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Math Help - Find an equation of the line tangent to the curve y1 given below that is parallel to

  1. #1
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    Find an equation of the line tangent to the curve y1 given below that is parallel to

    y1 = 2x√x

    y2 = 4 + 6x

    y=
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  2. #2
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    Quote Originally Posted by plstevens View Post
    y1 = 2x√x

    y2 = 4 + 6x

    y=
    The general equation of a line is y - y_1 = m(x - x_1) where m is the gradient and (x_1, y_1) is a known point on the line.

    In your question, parallel to y2 = 4 + 6x means that m = 6.

    To get a known point:

    x-coordinate: Solve \frac{dy_1}{dx} = 6 for x.
    y-coordinate: Substitute the value of x found above into y_1 = 2 x \sqrt{x} = 2 x^{3/2}.
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    ...parallel to the given line y2. thats the rest of the directions
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    Quote Originally Posted by plstevens View Post
    ...parallel to the given line y2. thats the rest of the directions
    So?? I've already said that means that m = 6. Just to be crystal clear ..... it means that m = 6 in the general equation <br />
y - y_1 = m(x - x_1)<br />
.
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    i don't get it so how does this help me to know what y1 is and what y2 is?
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    To get a known point: Mr F adds: That is, to get the (x_1 , y_1)

    x-coordinate: Solve \frac{dy_1}{dx} = 6 for x.
    y-coordinate: Substitute the value of x found above into y_1 = 2 x \sqrt{x} = 2 x^{3/2}.
    Quote Originally Posted by plstevens View Post
    i don't get it so how does this help me to know what y1 is and what y2 is?
    x-coordinate: Solve \frac{dy_1}{dx} = 6 for x:

    \frac{dy_1}{dx} = \frac{3}{2} \times 2x^{\frac{3}{2} - 1} = 3 x^{1/2}.

    So solve 3 x^{1/2} = 6:

    \Rightarrow x^{1/2} = 2

    \Rightarrow x = 4.

    Now substitute x = 4 into y_1 = 2 x^{3/2} = 2 x x^{1/2} = 2 x \sqrt{x}:

    y = 2 (4) \sqrt{4} = 2 (4) (2) = 16.

    So a known point is (4, 16) ........

    So x_1 = 4 and y_1 = 16 .......
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