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Find tangent

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,

Find and an equation of the tangent(s) to the curve at the given point
\(\displaystyle x=2\sin(2t)\), \(\displaystyle y=2\sin(t)\) \(\displaystyle \left(\sqrt{3},1 \right)\)
first we need to find the slope so we derivate
\(\displaystyle \frac{dy}{dt}=2\cos(t)\), \(\displaystyle \frac{dx}{dt}=4\cos(2t)\)
so we got \(\displaystyle \frac{dy}{dx}= \frac{2\cos(t)}{4cos(2t)}\)
we need to solve t for the given point but \(\displaystyle \frac{dx}{dt} \neq 0\)
so if we solve \(\displaystyle \sqrt{3}=2\sin(2t)\) we get that \(\displaystyle t \neq \frac{1}{2} \sin^{-1} \left(\frac{3}{2} \right)\) now we got that \(\displaystyle t= \sin^{-1}\left(\frac{1}{2} \right) = \frac{\pi}{6}\) so we get that our slope is \(\displaystyle \frac{\sqrt{3}}{2}\)
and if we put that in \(\displaystyle y-y_1=m(x-x_1)\) we get that equation of the tangent is
\(\displaystyle y=\frac{\sqrt{3}x}{2}-\frac{1}{2}\)
Is this correct?

Regards,
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hello MHB,

Find and an equation of the tangent(s) to the curve at the given point
\(\displaystyle x=2\sin(2t)\), \(\displaystyle y=2\sin(t)\) \(\displaystyle \left(\sqrt{3},1 \right)\)
first we need to find the slope so we derivate
\(\displaystyle \frac{dy}{dt}=2\cos(t)\), \(\displaystyle \frac{dx}{dt}=4\cos(2t)\)
so we got \(\displaystyle \frac{dy}{dx}= \frac{2\cos(t)}{4cos(2t)}\)
we need to solve t for the given point but \(\displaystyle \frac{dx}{dt} \neq 0\)
so if we solve \(\displaystyle \sqrt{3}=2\sin(2t)\) we get that \(\displaystyle t \neq \frac{1}{2} \sin^{-1} \left(\frac{3}{2} \right)\) now we got that \(\displaystyle t= \sin^{-1}\left(\frac{1}{2} \right) = \frac{\pi}{6}\) so we get that our slope is \(\displaystyle \frac{3}{2}\)
and if we put that in \(\displaystyle y-y_1=m(x-x_1)\) we get that equation of the tangent is
\(\displaystyle y=\frac{\sqrt{3}x}{2}-\frac{1}{2}\)
Is this correct?

Regards,
Hi Petrus, :)

You don't need to find \(t\), although doing that would also solve the problem. I think it would be a little elegant(and less tedious) if you try to find \(\cos{t}\) instead. Starting from the equation that you obtained for the derivative,

\[\frac{dy}{dx}= \frac{2\cos(t)}{4cos(2t)}=\frac{\cos{t}}{2(2\cos^2{t}-1)}\]

Note that by finding \(\cos{t}\) you can find the derivative. So how can you find \(\cos{t}\) starting from, \(x=2\sin{2t}\) and \(y=2\sin{t}\) ? Well divide \(x\) by \(y\) and use the double angle formula. According to my algebra you should get \(\displaystyle \frac{\sqrt{3}}{2}\) for the slope. Check your calculations again. :)
 

Petrus

Well-known member
Feb 21, 2013
739
Hi Petrus, :)

You don't need to find \(t\), although doing that would also solve the problem. I think it would be a little elegant(and less tedious) if you try to find \(\cos{t}\) instead. Starting from the equation that you obtained for the derivative,

\[\frac{dy}{dx}= \frac{2\cos(t)}{4cos(2t)}=\frac{\cos{t}}{2(2\cos^2{t}-1)}\]

Note that by finding \(\cos{t}\) you can find the derivative. So how can you find \(\cos{t}\) starting from, \(x=2\sin{2t}\) and \(y=2\sin{t}\) ? Well divide \(x\) by \(y\) and use the double angle formula. According to my algebra you should get \(\displaystyle \frac{\sqrt{3}}{2}\) for the slope. Check your calculations again. :)
Hello Sudharaka,
It was typo :eek:
ehmm I get \(\displaystyle \frac{2(2\sin(t)\cos(t))}{2\sin(t)}\)
that means \(\displaystyle \cos(t)=\frac{2\sin(t)}{4\sin(t)}\)
so this work :) But could you possible link me or tell why does this work :)? I never done this before(with solving for cos)! Elegant and very intressting!

Regards,
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hello Sudharaka,
It was typo :eek:
ehmm I get \(\displaystyle \frac{2(2\sin(t)\cos(t))}{2\sin(t)}\)
that means \(\displaystyle \color{red}{\cos(t)=\frac{2\sin(t)}{4\sin(t)}}\)
Yes you get,

\[\frac{x}{y}=\frac{2(2\sin(t)\cos(t))}{2\sin(t)}=2 \cos{t}\]

The equation highlighted in red is clearly wrong. I don't understand why/how you wrote it. :)

so this work :) But could you possible link me or tell why does this work :)? I never done this before(with solving for cos)! Elegant and very intressting!

Regards,
So finally we get,

\[\frac{x}{2y}=\cos t\]

Now find \(\cos{t}\) at the point \(\left(\sqrt{3},1 \right)\) and using that you can find \(\dfrac{dy}{dx}\).

I don't quite understand what you meant by "But could you possible link me or tell why does this work?". This method works because you can find the gradient of the tangent and hence the equation of the tangent line which is what you need to find for this question.