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agentx5

  • 2 years ago

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. :-)

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  1. agentx5
    • 2 years ago
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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?

  2. agentx5
    • 2 years ago
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    \[\huge slope = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}\]

  3. agentx5
    • 2 years ago
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    What I'm not understanding is where my substitution is coming from. Maybe a simple example is in order, let me go find one...

  4. agentx5
    • 2 years ago
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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)

  5. malevolence19
    • 2 years ago
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    By (6,7) do you mean from 6 to 7?

  6. agentx5
    • 2 years ago
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    Ah nope @malevolence19, it's the point there

  7. agentx5
    • 2 years ago
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    I presume it means the graph contains that point, or the tangent line & the graph both do more specifically.

  8. agentx5
    • 2 years ago
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    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

  9. agentx5
    • 2 years ago
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    It's like, oh hey you got any questions on this? No? Oh ok let's try something twenty times harder, ok jump! :D

  10. malevolence19
    • 2 years ago
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    Do you know how to do with "eliminating the parameter"?

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