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Find the slope of R= -1+sin theta at the points theta= 0 and theta= pie
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yes, polar
oops\[\frac{dy}{dx}=\frac{r'\sin(\theta)+r\cos(\theta)}{r'\cos(\theta)-r\sin(\theta)}\]
in your case \(r=-1+\sin(\theta)\) and \(r'=\cos(\theta)\)
i got the derivative part, it is just the limits are confusing me, what am i suppose to do with me?
Plug them in.
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Those are not limits; they are distinct points.
what @primeralph said, evaluate unless you get \(\frac{0}{0}\)
oh, so plug both of them into the tangent equation and get two separate tangents??
Yes.
Tangents are at points; integrals are over bounds (limits).
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thanks!!
Well, you're easy to teach. You seem smart already, so keep it up.
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