Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have this implicit function $$y=f(x) \iff \sin(x+y)=k \sin(x), \quad$$ where $k>1$ is a constant.

I would like to know how a small variation in $x$ propagates on $y$.

I think I need to do an implicit differentiation but then it is not so clear to me how to solve the problem.

So the derivative of the LHS is

$$\frac{\mathrm{d}}{\mathrm{d}x}(\sin(x+y)) = \cos(x+y)(1+\frac{\mathrm{d}y}{\mathrm{d}x})$$

and the derivative of the RHS is

$$\frac{\mathrm{d}}{\mathrm{d}x}(k\sin(x)) = k\cos(x)$$

And solving for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ gives

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{k\cos(x)-\cos(x+y)}{\cos(x+y)}$$

and now how can I continue?

Thank you.

share|improve this question
1  
How about just rewriting this as $y=\sin^{-1}(k\sin(x)) - x$ and differentiating the right side? –  Thomas Andrews Jul 6 '11 at 15:55
    
@Thomas Andrews I do not know how to explain it but I am not so comfortable with the inverse trigonometric functions... :-) –  uvts_cvs Jul 6 '11 at 16:05
add comment

2 Answers

up vote 3 down vote accepted

It depends on the result you would like to obtain. First of all, $$ dy = \left(\frac{k\cos{x}}{\cos(x+y)}-1\right)dx $$ so the propagation of $dx$ on $dy$ depends on $x$ as well as on $y(x)$. You can make better if recall that $\cos^2(x+y) + k^2\sin^2x = 1$ which can help you to rewrite the denominator and obtain the dependence only on $x$.

share|improve this answer
    
+1, thank you, it was the dependency on y(x) that puzzles me a bit. –  uvts_cvs Jul 6 '11 at 9:31
    
@uvts_cvs: note that in general this trick will not work and you will have $dy = f(x,y)dx$. What puzzled you? –  Ilya Jul 6 '11 at 9:41
    
I was puzzled because I was not expecting the full dependency on $x$ and $y(x)$ but you are obviously right. –  uvts_cvs Jul 6 '11 at 9:47
    
Let's say I am interested at $x=0$ where we have $y=0$. How can I plug in these numbers into the $\frac{dy}{dx}$? I get $\frac{dy}{dx}=k-1$ and now how can I interpret this result? A small (how much small?) change in $x$ will be amplified by $k-1$? –  uvts_cvs Jul 6 '11 at 16:10
    
I have an advise for you to open $\sin(x+y)$, denote $z = \sin y$ and find $z(x)$ if you like to deal with explicit dependencies. Note that for $x=0$ you have $y = \pi k$. –  Ilya Jul 6 '11 at 16:25
add comment

Rewrite your equation as:

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{k\cos(x)}{\cos(x+y)}-1$$

Then notice that:

$$\cos(x+y) = \pm\sqrt{1-\sin^2(x+y)} = \pm \sqrt{1-k^2\sin^2(x)}$$

So:

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\pm \frac{k\cos(x)}{\sqrt{1-k^2\sin^2(x)}}-1$$

Where the sign is determined by the sign of $\cos(x+y)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.