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Math Help - Equation of line & Gradient qus.

  1. #1
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    Post Equation of line & Gradient qus.

    Sorry for so many questions. Im just stuck on these and they are meant to be simple!

    ------------------------------------------------------

    1) Find the equation of the line through the origin which is parallel to the line 4x - 27 - 5 = 0

    ------------------------------------------------------

    2) The equation of a curve is y = (x-3)(x+4). Find the gradient of the curve:
    a) at the point the curve crosses the y-axis
    b) at each of the points where the curve crosses the x-axis

    ------------------------------------------------------

    3) Find the coordinates of the point(s) on the curve y=3x^3 - x + 8 at which the gradient is
    a) 8
    b) 0

    ------------------------------------------------------

    4) The equation of a curve s=4t^2 +5t. Find the gradient of the normal at each of the points where the curve crosses the t-axis.

    ------------------------------------------------------

    5) If the gradient of a curve is y=2x^2 -3x -2. Find
    a) the gradient at the point where x=0
    b) the co-ordinates of the point where y=0
    c) the gradient at each of the points found in b)
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  2. #2
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    Quote Originally Posted by dadon
    Sorry for so many questions. Im just stuck on these and they are meant to be simple!

    ------------------------------------------------------

    1) Find the equation of the line through the origin which is parallel to the line 4x - 27 - 5 = 0

    ------------------------------------------------------
    This is a basic line x=k thus, any parallel line also has the same equation, wince it passes through the origin it must be x=0
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    Quote Originally Posted by dadon

    2) The equation of a curve is y = (x-3)(x+4). Find the gradient of the curve:
    a) at the point the curve crosses the y-axis
    b) at each of the points where the curve crosses the x-axis

    ------------------------------------------------------
    Thus, we have,
    y=(x-3)(x+4)=x^2+x-12
    If the gradient means derivative then,
    y'=2x+1

    It passes the y-axis when x=0 thus
    y=0^2+0-12=-12

    It passes the x-axis when y=0 thus,
    0=x^2+x-12=(x-3)(x+4)
    Thus,
    x=\left\{ \begin{array}{c}-4\\3\end{array}\right
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  4. #4
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    Quote Originally Posted by dadon

    3) Find the coordinates of the point(s) on the curve y=3x^3 - x + 8 at which the gradient is
    a) 8
    b) 0
    If,
    y=3x^3-x+8
    Then,
    y'=9x^2-1
    Yuu need it to be zero thus,
    0=9x^2-1
    Thus,
    9x^2=1
    Thus,
    x^2=1/3
    Thus,
    x=\pm \sqrt{1/3}=\pm \frac{\sqrt{3}}{3}
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  5. #5
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    Quote Originally Posted by dadon
    1) Find the equation of the line through the origin which is parallel to the line 4x - 27 - 5 = 0
    You probably have a typo here, given the relative location of the y and 7
    keys I will guess this should have been:

    " Find the equation of the line through the origin which is parallel to the line 4x - 2y - 5 = 0"

    This may be rewritten:

    y=2x-5/2,

    and hence has slope 2, and so the equation through the origin
    parallel to this also has slope 2 and so has equation:

    <br />
y=2x<br />
.

    RonL
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  6. #6
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    Quote Originally Posted by dadon
    2) The equation of a curve is y = (x-3)(x+4). Find the gradient of the curve:
    a) at the point the curve crosses the y-axis
    b) at each of the points where the curve crosses the x-axis
    Expanding the terms on the RHS and then differentiation gives:

    <br />
y=x^2+x-12<br />
,

    so:

    <br />
\frac{dy}{dx}=2x+1<br />

    a)The curve crosses the y axis when x=0, hence the gradient
    of the curve is obtained by plugging this value of x into the expression
    for the derivative and so the required gradient is 1

    b)The curve crosses the x axis at x=3 and x=-4.
    Plugging these into the expression for the derivative give gradients of 7 and -7 respectively.

    RonL
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  7. #7
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    Quote Originally Posted by dadon
    3) Find the coordinates of the point(s) on the curve y=3x^3 - x + 8 at which the gradient is
    a) 8
    b) 0
    The derivative of y=3x^3-x+8 is:

    <br />
\frac{dy}{dx}=9x^2-1<br />

    a) Now this is 8 when:

    <br />
9x^2-1=8<br />
,

    or:

    <br />
x^2=1<br />

    or x=\pm 1, which may be plugged into the equation for y to
    give y=10 when x=1 and y=6 when x=-1.
    So the required points are (1,10) and (-1,6).

    b)Now this is 0 when:

    <br />
9x^2-1=0<br />
,

    or:

    <br />
x^2=1/9<br />

    or x=\pm 1/3.

    Plugging these values back into the equation for y gives the other
    coordinate, and so we find the required points are: (1/3,8) and (-1/3,8).

    RonL
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  8. #8
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    Quote Originally Posted by dadon
    4) The equation of a curve s=4t^2 +5t. Find the gradient of the normal at each of the points where the curve crosses the t-axis.
    The points at which the curve crosses the t-axis correspond to the roots of:

    <br />
4t^2+5t=0<br />

    which are t=0 and t=-5/4.

    Now:

    <br />
\frac{ds}{dt}=8t+5<br />
,

    so the slopes at the two points are 5 and -5 respectively.

    Now the gradient of the normal is -1 over the gradient of the tangent, and so
    the gradients of the two normals are -1/5 and 1/5 respectively.

    RonL
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  9. #9
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    Quote Originally Posted by dadon
    5) If the gradient of a curve is y=2x^2 -3x -2. Find
    a) the gradient at the point where x=0
    b) the co-ordinates of the point where y=0
    c) the gradient at each of the points found in b)
    This has a confusing use of y for the gradient of the curve, and assuming
    this is right there is sufficient information to find only the x-coords for the
    answer to part b), and c) is redundant as it is given as 0.

    RonL
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  10. #10
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    This has a confusing use of y for the gradient of the curve, and assuming this is right there is sufficient information to find only the x-coords for the answer to part b), and c) is redundant as it is given as 0.
    Sorry the question should be:

    5) If the equation of a curve is y=2x^2 -3x -2. Find
    a) the gradient at the point where x=0
    b) the co-ordinates of the point where y=0
    c) the gradient at each of the points found in b)
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