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anonymous
 one year ago
Find the slope of the cardioid r=2+2cos(theta) at the point corresponding to (theta)=pi/4
anonymous
 one year ago
Find the slope of the cardioid r=2+2cos(theta) at the point corresponding to (theta)=pi/4

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3\(r=2+2cos(\theta)\) the slope takes you back into cartesian, ok? so you want \(\frac{dy}{dx}\) yep?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This one specifically is nothing like the ones I've done in my previous assignment. So I'm confused as to how to even begin to tackle it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wouldn't the tangent of pi/4 in this case be the slope?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3do you know how to do that?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3this is it https://www.desmos.com/calculator/qysgdfnvnr

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0totally makes sense now that I have the graph! thanks!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have one more question. Maybe you can help with this one too, because MacLaurin Series are my weakest topic in this course so far.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3just to complete this thread, the formula to get the slope in cartesian is \(\large \frac{dy}{dx} = \frac{\frac {d r}{d \theta} \ sin \theta + r \ cos \theta}{\frac {d r}{d \theta} \ cos \theta  r sin \theta}\) very deriveable....from the basic premises that \(x = r cos \theta \) etc

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3maclaurin no worries stick it in a new thread first, though
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