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# System prepended metadata

title: Modified Policy
tags: [teaching]

---

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tags: teaching
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# Modified Policy

This site can be visited by http://mark.honeymath.com


# Table of contents

- [Changes -- (changes of marking policy)](#Changes)
- [Pattens --（Some other correct approach by students）](#Parterns)
- [Error Patterns -- (Some common error and my grading )](#Error-patterns)
- [Simple ways of grading -- (For our convenience)](#Simple-ways-of-grading)





## Changes

**[🔼Back to top](#Modified-Policy)**
#### Not take off points for Units

I ignored unit error since test says confusing units like

$$0.0867 gm/cm^3$$

 It may understood it as $mg$ or $gram$, or $g\times m$. So I give them marks even if they do not write unit or write wrong unit.

#### For 2 Point Final answer -- I Change to --> 2 point of plug in correct value

I see lots of incorrect final answer due to only forgot to derive with t. It is unfair to take 4 points off. So I change this part of <span style="color:red">Evaluation</span>

There are two values, one is t, the other is l. each variable is 1 point. **Typically I only take 1 point off if they forgot plugin value or plugin wrong value**

## Parterns

**[🔼Back to top](#Modified-Policy)**
### Approach the problem by $55+2t$

Some student says

$$
m(t)=0.0867(55+2t)^3
$$

and then they take derivative by $t$. 
**It got at least 4 marks** because it is a substitution of our step of finding $dm/dl$ and $dl/dt$ and chain rule.

**Usual error**: Seems they always plug $t=48$ into the expression, which I took off 1 point because they should plug in $t=0$.

### Implicit method

Some says
$$
0.0867 = \frac{m}{l^3}
$$
and take derivative on both side to get
$$
0=\frac{\frac{dm}{dt}l^3-2l^2m\frac{dl}{dt}}{l^6}
$$

This is a correct approach. It uses the quotient rule. **It got at least 4 marks** because it is a substitution of our step of finding $dm/dl$ and $dl/dt$ and chain rule.

### Clever Chain rule

Some clever people says
$$
\frac{dm}{dt}=\frac{dm}{dl^3}\frac{dl^3}{dt}
$$
This is correct, it is considered to uses the chain rule. 
**Usual error**: Seems they did not realize how to calculate $\frac{dl^3}{dt}$, I took **2 points off** and mark that $\frac{dl^3}{dt}=3l^2\frac{dl}{dt}$

## Error patterns

**[🔼Back to top](#Modified-Policy)**

### Does not read correct information to obtain relation of m and l

Students write

$$
m=0.0867 l^2 \qquad \text{or} \qquad m=0.0867l \qquad \text{or} \qquad m=\frac{0.0867}{l^3}
$$

I forgave them, **I combined this error as compuational error**, and just assume this expression as the question and check their steps.

### Other forgiveness

Because I did not grade their units, any answer as a multiple of 10 power is considered correct. Like 1570, 15.7......

## Simple ways of grading

**[🔼Back to top](#Modified-Policy)**

- If there is no derivative appears. At most 2 points. Only: **1 point if they set up variable**, **1 point for they realize $m=0.0867l^3$**

- If 157 appears, it is highly possible is a **7 marks** exam.

- If there is no $t$ appears, then they must forgot the chain rule so took **2 points** off


### Warning

I will write more if I remember more.