??Parametric equations??

[ziddy83]

hey whats up,
ok i have a question on differentiating parametric equations. The question says to find the equation of the two tangent lines to the curve. here's the equations...

x = 6 cos t
y= 2sin2t

Now, after i differentiate them, do i just plug in the corresponding point? , here's what i got after i differentiated them...

\frac {dy}{dx} = \frac {2cos2t} {(3)(-sint)}

after plugging in the corresponding point for t, i will have the slope, but is that slope for both of the equations? :confused:

[maverick280857]

Your reasoning is quite correct except for the fact that you will get not a 3 in the denominator but a 6. Yes after you apply the chain rule, all you have to do is substitute t.

[daster]

He'll get a 3 since y'=4cos2t, x'=-6sint and 4/6=2/3.

[ehild]

hey whats up,
ok i have a question on differentiating parametric equations. The question says to find the equation of the two tangent lines to the curve. here's the equations...

x = 6 cos t
y= 2sin2t

Now, after i differentiate them, do i just plug in the corresponding point? , here's what i got after i differentiated them...

\frac {dy}{dx} = \frac {2cos2t} {(3)(-sint)}

after plugging in the corresponding point for t, i will have the slope, but is that slope for both of the equations? :confused:


It is the slope of the tangent line of the CURVE.

ehild

[ziddy83]

How do i find the corresponding points? lol...

[lalbatros]

Hello

Now that you have the slope, you still need to write down the equation for the tangent:

y = slope*(x - 6cost) + 2sin2t
I guess this will answer this question fully.

I made a drawing of the curve, a choosen point and its tangent, in MS Excel.
I attach it to this post, compressed in a zip file.
By changing the reference point, you can see that the equation for the tangent is ok: it touches the curve smoothly.
There is a small difficulty to t=0 or pi because the slope becomes infinite (the tangent is vertical then).