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#### 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

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,

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,

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