Tangent Line on a Parametric Curve

suchgreatheig
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Homework Statement


A curve is defined parametrically by x=sin3t, y=cos3t, 0≤ t ≤ 2pi. Find the equation of the line tangent to the curve at the point defined by t=2pi/9.


Homework Equations





The Attempt at a Solution


?
 
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suchgreatheig said:

The Attempt at a Solution


?

Since you need to find the equation of a tangent, what do you need to find this equation?
 
I know that I need a derivative which I got as -3sinst/3cos3t, but I'm stuck from there.
 
suchgreatheig said:
I know that I need a derivative which I got as -3sinst/3cos3t, but I'm stuck from there.

so if dy/dx=-sin(3t)/cos(3t), what is the gradient when t=2π/9 ?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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