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Math Help - [SOLVED] Differentiation & Intergration Question

  1. #1
    Member looi76's Avatar
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    [SOLVED] Differentiation & Intergration Question

    Question:



    The diagrams shows the curve y = x(x-1)(x-2), which crosses the x-axis at the points O(0,0), A(1,0) and B(2,0).

    (i) The tangents to the curve at the points A and B meet at the point C. Find the coordinates of C.
    (ii) Show by integration that the area of the shaded region R_1 is the same as the area of the shaded region R_2.


    Attempt:

    I can't see the point C on the graph!
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  2. #2
    Member Danshader's Avatar
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    Point C is the intersection between the tangent of the curve at A and the tangent of the curve at B:
    Attached Thumbnails Attached Thumbnails [SOLVED] Differentiation & Intergration Question-asd.jpg  
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  3. #3
    Member looi76's Avatar
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    Quote Originally Posted by Danshader View Post
    Point C is the intersection between the tangent of the curve at A and the tangent of the curve at B:
    Thanks Danshader!

    Can you also explain to me how to get the tangent from the equarion of the curve?
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  4. #4
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    Isomorphism's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:

    The diagrams shows the curve y = x(x-1)(x-2), which crosses the x-axis at the points O(0,0), A(1,0) and B(2,0).

    (i) The tangents to the curve at the points A and B meet at the point C. Find the coordinates of C.
    First of all get the equations of the tangents.
    y - y_{A}= \left(\frac{dy}{dx}\right)_{A}(x - x_{A})
    y - y_{B}= \left(\frac{dy}{dx}\right)_{B}(x - x_{B})
    After this solve the above two equations simultaneously to get C. You do this because C lies on both of them.

    Quote Originally Posted by looi76 View Post
    (ii) Show by integration that the area of the shaded region R_1 is the same as the area of the shaded region R_2.
    Where are you stuck here? Try writing the integrals...

    Good luck
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  5. #5
    Member Danshader's Avatar
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    well first you need to find the gradient of the lines AC and BC, remember that differentiating the equation of the curve will give you the gradient at a particular point:

    <br />
\frac{d}{dx}y = x(x-1)(x-2)<br />
    <br />
\frac{d}{dx}y = x^3 - 3x^2 +2x<br />
    <br />
\frac{dy}{dx} = 3x^2 - 6x +2<br />

    substituting the coordinates of x given in A(1,0) and B(2,0) you can get the gradient at A and B respectively.

    then using the general line equation:

    y=mx+c<br />

    obtain the line equation for AC and BC

    then solving the simultaneous equation(equation of line AC and BC) you can get the coordinates of C
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  6. #6
    Member looi76's Avatar
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    Quote Originally Posted by Isomorphism View Post
    First of all get the equations of the tangents.
    y - y_{A}= \left(\frac{dy}{dx}\right)_{A}(x - x_{A})
    y - y_{B}= \left(\frac{dy}{dx}\right)_{B}(x - x_{B})
    After this solve the above two equations simultaneously to get C. You do this because C lies on both of them.



    Where are you stuck here? Try writing the integrals...

    Good luck
    Quote Originally Posted by Danshader View Post
    well first you need to find the gradient of the lines AC and BC, remember that differentiating the equation of the curve will give you the gradient at a particular point:

    <br />
\frac{d}{dx}y = x(x-1)(x-2)<br />
    <br />
\frac{d}{dx}y = x^3 - 3x^2 +2x<br />
    <br />
\frac{dy}{dx} = 3x^2 - 6x +2<br />

    substituting the coordinates of x given in A(1,0) and B(2,0) you can get the gradient at A and B respectively.

    then using the general line equation:

    y=mx+c<br />

    obtain the line equation for AC and BC

    then solving the simultaneous equation(equation of line AC and BC) you can get the coordinates of C
    Thanks Isomorphism & Danshader

    y = x(x - 1)(x - 2)
    y = (x^2 - x)(x - 2)
    y = x^3 - 2x^2 + -x^2 + 2x
    y = x^3 - 3x^2 + 2x

    \frac{dy}{dx} = 3x^{3-1} - 3 \times 2x^{2-1} + 2

    \frac{dy}{dx} = 3x^2 - 6x + 2

    \frac{dy}{dx}_A | x = 1███ \frac{dy}{dx}_A = 3(1)^2 - 6(1) + 2 = -1

    \frac{dy}{dx}_B | x = 2███ \frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = -4

    Equation of tangent to the curve at the point A(1,0)
    y - y_1 = m(x - x_1)
    y - 0 = -1(x - 1)
    y = -x + 1

    Equation of tangent to the curve at the point B(2,0)
    y - y_1 = m(x - x_1)
    y - 0 = -4(x - 2)
    y = -4x + 8

    y + x = 1 \longrightarrow equation 1
    y + 4x = 8 \longrightarrow equation 2
    -3x = -7
    x = \frac{7}{3}

