Recent content by Pawnag3
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Second Order Linear Differential Equation
Sorry! I want to solve for Θ, and Θ is a function of ε, β is just a constant. I just want a general solution please.- Pawnag3
- Post #3
- Forum: Calculus and Beyond Homework Help
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Second Order Linear Differential Equation
Hey, I'm not sure how to even approach this problem. It's not a simple ODE. Basically, I want to find the solution for Θ in terms of ε. The equation is \frac{1}{ε}*\frac{d}{dε}*(ε*\frac{dΘ}{dε})-β^{2}Θ=0 I tried to move the B^2 to the other side and I wasn't able to solve it that way. I...- Pawnag3
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- Differential Differential equation Linear Second order Second order linear
- Replies: 4
- Forum: Calculus and Beyond Homework Help
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Angle Help15 = arctan(2/x) - arctan (1/x)
Alright, sorry guys to waste your time, but I believe I figured it out. Thanks for the hint of "tanning" both sides. Instead of 15, it's supposed to be tan 15. So that, x/(x^2+2) = tan 15 x = 0.64 x = 3.08 (approximately) Thanks for the help!- Pawnag3
- Post #4
- Forum: Precalculus Mathematics Homework Help
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Angle Help15 = arctan(2/x) - arctan (1/x)
Yeah, sorry. I just misclicked the first time- Pawnag3
- Post #3
- Forum: Precalculus Mathematics Homework Help
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Angle Help15 = arctan(2/x) - arctan (1/x)
Homework Statement Basically, solve for x 15 = arctan(2/x) - arctan (1/x) Homework Equations tan (A-B) = (Tan A -Tan B) / (1+Tan A*Tan B) The Attempt at a Solution I really tried everything. My first step was to: Let y = arctan (2/x) Therefore, tan y = 2/x Similarly, u = 1/x Then, tan (y-u) =...- Pawnag3
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- Angle
- Replies: 4
- Forum: Precalculus Mathematics Homework Help
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What Is the Phenomenon of Changing Direction Towards a Moving Target Called?
Hi, I am just wondering to what this phenomen is called and how I can graph equations for it. Basically, we have a reference point moving North from origin (with direction vector of 0, k). I'll call it "s" Then, at "d" distance on the y-axis, we have another object "r". Now, reference point...- Pawnag3
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- Change Direction Time
- Replies: 1
- Forum: Introductory Physics Homework Help
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Momentum and Pressure and Forces
Umm, thanks, that clarifies everything :D But I have one question: 1) How did you make the formatting look so nice? :D I couldn't figure it out (but then again, I didn't spend too much time, under which heading is it? :P)- Pawnag3
- Post #3
- Forum: Introductory Physics Homework Help
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Momentum and Pressure and Forces
This is a theoretical question that I have, it might be somewhat elementary, or I might be missing something. Basically, we have two formulas: 1) We know that P = F/A By multiplying by d/d, we get: P = Work/Volume P = Energy/Volume P = 1/2 mv2 / V Therefore F/A = 1/2 mv2 / V And...- Pawnag3
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- Forces Momentum Pressure
- Replies: 3
- Forum: Introductory Physics Homework Help