Calculus in Three Dimensions (21-259) — Feedback on Homework 1

Homework 1 was due on Tuesday 2nd September 2014. Questions 5, 6, 8, 9 and 13 were graded.

General comments. This homework was mostly done very well! A very prevalent phenomenon was people rounding their answers and then using their rounded answers in subsequent work. This is a bad bad bad thing to do; it's not just that I'm a pedantic mathematician, it's that the answers you get in this way are wrong! I didn't penalise people for rounding too early this time, but in the future I might, because this is the kind of error that can cause bridges to collapse. If you want to give a rounded final answer then that's fine, but leave an exact answer there too. (There may of course be exceptions, like an exact form might not be available, but this certainly wasn't the case on this homework.)

Question 5. Most errors in this question were algebraic, and I didn't penalise these errors very heavily. The big problems occurred when people couldn't work out how to begin the question. The translation of 'the distance from $P$ to $A$ is twice the distance from $P$ to $B$' is $$\left|\overrightarrow{PA}\right| = 2\left|\overrightarrow{PB}\right|$$ Letting $P=(x,y,z)$ and solving this equation yields the equation of a sphere, from which you can read off its centre and radius.

Question 6. This question was done correctly by most people. A common slip-up was people noticing that $|\mathbf{v}| = 6$ and then dividing by $\sqrt{6}$ instead: this is tempting, because normally the magnitude of a vector is a square root, but in this case it is not so!

Question 8. There were lots of arithmetic errors in parts (a) and (b) but no conceptual errors, so I'm not too worried about that. The main problem in part (c) was people finding the scalar projection, or $\mathrm{proj}_{\mathbf{b}}(\mathbf{a})$, instead of what is asked.

Question 9. This was mostly done well by everyone who used the magnitude of a cross product. A common error was people forgetting to divide by $2$: remember, the cross product gives you the area of the corresponding parallelogram.

Question 13. The main thing to notice in this question is that the direction of the force is given, but not its magnitude. So if $\mathbf{F}$ is the force vector, all we know is $\mathbf{F} = k\langle 0,3,{-4}\rangle$, where $k$ is some nonnegative scalar, and our job is, essentially, to find the value of $k$ (and hence the magnitude of $\mathbf{F}$). The correct formula to have in mind here is $$|\vec{\tau}| = |\mathbf{r}||\mathbf{F}|\sin\theta$$ We know $|\vec \tau| = 100$, we know $\mathbf{r} = \langle 0,{0.3},0\rangle$, and we can work out $\theta$.

Where most people messed up was in finding $\theta$. Lots of people found the value of $\tan \theta$ using a right-angled triangle, and then solved for $\theta$. The problem with this was that they found a rounded value of $\theta$, not an exact value, and this led to inaccurate final answers. In fact, you can read off from the right-angled triangle that $\sin \theta = \frac{4}{5}$; alternatively, you can use the dot product to find $\cos \theta = \frac{3}{5}$ and then use the fact that $\sin^2 \theta = 1-\cos^2\theta$. If you did this then you'll have got the correct answer.

Also some people seem to think force is measured in Newton-metres ($\mathrm{N}\,\mathrm{m}$); that's the unit of work, not of force, which is just Newtons ($\mathrm{N}$).

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