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How do you change the bounds/limits on the integrand for a parametric equation? I understand the integral-based area formula for Cartesian equations fine, but I'm a bit lost here and could definitely use some clarification help. :-)

Mathematics
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Cartesian mode: \[Area = \huge \int\limits_{a}^{b} f(x) dx\] Parametric mode? \[Area =\huge \int\limits_{\alpha}^{\beta} g(t) \cdot f'(t) \ dt\] The heck is this?
\[\huge slope = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}\]
What I'm not understanding is where my substitution is coming from. Maybe a simple example is in order, let me go find one...

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Example from the homework, verbatim: Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 6 + ln|t|, y = t^2 + 6, (6, 7)
By (6,7) do you mean from 6 to 7?
Ah nope @malevolence19, it's the point there
I presume it means the graph contains that point, or the tangent line & the graph both do more specifically.
And one of the easier looking questions... Hmpf! I dislike lesson plans that give you super-easy examples in the lecture notes but then kick your brain around like a can when you get to the homework :-P
It's like, oh hey you got any questions on this? No? Oh ok let's try something twenty times harder, ok jump! :D
Do you know how to do with "eliminating the parameter"?

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