.
use below to find the slope : \[\dfrac{dy}{dx} = \dfrac{dy}{dt} / \dfrac{dx}{dt}\]
click that link for details ^
I kinda looked at that before but I'm confused how to find the derivative of this equation; I've never found derivatives of parametric equations before
easy, x=cos(3t) dx/dt = ?
-3sin(3t)?
yep y=sin(5t) dy/dt = ?
5cos(5t)
and just divide them?
exactly !
and plug in pi/9 for t, right?
Because that's what I'm doing
yes !
and that's just going to be the slope, okay :) how do I find the equation after though? like what formula do I have to use? Is it still point-slope? Or is that different
\[\dfrac{dy}{dx} = \dfrac{dy}{dt} / \dfrac{dx}{dt} = -\dfrac{5\cos (5t)}{3\sin (3t)} \]
plugin t = pi/9 above to get the slope at t = pi/9
yup, I did that and got (10sin(pi/18))/3sqrt(3)
yes u wil still use the same point-slope formula
How will I use the point slope formula with this? Since there's only one coordinate?
how did u get (10sin(pi/18))/3sqrt(3) ?
http://www.wolframalpha.com/input/?i=%285cos%285*pi%2F9%29%29%2F%28-3sin%283*pi%2F9%29%29 is this like this?
is it*
lol okay, u have simplified it...
yes thats ur slope, what about point ?
Like, how do I plug it into the point slope formula?
What will the point slope formula look like for this
Like I know m would be the answer I got for the slope, but the other values I dont know
pont-slope form : \(y - y_1 = m(x-x_1)\)
u already knw \(m\)
to get the point, plugin \(t = \frac{\pi}{9}\) in : x=cos(3t), y=sin(5t)
Thanks! I got y=0.3x+0.8 for the tangent line (which I hope is correct)
I have to estimate it to the nearest 0.1 units
http://www.wolframalpha.com/input/?i=sin%285pi%2F9%29+-+%280.3cos%283pi%2F9%29%2B0.8+%29
not getting 0 ?
is that due to rounding ?
So I'm not sure
looks good :)
Why does wolfram alpha give a different answer though?
wolfram answer is same as urs
that link was just for checking... ur equation was off by 0.03 which is okay since u have rounded it to 0.1
Oh ok, good :) thanks
np :)
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