 + y^2 + z^2}} ] **Options:** * ○ continuous everywhere * ○ continuous except where (x^2 + y^2 + z^2 < 1) * ○ continuous except where (x^2 + y^2 + z^2 > 1) * ○ continuous except where (x^2 + y^2 + z^2 > 0) * ○ continuous except at ((0, 0, 0))
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Match the graph of the surface with one of the contour maps. f(x, y) = e^(1 − x^2 + y^2)
▶
10：占卜的后续步骤——变卦与解卦主要内容：根据三个数字确定“本卦”、“互卦”和“变卦”。确定“动爻”（变爻）的位置。 解卦的核心：以本卦的卦辞和变爻的爻辞为主，互卦和变卦为辅。
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第10课：占卜的后续步骤——变卦与解卦 主要内容： 如何根据三个数字确定“本卦”、“互卦”和“变卦”。 确定“动爻”（变爻）的位置。 解卦的核心：以本卦的卦辞和变爻的爻辞为主，互卦和变卦为辅。
▶
Home 9：《易经》的运用——占卜的原则与方法 主要内容： 占卜的哲学基础：“寂然不动，感而遂通”。 占卜的三原则：不诚不占、不义不占、不疑不占。 介绍傅佩荣老师推荐的“数字占卜法”（简单、易行、无需道具）。
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第5课、所见即所学？ （哈伯和托伦的“感觉剥夺与图形识别”研究） 主要内容：让被试长时间佩戴特制的 distorting 眼镜，发现其知觉逐渐适应并恢复正常。但当摘掉眼镜后，其知觉又需要重新适应。这证明知觉不仅由感觉输入决定，更深受过去经验和学习的影响。
![Here’s the full markdown version of your provided images:
---
### (a) Adjacency List Representation
**Graph:**
Vertices: a, b, c, d
**Adjacency List:**
| Vertex | Adjacent Vertices |
| :----: | :---------------: |
| a | c |
| b | d |
| c | b, d |
| d | b |
---
### (b) Adjacency Matrix Representation
**Graph:**
Vertices: a, b, c, d
**Adjacency Matrix:**
[
M_b =
\begin{pmatrix}
& a & b & c & d \
a & 0 & 1 & 0 & 1 \
b & 1 & 0 & 1 & 1 \
c & 0 & 1 & 0 & 1 \
d & 1 & 1 & 1 & 0
\end{pmatrix}
]
---
### (c) Adjacency Matrix Representation
**Graph:**
Vertices: a, b, c, d
**Adjacency Matrix:**
[
M_c =
\begin{pmatrix}
& a & b & c & d \
a & 1 & 0 & 1 & 1 \
b & 0 & 1 & 0 & 1 \
c & 1 & 1 & 1 & 0 \
d & 0 & 1 & 0 & 1
\end{pmatrix}
]
---
注意使用 mathtex 显示公式
中文输出](https://manimvideo.explanation.fun/video/cover/578443240306839553.png)
▶
Here’s the full markdown version of your provided images: --- ### (a) Adjacency List Representation **Graph:** Vertices: a, b, c, d **Adjacency List:** | Vertex | Adjacent Vertices | | :----: | :---------------: | | a | c | | b | d | | c | b, d | | d | b | --- ### (b) Adjacency Matrix Representation **Graph:** Vertices: a, b, c, d **Adjacency Matrix:** [ M_b = \begin{pmatrix} & a & b & c & d \ a & 0 & 1 & 0 & 1 \ b & 1 & 0 & 1 & 1 \ c & 0 & 1 & 0 & 1 \ d & 1 & 1 & 1 & 0 \end{pmatrix} ] --- ### (c) Adjacency Matrix Representation **Graph:** Vertices: a, b, c, d **Adjacency Matrix:** [ M_c = \begin{pmatrix} & a & b & c & d \ a & 1 & 0 & 1 & 1 \ b & 0 & 1 & 0 & 1 \ c & 1 & 1 & 1 & 0 \ d & 0 & 1 & 0 & 1 \end{pmatrix} ] --- 注意使用 mathtex 显示公式 中文输出
![### 5. Verify that the following program segment is correct with respect to the initial assertion *T* and the final assertion: [ (x \le y \land \text{max} = y) \lor (x > y \land \text{max} = x) ] ```plaintext if x <= y then max := y else max := x ``` --- **Solution:** Initial assertion *T* means this segment will always run and everything is always correct at the beginning of the segment. If ( x < y ) initially, *max* is set equal to *y*, so the left side of the final assertion ∨: ((x \le y \land \text{max} = y)) is true. If ( x = y ) initially, *max* is set equal to *y*, so ((x \le y \land \text{max} = y)) is again true. If ( x > y ), *max* is set equal to *x*, so the right side of the final assertion ∨: ((x > y \land \text{max} = x)) is true. Since ∨ is true whenever one or the other or both sides are true, the final assertion is always true and the program segment is correct. 注意使用 mathtex 显示公式
中文解题](https://manimvideo.explanation.fun/video/cover/578435076945432577.png)
▶
### 5. Verify that the following program segment is correct with respect to the initial assertion *T* and the final assertion: [ (x \le y \land \text{max} = y) \lor (x > y \land \text{max} = x) ] ```plaintext if x <= y then max := y else max := x ``` --- **Solution:** Initial assertion *T* means this segment will always run and everything is always correct at the beginning of the segment. If ( x < y ) initially, *max* is set equal to *y*, so the left side of the final assertion ∨: ((x \le y \land \text{max} = y)) is true. If ( x = y ) initially, *max* is set equal to *y*, so ((x \le y \land \text{max} = y)) is again true. If ( x > y ), *max* is set equal to *x*, so the right side of the final assertion ∨: ((x > y \land \text{max} = x)) is true. Since ∨ is true whenever one or the other or both sides are true, the final assertion is always true and the program segment is correct. 注意使用 mathtex 显示公式 中文解题
![### 5. Verify that the following program segment is correct with respect to the initial assertion *T* and the final assertion:
[
(x \le y \land \text{max} = y) \lor (x > y \land \text{max} = x)
]
```plaintext
if x <= y then
max := y
else
max := x
```
---
**Solution:**
Initial assertion *T* means this segment will always run and everything is always correct at the beginning of the segment.
If ( x < y ) initially, *max* is set equal to *y*, so the left side of the final assertion ∨:
((x \le y \land \text{max} = y)) is true.
If ( x = y ) initially, *max* is set equal to *y*, so ((x \le y \land \text{max} = y)) is again true.
If ( x > y ), *max* is set equal to *x*, so the right side of the final assertion ∨:
((x > y \land \text{max} = x)) is true.
Since ∨ is true whenever one or the other or both sides are true, the final assertion is always true and the program segment is correct.
注意使用 mathtex 显示公式](https://manimvideo.explanation.fun/video/cover/578434488337780737.png)
▶