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Math Help - Circle & derivative qus.

  1. #1
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    Question Circle & derivative qus.

    hey

    I have done part a) of the question which need checking & need help on part b)

    a) Find the centre and radius of the following circle:

    2x^2 + 2y^2 + -4x + 10y = 8

    radius = 3.3 centre = (1, -2.5) << this is what i got

    b) Find the equation of the tangent to the circle when the circle intersects the x-axis in the positive quadrant.



    The derivative of the following

    f(x) = Square root((x+1)/(x-1))

    I know the answer to this.. which is -1/2(x+1)^1/2 (x+1)^-3/2

    Do not know how to get to the answer, gets abit confusing when i applied the quotient rule as my numerator was zero but don't know if I was doing it right.
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  2. #2
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    Quote Originally Posted by dadon
    hey

    I have done part a) of the question which need checking & need help on part b)

    a) Find the centre and radius of the following circle:

    2x^2 + 2y^2 + -4x + 10y = 8

    radius = 3.3 centre = (1, -2.5) << this is what i got

    b) Find the equation of the tangent to the circle when the circle intersects the x-axis in the positive quadrant.

    The derivative of the following

    f(x) = Square root((x+1)/(x-1))

    I know the answer to this.. which is -1/2(x+1)^1/2 (x+1)^-3/2

    Do not know how to get to the answer, gets abit confusing when i applied the quotient rule as my numerator was zero but don't know if I was doing it right.
    I see you have two separate questions/problems here. The first ones is about the circle with parts a and b. The second one is about the derivative of f(x).

    Here are some ways to solve them.

    ------------------------------
    A) Circle 2x^2 +2y^2 -4x +10y = 8 -----------(i)

    a)find the center and radius.

    To find the center and radius, we need to convert the general form, (i), into its standard form
    (x-h)^2 +(y-k)^2 = r^2 -------------(ii)
    where
    (h,k) is the center
    r is the radius.

    We do that by applying the "complete the square" on (i):
    2x^2 +2y^2 -4x +10y = 8 --------(i)
    [2x^2 -4x] +[2y^2 +10y] = 8
    We do completing squares only when the squared variable has 1 as its coefficient, so,
    2[x^2 -2x] +2[y^2 +5y] = 8
    [x^2 -2x] +[y^2 +5y] = 4
    [x^2 -2x +(2/2)^2 -(2/2)^2] +[y^2 +5y +(5/2)^2 -(5/2)^2] = 4
    [(x-1)^2 -1] +[(y +2.5)^2 -6.25] = 4
    (x-1)^2 +(y+2.5)^2 = 11.25 ----------(iia)----in the form of (ii).

    Therefore, center is (1,-2.5), and radius is sqrt(11.25) = 3.3541 units. ----answer.

    b) I assume that you mean the tangent "where", not "when" as posted, the circle intersects the x-axis in the first quadrant.
    Because it it is "when" then there infinitely many possible tangent lines.

    Where the circle intersects the x-axis, the y is 0, so, using (iia),
    (x-1)^2 +(0+2.5)^2 = 11.25
    (x-1)^2 +6.25 = 11.25
    (x-1)^2 = 5
    x-1 = +,-sqrt(5)
    x = 1 +sqrt(5) ----------in the 1st quadrant.
    So the point in question is (1+sqrt(5),0). ----------***

    The slope of the tangent line at any point is dy/dx or y', so differentiating both sides of (iia) with respect to x,
    (x-1)^2 +(y+2.5)^2 = 11.25 ----------(iia)
    2(x-1) +2(y+2.5)*y' = 0
    y' = -(x-1)/(y+2.5)
    Plugging the coordinates of the point in question,
    y' = -(1+sqrt(5) -1) / (0 +2.5)
    y' = -sqrt(5) /2.5 -------------the slope of the tangent line.
    Then, using the point-slope form of the equation of a line, the tangent line is
    y -0 = -sqrt(5)[x -(1+sqrt5)]
    y = -sqrt(5)[x -1 -sqrt(5)] ------------answer.
    Or, in its y = mx +b form,
    y = -sqrt(5)*x +(5+sqrt(5)) -----------answer.

