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Solution of equation involving trig 
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#1
Jan2513, 06:57 PM

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Ok I have this equation which is [tex]x\sqrt{1y^2}dx = dy[/tex] that I have to solve.
I solved it and got [tex]\frac{x^2}{2} = arcsiny + c[/tex] I think that's right. Now the next part of the equation requires me to find the solution of this equation that passes through the point (0,1). That means I plug 1 in for y and 0 in for x? But [tex]arcsin(1)[/tex] would give me an answer in degrees or radians, wouldn't it? I'm solving for C, right? Shouldn't I just get a number and not an angle? Thanks. 


#2
Jan2513, 09:52 PM

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Well your general solution is saying that x is an angle isn't it?
(rearrange so that y is written as a function of x and you'll see).  aside: you can typeset trig and inversetrig functions in latex by putting a \ in front of the name, so \arcsin y gets you ##\arcsin y## instead of ##arcsin y##. Just sayin. 


#3
Jan2613, 03:35 PM

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Is there a keyword that you know of that I could Google that would bring up results showing me that process? 


#4
Jan2613, 07:11 PM

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Solution of equation involving trig
if u=arcsin(v) then (taking the sine of both sides) sin(u)=v. 


#5
Jan2613, 10:25 PM

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You should have learned long before differential equations that you do NOT use "degrees" for problems like these! Back in Calculus you learned that the derivative of sin(x) is cos(x) and that the derivative of cos(x) is sin(x). But you SHOULD have learned, also, that those are only true as long as x is in "radians".
What is really true is that the "x" in sin(x) or cos(x) is NOT an angle at all. When we are working with functions, the variables do NOT, except in specific applications, have any "units" at all they are just numbers. 


#6
Jan2613, 11:07 PM

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Thanks HallsofIvy: I was having a debate with myself whether I should point that stuff out now or later.
I think the uncertainty is understandable: if this function were derived from a physical situation where x is a measurement with units then there may be a problem with the derivation (like a hidden scale factor). There's no indication that this is the case here I know  but I think it's a good instinct. But I'd have to deal with the issue sooner or later. 


#7
Jan2713, 11:21 AM

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And in none of my calc classes did we even get to hyperbolic trig functions. So instead of degrees, finding the radian value of arcsin(1) and subtracting that from both sides would give me the true answer for C? Thanks for the responses. 


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