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Math Help - Need Help (tangent to parameterized curves)

  1. #1
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    Need Help (tangent to parameterized curves)

    I have 2 questions:

    The first says: find an equation for the line tangent to the ccurve at the point defined by the given value of t. Also find the value of d^2y/dx^2.

    x= t y= sq root(t) t= 1/4

    (I am not sure how to do this problem so an explanation would be great)

    Also I have a question that says: The position of a particle moving along a coordinate line is s= sq root (1+4t) with s in meters and t inseconds. Find the particles velocity and acceleration at t=6

    For this problem I just found the derivative which is

    (1/2(1+4t)^-1/2) x (4t) and I got an answer of 6 for the velocity and the back of my book has 2/5 as the answer. What am I doing wrong?
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  2. #2
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    Quote Originally Posted by KarlosK View Post
    I have 2 questions:

    The first says: find an equation for the line tangent to the ccurve at the point defined by the given value of t. Also find the value of d^2y/dx^2.

    x= t y= sq root(t) t= 1/4

    (I am not sure how to do this problem so an explanation would be great)

    Also I have a question that says: The position of a particle moving along a coordinate line is s= sq root (1+4t) with s in meters and t inseconds. Find the particles velocity and acceleration at t=6

    For this problem I just found the derivative which is

    (1/2(1+4t)^-1/2) x (4t) and I got an answer of 6 for the velocity and the back of my book has 2/5 as the answer. What am I doing wrong?
    (1) for a tangent line, you need two things ... the point of tangency (x,y) and the slope at that point , \frac{dy}{dx}<br />
.

    the point comes straight from evaluating x(t) and y(t) at the desired value of t.

    slope ... m = \frac{dy}{dx} =  \frac{y'(t)}{x'(t)}


    for the 2nd derivative, note that \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)

    \frac{d}{dx}\left(\frac{y'(t)}{x'(t)}\right)  = \frac{x'(t) \cdot y''(t) \cdot \frac{dt}{dx} - y'(t) \cdot x''(t)  \cdot \frac{dt}{dx}}{[x'(t)]^2} = \frac{x'(t) \cdot y''(t) - y'(t) \cdot x''(t)}{[x'(t)]^3}


    s = \sqrt{1+4t} = (1+4t)^{\frac{1}{2}}<br />

    v = s'(t) = \frac{1}{2}(1+4t)^{-\frac{1}{2}} \cdot 4 = \frac{2}{\sqrt{1+4t}}

    v(6) = \frac{2}{\sqrt{25}} = \frac{2}{5}

    note that a = v'(t) = s''(t)
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