    Substitute x in equation 1
    y + x = 1
    y + \frac{7}{3} = 1
    y = 1 - \frac{7}{3}
    y = -\frac{4}{3}

    C\left(\frac{7}{3},-\frac{4}{3}\right)

    Is it correct?
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  7. #7
    Member Danshader's Avatar
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    well you got the idea there but you made a little mistake in calculations

    \frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = 2
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  8. #8
    Member looi76's Avatar
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    Quote Originally Posted by looi76 View Post
    Thanks Isomorphism &amp; Danshader

    y = x(x - 1)(x - 2)
    y = (x^2 - x)(x - 2)
    y = x^3 - 2x^2 + -x^2 + 2x
    y = x^3 - 3x^2 + 2x

    \frac{dy}{dx} = 3x^{3-1} - 3 \times 2x^{2-1} + 2

    \frac{dy}{dx} = 3x^2 - 6x + 2

    \frac{dy}{dx}_A | x = 1███ \frac{dy}{dx}_A = 3(1)^2 - 6(1) + 2 = -1

    \frac{dy}{dx}_B | x = 2███ \frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = -4

    Equation of tangent to the curve at the point A(1,0)
    y - y_1 = m(x - x_1)
    y - 0 = -1(x - 1)
    y = -x + 1

    Equation of tangent to the curve at the point B(2,0)
    y - y_1 = m(x - x_1)
    y - 0 = -4(x - 2)
    y = -4x + 8

    y + x = 1 \longrightarrow equation 1
    y + 4x = 8 \longrightarrow equation 2
    -3x = -7
    x = \frac{7}{3}

    Substitute x in equation 1
    y + x = 1
    y + \frac{7}{3} = 1
    y = 1 - \frac{7}{3}
    y = -\frac{4}{3}

    C\left(\frac{7}{3},-\frac{4}{3}\right)

    Is it correct?
    Quote Originally Posted by Danshader View Post
    well you got the idea there but you made a little mistake in calculations

    \frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = 2
    \frac{dy}{dx}_A | x = 1███ \frac{dy}{dx}_A = 3(1)^2 - 6(1) + 2 = -1

    \frac{dy}{dx}_B | x = 2███ \frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = 2

    Equation of tangent to the curve at the point A(1,0)
    y - y_1 = m(x - x_1)
    y - 0 = -1(x - 1)
    y = -x + 1

    Equation of tangent to the curve at the point B(2,0)
    y - y_1 = m(x - x_1)
    y - 0 = 2(x - 2)
    y = 2x - 4

    y + x = 1 \longrightarrow equation 1
    y + 2x = -4 \longrightarrow equation 2
    2x = -4
    x = \frac{-4}{2}
    x = -2

    Substitute x in equation 1
    y + x = 1
    y + (-2) = 1
    y = 1 + 2
    y = 3

    C(-2,3)

    Is it correct?
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  9. #9
    Member Danshader's Avatar
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    y + x = 1 \longrightarrow equation 1
    y + 2x = -4 \longrightarrow equation 2
    2x = -4 <======== where did this came from??
    x = \frac{-4}{2}
    x = -2


    this part is wrong.
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  10. #10
    Member looi76's Avatar
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    Quote Originally Posted by Danshader View Post
    y + x = 1 \longrightarrow equation 1
    y + 2x = -4 \longrightarrow equation 2
    2x = -4 <======== where did this came from??
    x = \frac{-4}{2}
    x = -2


    this part is wrong.
    y + x = 1 \longrightarrow equation 1
    y - 2x = -4 \longrightarrow equation 2
    -3x = 5
    x = \frac{5}{-3}
    x = -\frac{5}{3}

    Substitute x in equation 1
    y + x = 1
    y + -\frac{5}{3} = 1
    y = 1 -\frac{5}{3}
    y = -\frac{2}{3}

    C(-\frac{5}{3},-\frac{2}{3})

    Is it correct?
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  11. #11
    Member Danshader's Avatar
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    lol you missed out on a small negative sign

    equation 1
    equation 2



    you should get 5/3 for x and -2/3 for y
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:



    The diagrams shows the curve y = x(x-1)(x-2), which crosses the x-axis at the points O(0,0), A(1,0) and B(2,0).

    (i) The tangents to the curve at the points A and B meet at the point C. Find the coordinates of C.
    (ii) Show by integration that the area of the shaded region R_1 is the same as the area of the shaded region R_2.

    Attempt:

    I can't see the point C on the graph!
    For part two show that \int_0^{1}x(x-1)(x-2)dx=\bigg|\int_1^{2}x(x-1)(x-2)dx\bigg|
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