Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = 1 + 2√t, y

Question

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.
x = 1 + 2√t, y = t3 – t, z = t3 + t; (3,0,2)

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Adela 5 years 2021-08-30T00:54:03+00:00 1 Answers 19 views 0

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  1. Vodka
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    2021-08-30T00:55:37+00:00
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    Solution :

    Given parametric equation for :

    $x=1+2 \sqrt t$

    $y=t^3-t$

    z=t^3+t

    The point is (3, 0, 2)

    The vector equation is equal to :

    $r(t) = \left<1+2 \sqrt t, t^3 -t, t^3+t \right>$

    Solving for r'(t) by differentiating each of the components of r(t) w.r.t. to t,

    $r'(t)= \left< \frac{1}{\sqrt t}, \ 3t^2-1, \ 3t^2+1 \right>$

    The parameter value corresponding to (3, 0, 2) is t = 1. Putting in t=1 into r'(t) to solve for r'(t), we get

    $r'(1) = \left< \frac{1}{\sqrt 1}, \ 3(1)^2-1, \ 3(1)^2+1 \right>$

    We know that parametric equation for  line through the point $(x_0, y_0, z_0)$  and parallel to the direction vector <a, b, c > are

    $x=x_0+at$

    $y=y_0+bt$

    z=z_0+ct

    Now substituting the  $(x_0, y_0, z_0)$   = (3, 0, 2) and  <a, b, c > into x, y and z, respectively to solve for the parametric equation of the tangent line to the curve, we get:

    $x=3+(1)t$

    x  = 3 + t

    y = (0) + (2)t

    y = 2t

    z = (2) + (4)t

    z = 2 + 4t

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )