Archived May, 2026.

Questions on Review for Exam 1

mark

You can ask questions about the review for exam 1 by posting them here!

User 008

For question nine we end up getting \pi^2, should we know what number this is or should we just leave it as \pi^2?

mark

Well, question nine asks you to compute the TNB frame for the parameterized motion defined by

\vec{p}(t) = \langle t,\cos(\pi t), \sin(\pi t) \rangle

Certainly, the final answer is not just \pi^2 - right?

I guess you mean that you stumble upon a \pi^2 along the way and you're not sure what to do with it. My response to that would be that yes, you should just leave \pi^2 as \pi^2. It's a perfectly good, irrational number that's just a little less than 10.

User 008

That's what I meant! Thank you!

User 009

If anyone happens to know this late, is question 2 talking about the total change, the rate of change, or something else? (the first two are what I found about displacement vectors in the book, mostly)
Or is it just finding what other vector when combined with F makes d?

mark

Problem #2 on the review sheet asks the following:

Suppose that a force vector \vec{F} = \langle 1,3,-1 \rangle induces a displacement \vec{d} = \langle 2,2,2 \rangle. Find the work done by \vec{F} in this process.

The formula presented in class, as described also in section 11.3 of our text, is that

W = \vec{F}\cdot\vec{d}.

Thus, this problem is literally a long winded way of asking you to compute the dot product of the two vectors \vec{F} and \vec{d}.

User 009

Ah, ok, thank you! Sorry - I should've tried looking for 'force' instead of 'force vector' in the textbook too.

mark

No problem at all!!!

User 035

For number one, I feel as if I might be mixing up the projection equation. Is this correct?

\operatorname{proj}_{v} u = \frac{u \cdot v}{v \cdot v} \, v

I always remember it being u at the bottom and not v? Is it because we are trying to find the projection u onto v?

mark

Well, you could project u onto v or you could project v onto u so it depends which one you're doing. The one that you're projecting onto is the one in the subsript of the proj. Thus, the formula you wrote projects u onto v. If you wanted to project v onto u, then you'd have to switch all the u's and v's.

The problem, of course, asks you to break u up into two vectors - one in the direction of v and one perp. So I guess you need to project u onto v, as you've written.

mark