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Math Help - Define Integral Problem

  1. #1
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    Define Integral Problem

    Diagram shows the straight line y = -x + k touching the curve y = 3x - x^2 at point A .


    a)Find the value of k

    b)the coordinates of point A .
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  2. #2
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    Quote Originally Posted by faraday View Post
    Diagram shows the straight line y = -x + k touching the curve y = 3x - x^2 at point A .


    a)Find the value of k

    b)the coordinates of point A .
    Let the x-coordinate of A be x = a. Now use the fact that 3 - 2a = -1 (why?). Therefore y = -a + k (why?). Therefore -a + k = 3a - a^2 (why?). Hence solve for k.
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  3. #3
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    i dont understand..

    \frac{dy}{dx}=3 - 2x

    how can u get \frac{dy}{dx} = 3 - 2x = -1?
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  4. #4
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    Hi!

    Quote Originally Posted by faraday View Post
    i dont understand..

    \frac{dy}{dx}=3 - 2x

    how can u get \frac{dy}{dx} = 3 - 2x = -1?
    The point A has the x-coordinate a.

    You do have the straight line y_1 = -x+kand the curve 3x-x^2 .

    you know the straight line touches the curve at point A with the x-coordinate a

    So that means at x = a (point A) the straight line has the same increase at point A as the curve.

    That means

    y_1 ' = y_2 '

    \frac{dy_1}{dx}= \frac{dy_2}{dx}

    more precisely:

    \frac{d}{dx}(-x+k) = \frac{d}{dx}(3x-x^2)

    -1 = \frac{d}{dx}(3x-x^2)

    -1 = 3-2x
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  5. #5
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    thus, the value of k = 2 and the coordinates of point A = (2,0) right ?
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  6. #6
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    if the question want the area of shaded region

    at the curve y = 3x-x^2 the x-intercept is

    0=3x - x^2

    x(3 - x)=0

    x = 3


    hence, y=-x^2 + 3x

    \int_{2}^{3} {y} \ dx

    \int_{2}^{3} (-x^2 + 3x) dx

    \left[ \frac{-x^3}{3} + \frac {3x^2}{2} \right]_{2}^{3}

    \left[\frac{-(3)^3}{3} + \frac{3(3)^2}{2} \right] - \left[\frac{-(2)^3}{3} + \frac{3(2)^2}{2} \right]

    \left[-9 + \frac{27}{2} \right] - \left[\frac{-8}{3} + 6 \right]

    =\frac{7}{6} unit^2

    am i right?
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  7. #7
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    Quote Originally Posted by faraday View Post
    thus, the value of k = 2 and the coordinates of point A = (2,0) right ?
    No! It is A=(2,2) and k =4.

    As you know we shall solve

    3 - 2a = -1

    solving for a :

    -2a = -1 -3

    -2a = -4

    a = 2

    now substitute a = 2 in -a + k = 3a - a^2 (as mr fantastic was saying)

    -2+k = 3*2 - 2^2

    -2+k = 6-4

    -2+k = 2

    k = 2+2 = 4

    So the straight line is y = -x+4

    now substitute x = 2 and you have

    y = -2+4 =2

    So A = (2,2)

    Quote Originally Posted by faraday View Post
    if the question want the area of shaded region

    at the curve y = 3x-x^2 the x-intercept is

    0=3x - x^2

    x(3 - x)=0

    x = 3


    hence, y=-x^2 + 3x

    \int_{2}^{3} {y} \ dx

    \int_{2}^{3} (-x^2 + 3x) dx

    \left[ \frac{-x^3}{3} + \frac {3x^2}{2} \right]_{2}^{3}

    \left[\frac{-(3)^3}{3} + \frac{3(3)^2}{2} \right] - \left[\frac{-(2)^3}{3} + \frac{3(2)^2}{2} \right]

    \left[-9 + \frac{27}{2} \right] - \left[\frac{-8}{3} + 6 \right]

    =\frac{7}{6} unit^2

    am i right?
    Yes, you are! GJ
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  8. #8
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    Ok..I very understand sir..Thanks!

    lastly,if the question want the volume generated,in term of \pi,
    when the region bounded by the curve y=3x - x^2 and the x-axis is revolved through 360^0 about the x - axis

    i get,

    =\int_{2}^{3} \pi{y^2} dx

    =\int_{2}^{3} \pi(3x-x^2)^2 dx

    =\pi\int_{2}^{3} (9x^2-6x^3+x^4) dx

    =\pi \left[\frac{9x^3}{3}-\frac{6x^4}{4}+\frac{x^5}{5} \right]_{2}^{3}

    =\pi\left[\frac{9(3)^3}{3}-\frac{6(3)^4}{4}+\frac{(3)^5}{5} \right] - \left[\frac{9(2)^3}{3}-\frac{6(2)^4}{4}+\frac{(2)^5}{5} \right]

    =\pi\left(81-\frac{243}{2} + \frac{243}{5} \right) - \left(24-24 + \frac{32}{5}\right)

    =\frac{17}{10} \pi unit^3

    right?
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  9. #9
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    Quote Originally Posted by faraday View Post
    lastly,if the question want the volume generated,in term of \pi,
    when the region bounded by the curve y=3x - x^2 and the x-axis is revolved through 360^0 about the x - axis

    i get,

    =\int_{2}^{3} \pi{y^2} dx
    okay, that's the formula for the volume generated between x=2 and x=3

    Quote Originally Posted by faraday View Post
    =\frac{17}{10} \pi unit^3

    right?
    It sure is.
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  10. #10
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    thanks very much sir..sorry for wasting your time..but i really appreciate with you..thanks!
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  11. #11
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    You're welcome!

    sorry for wasting your time
    Never mind. That's what the forums is for. Asking questions and giving/getting answers.
    Have a nice day, mate.
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