When dealing with several numbers and long equations, it's common to make careless arithmetic mistakes that give the wrong answer. I was wondering if anyone had tips to catch these mistakes, or even better avoid them more often. (aside from the obvious checking your work- that's a must)
[Math] How to avoid arithmetic mistakes
arithmetic
Related Solutions
It is hard to give anything more than general advice, which is to go through practice papers and calculus type questions. Doing calculus type questions refreshes you on the techniques and doing enough of them gets you faster, while doing practice papers also gets you used to using in-exam methods, taking advantage of the nature of these multiple choice questions.
For example, rather than finding the actual equation of the plane and plugging them in, I would have computed the gradient vector, which is $$ ( -2z^2+6xy, 3x^2+2zy, y^2-4xz)\mid_{(1,1,1)}= (4,5,-3).$$ Any point $P$ that lines on the plane in question must satisfy $(P- (1,1,1))\cdot (4,5,-3)=0$ so subtract 1 from each coordinate and just compute dot products until you find one that is $0.$ You wouldn't even have to compute (a) or (d) properly, you'll notice by signs that the dot product will be positive. It is quite similar to finding the equation of the plane first and plugging them in, but seems to save at least a little bit of time.
General Tips:
- Don't spend more than 2 minutes on any question, after your two minutes if you haven't got the answer, you should at least be able to rule out 2 answers. Mark off those 2 answers for now.
- Dimensional analysis is a good trick. There are some calculus type rates questions (A car uses this much fuel per mile when travelling at some speed...). The options might be some integrals with various integrands, and only one of them even has the correct units.
- If you can rule out 2 options, good, and if you can rule out 3 even better, guessing is benefical. Don't guess if you can't rule anything out, because there is only 1/5 chance of getting the right answer, and an incorrect answer subtracts 0.25.
- Some things they seem to like testing, so worth revising are: Green's theorem, and using it to find areas, Volumes of surfaces of revolution, finding volumes by setting up double integrals, Decompositions of finite abelian groups, basic statistics and probability was something I had to revise, Residue theorem, Cauchy-Riemann equations. Also, many of the incorrect answer choices for questions on Lebesgue theory seem to come from incorrectly thinking uncountable implies 0 measure. Keep in mind the Cantor set for these questions.
Good luck!
A calculus exam is designed to test whether you know calculus, not its prerequisites. Your professor probably assumes you are well-versed enough in arithmetic and algebra that you will not make substantial "simple" errors. If you know the calculus,[*] then, but you're consistently making errors in arithmetic and algebra on the exam, then I'd suggest you need to change the focus of your studying.
Try digging up an old arithmetic or algebra textbook and working through the exercises. Do 10 or 20 at a time, then stop and grade yourself on accuracy. Keep doing this until you consistently score perfectly on these mini-exams. You're essentially drilling these skills into your subconscious so that you intuitively know how these manipulations work, and if you make a mistake, you immediately catch it.
When you're studying calculus proper, you probably go over old exam problems and homework exercises and do them. When you check your solutions to these and run across arithmetic errors, don't assume they were "just" silly mistakes and move on. You've identified that your problem on exams is exactly this, so you need to figure out where you're going wrong with the algebra/arithmetic in your problem-solving process and start working to correct it.
Here's a sports analogy. Imagine you're on a basketball team and you've got a great grasp of strategy and court sense --- you know where your teammates are, where the defenders are, who's open, and who's not. However, when you have the ball, you can't seem to control it. Your dribbling is wild and when you pass or shoot, it doesn't go where you thought you put it. In practice, what do you do? You spend hours just dribbling and then you line up and take a thousand shots at the basket. When you're playing scrimmage and the ball slips out of your hands, do you ignore it? No, you think about why you lost control and how to do better the next time. In the same way, you need to drill yourself on the prerequisites for calculus and be more self-critical when you don't understand the algebra in a problem.
[*] By this I mean that when you see a problem, you know very quickly how to solve it. For instance, if you see "Find the area between $f$ and $g$," you know immediately that you need to find the intersection points of $f$ and $g$ and the integrate $|f-g|$ between those intersection points.
Best Answer
There are certain quick methods called sanity checks which will catch most (but not all) arithmetic errors. One common one is to replace each number with the sum of its digits, which is the "casting out nines" method mentioned in Robert Israel's answer. To check a computation, say $567\times 894=506,898$, we replace $567$ with $5+6+7=18$ and $894$ with $8+9+4=21$, and then replace each of these with the sum of their digits to get $9$ and $3$ (in general we keep doing this until we get down to $1$ digit), while on the other side we get $5+0+6+8+9+8=36$ and then $3+6=9$. We then check that $9\times 3=9$ after casting out nines on both sides, and so our answer is probably right (though not necessarily). This method is called "casting out nines" because it ensures that whatever answer you are checking differs from the actual answer by a multiple of nine (hopefully $0\times 9$).
However, this method has a serious drawback: if the answer you are checking is correct except for having the digits switched around (a relatively common error) the method will not catch the error. A remedy for this is to use "casting out elevens" where you take the alternating sums of the digits instead of the sums, such that the last digit is always added rather than subtracted. In our previous example, this becomes $5-6+7=6$, $8-9+4=3$ and $5+0-6+8-9+8=-4$. Here we have to be a little careful: we want to take the equation $6\times 3-(-4)=0$, cast out elevens (take the alternating sum) and verify that the the resulting equation holds, which in this case it does. We move everything to one side so that we don't have to work with numbers of different signs ($6\times 3=-4$ is true $\bmod 11$, which is what matters, but it is not obvious how to cast out elevens to show this). This ensures that whatever answer you are checking differs from the actual answer by a multiple of eleven (hopefully $0\times 11$), hence the name.
Edit: These methods can both be made rigrous with modular arithmetic. The first simple checks that an equation holds $\bmod 9$, and adding the digits comes from the fact that $$\begin{eqnarray} d_n\cdots d_1d_0 &=& \sum\limits_{i=1}^n 10^id_i\\ &\equiv& \sum\limits_{i=1}^n 1^id_i (\bmod 9)\\ &=&\sum\limits_{i=1}^n d_i \end{eqnarray}$$ while the second checks that an equation holds $\bmod 11$, and the alternating sum of the digits comes from the fact that $$\begin{eqnarray} d_n\cdots d_1d_0 &=& \sum\limits_{i=1}^n 10^id_i\\ &\equiv& \sum\limits_{i=1}^n (-1)^id_i (\bmod 11) \end{eqnarray}$$
Credit where credit is due: I believe I read about this years ago in a question to Dr. Math from an elementary school teacher who had been teaching the method and wanted to know how it worked.