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...
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?
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}...
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...
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...