Recent content by ciubba

D'oh! It should be \frac {102}{9.8} a= -0.25*102*cos(40)+32- \frac {32}{9.8}a -102sin(40) \rightarrow a=-3.82 Thanks!

Homework Statement Body A in Fig. 6-33 weighs 102 N, and body B weighs 32 N. The coefficients of friction between A and the incline are μs 0.56 and μk 0.25. Angle θ is 40. Let the positive direction of an x-axis be up the incline. In unit-vector notation, what is the acceleration of A if A is...

My calc. 2 book more or less only mentioned the hyperbolic functions to make integration easier, so, now that I have some free time, I'd like to explore the area further. Could someone recommend a good book on the subject or do I need to take more math first? A quick google search revealed...

Doing that gives me the right answer-- thanks!

Okay, I get D_1=4+y^2, \; D_2=(\sqrt{2}-y)^2, \; D_3=4+y^2 Thus, the objective function, their sum, is D_t=(\sqrt{2}-y)^2+8+y^2 D_t'=6 y-2 \sqrt{2} Which has a root at y=\frac{\sqrt{2}}{3} Unfortunately, that is the reciprocal of the book's answer. Where did I mess up?

Hmm, that's a good idea. So, v1=(-2,0) , v2=(0,sqrt(2)) , v3=(2,0) Edit: Still number crunching

Homework Statement An isosceles triangle has a base of length 4 and two sides of length 2sqrt(2). Let P be a point on the perpendicular bisector of the base. Find the location P that minimizes the sum of the distances between P and the three vertices. Homework Equations N/A The Attempt at a...

Why doesn't the root 25 change sign in the following picture: http://s10.postimg.org/qb9mxcsmf/1243124.png It's the same idea except that instead of \sqrt{x^4}=-x^2 we have \sqrt{x^6}=-x^3

I won't be learning about power series until the end of next semester, so it's likely that your explanation is a bit above my level for the time being. I'm still a bit confused by why it's - root 3. The only example problem in my book defines (x^6)^1/2= - x^3, but when they factor \sqrt{25x^6}...

Where did the O(v^2) come from? The formula seems to stop at 1+rv.

Can you link me the formula for that expansion? I've never seen it before.

Ah, you're right. it should be MINUS root 3 because \sqrt{x^4}=-x^2 I don't understand how you arrived at this part. If I factored out positive x^2, how would I evaluate x at - infinity? Wouldn't I just evaluate at positive infinity, which happens to produce the same result in this case?

Homework Statement Find lim_{x->- \infty} \; \frac{(x^6+8)^{1/3}}{4x^2+(3x^4+1)^{1/2}} Homework Equations N/A The Attempt at a Solution Factoring out \frac {(-x^6)^{1/3}}{-x^2} leaves me with \frac{(-1-8x^{-6})}{-4+(3+x^{-4})^{1/2 }} Taking the limit at infinity gives me...

If you decide to do general cases, a useful trick is to prove that as epsilon becomes small, so do the "bounds" of the inequalities. Good luck!

These are one of the hardest parts of calc, so don't feel bad. The pre-delta/epsilon definition of a limit of the form \lim_{x->a}F(x)=L is: F(x) is arbitrarily close to L for any x sufficiently close to a. The arbitrarily close part is |F(x)-L|<ϵ and it is arbitrary because we define epsilon...