    -------------------------------------------------------
    B) The derivative of
    f(x) = Square root((x+1)/(x-1)) ------(1)

    To avoid the quotient rule, if you are having trouble with it, is to use the product rule.
    Eq.(1) can be expressed as
    f(x) = [(x+1)*(x-1)^(-1)]^(1/2)
    f(x) = (x+1)^(1/2) *(x-1)^(-1/2)
    Then,
    f'(x) = [(x+1)^(1/2)]*{(-1/2)*(x-1)^(-3/2)} +[(x-1)^(-1/2)]*{(1/2)*(x+1)^(-1/2)}
    Simplifying that, you should get
    f'(x) = -(x+1)^(-1/2) *(x-1)^(-3/2) ---------answer.
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  3. #3
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    Post Re:

    thank you! been a great help!

    but looking back at the question it is "when"

    looking at the solution to this problem it is

    y = 2.82 - 0.88x

    so im quite confused only on part b)

    would drawing the circle help?
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  4. #4
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    Quote Originally Posted by dadon
    thank you! been a great help!

    but looking back at the question it is "when"

    looking at the solution to this problem it is

    y = 2.82 - 0.88x

    so im quite confused only on part b)

    would drawing the circle help?
    "When" the circle intersects the x-axis at the positve quadrant....
    (Here, again, I assume you mean "positive half", which is the 1st and 4th quadrants, relative to the x-axis.)
    But the circle really intersects the x-axis at the first half. It is not a question of when. It is reality.

    Say it is "when".
    Then the tangent line you want to find is any tangent line to the circle. Whether in the 1st, 2nd, 3rd or 4th quadrants.
    There are gogazillion of them.

    Or, yes, show a sketch/drawing of the tangent line you want to find, then we can solve for it.

    Also, sorry I made a mistake on the algebra in my answer. Let me correct it here.
    y' = -sqrt(5) /2.5 -------------the slope of the tangent line.
    Then, using the point-slope form of the equation of a line, the tangent line is
    y -0 = (-sqrt(5) /2.5)*[x -(1+sqrt5)]
    y = (-sqrt(5) /2.5)*[x -1 -sqrt(5)]
    y = -0.894[x -3.236] ----------------answer.
    Or, in its y = mx +b form,
    y = -0.894x +2.893 -----------answer

    There.
    Last edited by ticbol; April 13th 2006 at 04:04 PM.
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  5. #5
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    Hi thanks agian!

    Do you know how to use this method?

    I always use the equation: x^2 + y^2 + 2gx + 2fy + c = 0

    Using the equation and comparing it to the circle one to find g, f and c.

    Then use this to find centre and radius.

    Centre = (-g, -f) Radius = (g^2 + f^2 - c)^1/2

    As this way it is much easier to find the centre and radius. But I wanted to know if there is a way to use this method to find the equation of the tangent?

    Regards
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  6. #6
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    Quote Originally Posted by dadon
    Hi thanks agian!

    Do you know how to use this method?

    I always use the equation: x^2 + y^2 + 2gx + 2fy + c = 0

    Using the equation and comparing it to the circle one to find g, f and c.

    Then use this to find centre and radius.

    Centre = (-g, -f) Radius = (g^2 + f^2 - c)^1/2

    As this way it is much easier to find the centre and radius. But I wanted to know if there is a way to use this method to find the equation of the tangent?

    Regards
    I see how you use your method. It is simlar to the "completing the square" method. Yours in in formula form, so you need to memorize that.

    As for the equation of the tangent line, yes, you can always use the general form of the equation of the circle without converting it to its standard form first.
    Let us do the example.

    2x^2 +2y^2 -4x +10y = 8

    We find the slope of the tangent lines, which is dy/dx,
    Divide both sides of the equation by 2,
    x^2 +y^2 -2x +5y = 4
    Differentiate both sides with respect to x,
    2x +2y*dy/dx -2 +5dy/dx = 0
    (dy/dx)(2y +5) = 2 -2x
    dy/dx = (2 -2x)/(2y +5)
    dy/dx = 2(1 -x) / 2(y +2.5)
    dy/dx = -(x-1)/(y+2.5) ------------same as that found before.

    We find the point of tangency at the positive half of the x-axis, where y-0,
    x^2 +0^2 -2x +5(0) = 4
    x^2 -2x -4 = 0
    Using the Quadratic Formula,
    x = {-(-2) +,-sqrt[(-2)^2 -4(1)(-4)]} / (2*1)
    x = {2 +,-sqrt[20]} /2
    x = {2 +,-2sqrt(5)} /2
    x = 1 +,-sqrt(5)

    So, x = 1 +sqrt(5) in the positive half of the x-axis, and the point of tangency there is (1+sqrt(5),0). -----same again as was found before.
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  7. #7
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    Smile

    Thank you! Explained perfectly!
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