Solution of equation involving trig

[leroyjenkens]

Ok I have this equation which is x\sqrt{1-y^2}dx = dy that I have to solve.

I solved it and got \frac{x^2}{2} = arcsiny + c

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 arcsin(1) 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.

 

[Simon Bridge]

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).

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aside: you can typeset trig and inverse-trig functions in latex by putting a \ in front of the name, so \arcsin y gets you $\arcsin y$ instead of $arcsin y$. Just sayin.

 

[leroyjenkens]

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 inverse-trig functions in latex by putting a \ in front of the name, so \arcsin y gets you $\arcsin y$ instead of $arcsin y$. Just sayin.

You mean rearrange it so that I have y on one side and the rest of it on the other? I thought about doing that, but I have no clue how to.

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

 

[Simon Bridge]

You mean rearrange it so that I have y on one side and the rest of it on the other?

That's right.

I thought about doing that, but I have no clue how to.

um ... you use the fact that arcsine is the inverse function of sine.

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

 

[HallsofIvy]

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.

 

[Simon Bridge]

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.

 

[leroyjenkens]

um ... you use the fact that arcsine is the inverse function of sine.

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

Oh yeah, duh. I took a precalc and trig combined class, so I think that's why my trig is weak.

You should have learned long before differential equations that you do NOT use "degrees" for problems like these!

I don't think we ever had a constant argument in our trig functions in any of my calculus classes. But I figured having an answer in degrees made no sense.

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".

Actually I don't remember that at all. I just remember learning what derivatives were, and that the derivative of cos(x), for example, is -sin(x), but I don't remember any teacher specifically mentioning that is only true when x is in radians. Well, at least now I know.

And in none of my calc classes did we even get to hyperbolic trig functions.

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.

Yes, the 1 in arcsin(1) is just a number, but the answer to "what does arcsin(1) equal?" would not just be a number, and that's where my problem lies. I didn't want to just say "well it equals 90 degrees, so subtract 90 degrees from both sides", because that seemed wrong.
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.