Calculus ?

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter.

x = 5 + ln(t), y = t^2 + 2, (5, 3)

so y=?

1 Answer

Relevance
  • 2 months ago

    dx/dt = 1/t and dy/dt = 2t

    so, dy/dx = (2t)(t) => 2t²

    Hence, y - 3 = (2t²)(x - 5)

    Now, x = 5 and y = 3 when t = 1

    so, y - 3 = 2(x - 5)

    i.e. y = 2x - 7

    Now, lnt = x - 5 and t² = y - 2

    i.e. t = eˣ⁻⁵

    Then, e²⁽ˣ⁻⁵⁾ = y - 2....parameter, t eliminated

    or, y = e²ˣ⁻¹⁰ + 2

    so, dy/dx = 2e²ˣ⁻¹⁰

    With x = 5, dy/dx = 2e⁰ => 2

    This again gives, y - 3 = 2(x - 5)

    i.e. y = 2x - 7

    :)>  

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