? asked in Science & MathematicsMathematics · 1 decade ago

Calc- parametric curves- finding tangents?

Hi I do not understand how to find the tangent from complex numbers so I was wondering if you could help?

1. find an equationof the tangent to the curve at the point corresponding to the given value of the parameter.

x= e^sq root of t , y=t-lnt^2 , t=1

2. find an equation of the tangent to the curve at the given point by (a) without eliminating the parameter (b) by first eliminating the parameter

x=tanx, y=secx; (1, sq root of 2)

3. find the length:

x=(e^t) - t , y=4(e^t/2) , t is between -8 and 3

Thanks for any help! Even if you can only answer 1 or 2 questions!

2 Answers

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  • Ed I
    Lv 7
    1 decade ago
    Favorite Answer

    1. x = e^√t, y = t - ln t^2

    dx/dt = e^√t/(2√t)

    dy/dt = 1 - 2 ln t

    dy/dx = (dy/dt)/(dx/dt) = (1 - 2 ln t)/(e^√t/(2√t))

    At t = 1, dy/dx = (1 - 2 ln 1)/(e^√1/2√1)) = (1 - 0)/(e/2) = e/2

    At t = 1, x = e^√1 = e, and y = 1 - ln 1^2 = 1 - ln 1 = 1

    (e, 1)

    y - 1 = (e/2)(x - e)

    2. (a) x = tan t

    y = sec t

    dx/dt = sec^2 t

    dy/dt = sec t tan t

    dy/dx = (sec t tan t)/(sec^2 t) = tan t/sec t = (sin t/cos t)/(1/cos t) = sin t

    x = tan t

    1 = tan t

    t = π/4

    At t = π/4, dy/dx = sin π/4 = √2/2

    y - √2 = (√2/2) (x - 1)

    (b) tan^2 t + 1 = sec^2 t

    x^2 + 1 = y^2

    ± √(x^2 + 1) = y

    Since (1,√2) is on the curve, y = √(x^2 + 1) = (x^2 + 1)^(1/2)

    dy/dx = (1/2)(x^2 + 1)^(-1/2) • 2x = x/√(x^2 + 1)

    At x = 1, dy/dx = 1/√(1^2 + 1) = 1/√(1 + 1) = 1/√2 = √2/2

    Same slope, same point ==> same equation for the tangent

    3. x = e^t

    y = 4e^(t/2)

    dx/dt = e^t

    dy/dt = 2e^(t/2)

    (dx/dt)^2 + (dy/dt)^2 = (e^t)^2 + (2e^(t/2))^2 = e^(2t) + 4e^t

    ds = √((dx/dt)^2 + dy/dt)^2) dt = √(e^(2t) + 4e^t)) dt

    s = ∫ √(e^(2t) + 4e^t) dt {from t = -8 to t = 3} = 28.10726288... ≈ 28.107

    Source(s): Honestly, I couldn't see an easy way to evaluate the integral in #3, so I used my TI-84 calculator to find the length.
  • 4 years ago

    it's possible yes

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