implicit differentiation: xsin(xy)=y crosses x-axis at 2 pts

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rae27

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A curve xsin(xy)=y crosses the x-axis at two points (the image is reflected around the y-axis). At these two points of intersection the tangent line to the curve is vertical. Find the x-coordinate at each of the two points the curve crosses thed x-axis.

I'm stuck. I tried implicit differentation and got y'=sin(xy)/1-xcos(xy), then tried to substitute it into the equation 0=mx+b. Zero because the I'm looking for the x coorinate when y=0 and b=0. I keep getting x=0. Where am I going wrong? Did I not even start the problem correctly. I'm studying immplicit defferentiation.

Thanks
 
Re: implicit differentiation help

rae27 said:
A curve xsin(xy)=y crosses the x-axis at two points (the image is reflected around the y-axis). At these two points of intersection the tangent line to the curve is vertical. Find the x-coordinate at each of the two points the curve crosses thed x-axis.

I'm stuck. I tried implicit differentation and got y'=sin(xy)/1-xcos(xy), then tried to substitute it into the equation 0=mx+b. Zero because the I'm looking for the x coorinate when y=0 and b=0. I keep getting x=0. Where am I going wrong? Did I not even start the problem correctly. I'm studying immplicit defferentiation.

Thanks
Click to expand...

When you have a vertical line - what is the slope of that line?
 
Re: implicit differentiation help

So that would make y'=0=sin(xy)/1-cos(xy). Right? I tried substituting and I still get x=0 which isn't right. So where do I go from 0=sin(xy)/1-cos(xy)?

Thanks
 
Re: implicit differentiation help

I think your derivative is off kilter a wee bit.

In your last post, what you have set to 0 would find the horizontal tangent.

To find the vertical, the denominator = 0.

\(\displaystyle xsin(xy)-y=0\)

\(\displaystyle x^{2}y'cos(xy)+xycos(xy)+sin(xy)-y'=0\)........[1]

\(\displaystyle y'=\frac{-sin(xy)-xycos(xy)}{x^{2}cos(xy)-1}\)

Now, to find the vertical tangents: In [1], eliminate all the terms that do not have y', then divide out the y' in what remains.

We have

\(\displaystyle x^{2}cos(xy)-1=0\)

Notice, this is the denominator in y'.

Solve for y:

\(\displaystyle y=\frac{cos^{-1}(\frac{1}{x^{2}})}{x}\)............[2]

Now, sub this back into the original and solve for x. This will be the x coordinates of the vertical tangents.

To find the y coordinates, sub those values into [2]. Of course, they'll be 0.

Can you see why this is?. As the slope 'approaches infinity', y' gets larger and larger making all but what remains insignificant.

We are taking the limit \(\displaystyle y'\to\infty\)

Your x values should be nice integer solutions.
 
Re: implicit differentiation help

[attachment=0:6m7vycrl]rt.jpg[/attachment:6m7vycrl]

This is the implicit graph of xsin(xy) =y from Maple 8. As one can see, the slope at (1,0) and (-1,0) is vertical.
galactus, I'm new at this game, any way to make the graph more appealing?
 

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Re: implicit differentiation help

No, Glenn. I ran the same thing in Maple 10 and it pretty much looks the same.

Maybe change the line to another type. That is, make it wider or another color.

Perhaps if we embolden the line it would look better.
 
Re: implicit differentiation help

Maple 10, huh. I'm just getting acclimated to Maple 8 and already it is obsolete; this modern technology is too fast for me.

As an example, I was at an electronics store the other day and the sales person told me that if you can buy it (without a hint

of candor) it is already obsolete. Perhaps that is why they call this "the information age". Anyways, thanks galactus.
 
First, thanks, you cleared a couple things up for me, but i'm still a bit confused. Why do we eliminate all the terms that don't have y' to get x^2cos(xy)-1? Also i was able to solve for y and get y=cos^-1(1/x^2)/x, but when I substituted into y=xsin(xy), i got as far as cos^-1(1/x^2)=x^2sin(cos^-1(1/x^2)) and got stuck. I would end up with a cos I couldn't get rid of. Where did I go wrong?

Thanks.
 
Note that \(\displaystyle sin(cos^{-1}(\frac{1}{x^{2}}))=\frac{\sqrt{x^{4}-1}}{x^{2}}\)

So, when we make the subs we eventually whittle it down to:

\(\displaystyle \frac{\sqrt{x^{4}-1}-cos^{-1}(\frac{1}{x^{2}})}{x}=0\)

\(\displaystyle \sqrt{x^{4}-1}-cos^{-1}(\frac{1}{x^{2}})=0\)

\(\displaystyle \sqrt{(x^{2}+1)(x+1)(x-1)}=cos^{-1}(\frac{1}{x^{2}})\)

What x value(s) make these 0?.

\(\displaystyle cos^{-1}(\frac{1}{x^{2}})=0\Rightarrow x=-1, \;\ x=1\)

\(\displaystyle \sqrt{(x+1)(x-1)}=0\Rightarrow x=-1, \;\ x=1\)

Those are your x coordinates for the vertical tangents.
 
You mean for the \(\displaystyle sin(cos^{-1}(\frac{1}{x^{2}}))=\frac{\sqrt{x^{4}-1}}{x^{2}}\)?.

Yes, that is correct.

\(\displaystyle sin(cos^{-1}(x))=\sqrt{1-x^{2}}\)

\(\displaystyle \sqrt{1-(\frac{1}{x^{2}})^{2}}\)

\(\displaystyle =\sqrt{1-\frac{1}{x^{4}}}\)

\(\displaystyle =\sqrt{\frac{x^{4}-1}{x^{4}}}\)

\(\displaystyle =\frac{\sqrt{x^{4}-1}}{x^{2}}\)
 
Another approach: We are given the equation xsin(xy)-y=0 and are also told that when the curve of the above equation crosses the x axis at two points, the slope at these two points of the equation is vertical.

Find the x coordinates of these two points.

First, we know that whatever the x coordinates are, y = 0.

The problem here is to find y', the derivative of the equation, using implicit differentation, which galactus has already done for us. Hence y' = [sin(xy)+xcos(xy)]/[x^2cos(xy)-1].

Now, in order for the slope to be vertical, it must approach infinity, which implies the demoninator of the slope must approach 0.

Ergo, x^2cos(xy)-1 = 0 implies y = [arccos(1/x^2)]/x, but y = 0 at the two points the curve intersects the x axis, so 0 = arccos(1/x^2), which implies x = + or - 1. (arccos(1) =0)

Hence, we are done, the points being (-1,0) and (1,0).
 
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