parametric equations please help sigh

[JadaPsherman]

a curve is parametrically defined by the equations x= sin (2t) and y = cos (t) where t is measured in radians .
a) determine dy/dx and the gradient of the curve when t= pi/3
b) determine the x, y point when t= pi/3
c) hence determine the equation of the tangent to the curve pi/3

idk what to do all i have is x= sin 2t .. dx/dt = 2cos (2t)
y= cos t .... dy/dt = - sint

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dy/dx= dy/dt /dx/dt

im lost please help

 

[Steven G]

a curve is parametrically defined by teh equations x= sin (2t) and y = cos (t) where t is measured in radians .
a) determine dy/dx and the gradient of the curve when t= pi/3
b) determine the x, y point when t= pi/3
c) hence determine the equation of the tangent to the curve pi/3

idk what to do all i have is x= sin 2t .. dx/dt = 2cos (2t)
y= cos t .... dy/dt = - sint

dy/dx= dy/dt /dx/dt
im lost please help​

I am totally confused. You found dy/dt and dx/dt. You know that dy/dx =(dy/dt)/(dx/dt). So why not divide the two results and be done with part a.
Part (b): You have x(t) and y(t) so just find x(...) and y(...) and be done with part b.
Show us your work with a and b AND your attempt at part c and we will help you from there

 

[JadaPsherman]

I am totally confused. You found dy/dt and dx/dt. You know that dy/dx =(dy/dt)/(dx/dt). So why not divide the two results and be done with part a.
Part (b): You have x(t) and y(t) so just find x(...) and y(...) and be done with part b.
Show us your work with a and b AND your attempt at part c and we will help you from there

i got - sint t/2cos2t for dy/dx
but anyway thanks im very much confused just trying to follow an example from my text book. the topic wasnt taught because university is independent learning. but yet im confused . But il figure it out

 

[Steven G]

i got - sint t/2cos2t for dy/dx
but anyway thanks im very much confused just trying to follow an example from my text book. the topic wasnt taught because university is independent learning. but yet im confused . But il figure it out

If you want dy/dx to just have x's then you can solve x=sin(2t) for t. arcsin(x) =2t so t = arcsin(x)/2.
x(t) = sin(2t), so x(pi/3) = sin(2pi/3)= sqrt(3)/2. Do the same for y(t)
Part c: after you get x(pi/3) and y(pi/3) you have one point in the x-y plane. You also have dy/dx in terms of t. Plug in pi/3 for t and then you have dy/dx at t=pi/3. That is yopu have the slope of the line which you want the eq for. Armed with a point on this line and the slope of this line you can find the eq of this line.