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. :-)

Question

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.  :-)

anonymous · Answer

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?

anonymous · Answer

$$\huge slope = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}$$

anonymous · Answer

What I'm not understanding is where my substitution is coming from.  Maybe a simple example is in order, let me go find one...

anonymous · Answer

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)

anonymous · Answer

By (6,7) do you mean from 6 to 7?

anonymous · Answer

Ah nope @malevolence19, it's the point there

anonymous · Answer

I presume it means the graph contains that point, or the tangent line & the graph both do more specifically.

anonymous · Answer

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

anonymous · Answer

It's like, oh hey you got any questions on this?  No?  Oh ok let's try something twenty times harder, ok jump!  :D

anonymous · Answer

Do you know how to do with "eliminating the parameter"?