Recent content by ardentmed

So just for the record, in a nutshell, I optimized it, figured out that it can't be minimum AND maximum at the same point, so I tested the endpoints as well (since, after all, it is a continuous function) at x=0 and x=2 (and x=~.7 for the optimized value).

Alright, so I found the endpoints and re-did the question from scratch, taking into account that the equilateral triangle's angle is $\pi$/3 . I also found the endpoints: I think I'm on the right track. What do you think, Mark? Thanks in advance.

Yes, I wrote it last week. Thanks to all the pointers you guys gave me with these problem sets, I'm sure I aced it. No doubt about it. Remember that 2m long string question? Something similar (but not exactly the same question) came up as one of the tough long answer problems and I'm sure I...

Thanks for the clarification. I checked all of my answers and here are my results: http://i.share.pho.to/d11a1ffa_o.png http://i.share.pho.to/a4cb5658_o.png http://i.share.pho.to/f5213c0d_o.png Thanks in advance.

I'm not too sure if I simplified these correctly. Am I on the right track? http://i.share.pho.to/c1134604_o.png Thanks again.

Edited for brevity.

Alright, so if T ~ 1.24, that gives us one of the areas, then, This represents the maximum total area, since this was the global maximum. As this function has no absolute maximum (or minimum?), the maximum (and minimum?) total area must occur at an endpoint (either L = 0 or L = 2). Checking...

Thanks for catching my mistake. I made a slight typo at the beginning of my work. I ended up getting: 0= (18T-36+√3*T)/36π T=36/(18+√3) T~1.824443. Am I on the right track? Thanks again.

Thank you so much for the thorough explanation. However, I still don't quite grasp why graphing the acceleration and velocity functions and finding which parts are both positive and which are negative doesn't work. What is the flaw in my logic? Thanks again.

Alright, so knowing that C=2-T, the following substitution can be made: A = (2-T)^2 / 4$\pi$ + (√3*[T^3])/36 Optimizing results in: A' = (T-2)/2$\pi$ + √3(T^2)/12 Thus, 0=T^2 * $\pi$√3 + 6T - 12 Using the quadratic formula, I computed ~1.032742 and ~ -2.135399. Am I on the right...

So it should be: Speeding up from (0,3) and slowing down from (-$\infty$,0)u(3,$\infty$). But since the function starts at t=0, it is slowing down from (3,$\infty$) Is that correct? Thanks again.

Given that: $$|v(t)|=\begin{cases}6t^2-18t, & t<0 \\ 18t-6t^2, & 0\le t\le3 \\ 6t^2-18t, & 3<t \\ \end{cases}$$ f′(t)=12t-18 for t<0 f′(t)=18-12t for 0<t<3 f′(t)=12t-18 for t<3 The first one, when substituting in an arbitrary value with the limitation in mind gives a negative velocity...

Knowing that the total perimeter of the two shapes is equal to 10, we can add the perimeters together and use the perimeter function as a system of equations. Thus, we can isolate either variable on one side, and use the resulting value and insert it into the area function and then proceed to...

If Pc = 2$\pi$r, then r= P/2$\pi$ Thus, Ac = $\pi$(P/2$\pi$)^2 And if Pt = 3L Then At = P/3 * 1/2 * √ (3)/2 Am I on the right track? Thanks again.

$f^{\prime}(t)$? = 6t(t + 3)Thus, t=3.