Recent content by Bear_B

Hurkyl, it just felt like the problem started "boring" and stayed "boring." Yeah, it ultimately seems to come down to a comparison of integer-value solutions for the box. Your way seems more "exciting" though.

Thanks. That's what I originally thought from my knowledge of what a derivative is, by definition. However, as previously stated, I have little to no experience formally working in discrete mathematics and was unsure of my comprehension of a derivative and if derivatives might apply to discrete...

I worked it out myself. Everything is correct. I would add a couple of notes about part c: You might want to switch your calculator to radian mode if you used a calculator to find arctan (5/12) just to be on the safe side. IMHO, there is never really any reason to choose to work in degree mode...

I have solved this problem but still have a question about it (problem and my solution posted below). What I wanted to do was express the problem as a function and use an optimization technique that would require taking the first derivative of the equation. The problem I ran into is: if using...

Thanks. Now I understand the concept that the Force (F) is being taken with respect to distance (r) determining the rate of change of F as r changes for two objects with mass m and M. Since dF/dr is asking how F changes when r changes it has to assume that the m and M are constants unless...

F = GmM/r2 dF/dr = GmM [ d/dr (r-2)] = GmM (-2r-3) = GmM (-2/r3) dF/dr = -2GmM / r3 ok, so here is the answer, but why are m and M constants?

Ok, I had seen that result worked out elsewhere when I couldn't get on here. Can you please tell me why m and M are constants? I thought the equation to determine the gravitational force was for an unknown mass m and an unknown mass M at a unknown distance r from each other...knowing that m...

Homework Statement Newton's Law of Gravitation says that the magnitude (F) of the force exerted by a body of mass (m) on a body of mass (M) is F = (GmM)/(r^2) where G is the gravitational constant and r is the distance between the bodies. Find dF/dr and explain it's meaning. What...

Simply put, if you can differentiate it, it has tangent lines. So you can have tangent lines for things that aren't functions too.

Yeah, I caught that mistake as I went back through my work...I idn't even realize that I had done it wrong so many times until I did it right 2X! There was only one guy in my Calculus class that was able to get this so he helped me out. I will post the solution once I have it typed up. The ace...

f'(x) is the instantaneous rate of change at any x on the graph of f(x). In other words, f'(x) is the exact slope at any point on the graph of f(x). First, take the derivative of f(x). Then from there you can evaluate for any x (in your case x=2) to find the slope of your tangent line.

Well, If my first derivative is wrong, then all subsequent work will be wrong. So starting from scratch, here is my work for the derivative of the original parametric equation y(x^3)-2(x^2)(y^2)+x(y^3)=192: dy/dx = y'(x) dy/dx [y(x^3)-2(x^2)(y^2)+x(y^3)=192] d/dx y(x^3) - 2 [d/dx...

HALP! I have already killed a forest trying to work this one out on paper. Homework Statement a. Use Implicit differentiation to find the equations of the horizontal tangent lines to the parametric equation: yx^3-2x^2y^2+xy^3=192. b. Use Implicit differentiation to find the equations...

Ok, I got the answer. The tip on substitution was what really allowed my to break this one open. Since I think this is a great problem that requires a creative strategy (and possibly multiple attempts) to get the simplest answer, I am going to post the rest of my work here. The result of...

Homework Statement Find y''(x) of the parametric equation 9x^2+y^2=9 using implicit differentiation. Homework Equations I already came up with y'(x) = -9x/y The Attempt at a Solution Here is what I have for y''(x) so far y''(x) = d/dx (-9xy^-1) =-9(d/dx)(xy^-1)...