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Math Help - high school / calculus

  1. #1
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    high school / calculus

    I have received help from many of you last time i requested help and i thank you all for helping me out and not making me look like a fool in front of my child but now he needs my help again and he's giving me a few problems to show him how to solve so once again i'm seeking your help... if you guys could please show me how to solve these problems so i could try and teach it to him after teaching it to myself that would be great thank you all in advance sorry if i mistyped anything i'm doing this really fast because i have to go thank you all

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  2. #2
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    Quote Originally Posted by philipsk View Post
    ...but now he needs my help again and he's giving me a few problems to show him how to solve...

    Hello,

    to 1.

    To calculate the slope of a tangent you need the gradient of the function. The 1rst derivative gives you the gradient for any value of x. Use quotient rule and chain rule:
    f(x)=\frac{x}{\sqrt{2x-1}},\ \ x>\frac{1}{2}
    f'(x)=\frac{1 \cdot \sqrt{2x-1}-x \cdot \frac{1}{2}(2x-1)^{-\frac{1}{2}} \cdot 2}{2x-1},\ \ x>\frac{1}{2}

    f'(x)=\frac{1 \cdot \sqrt{2x-1}-x \cdot \frac{1}{2}(2x-1)^{-\frac{1}{2}} \cdot 2}{2x-1},\ \ x>\frac{1}{2}. Simplify to

    f'(x)=\frac{x-1}{(\sqrt{2x-1})^3}. You are looking for a horizontal tangent which have the slope zero:

    f'(x) = 0 if x - 1 = 0. Thus x = 1

    Plug in this value into f(x) and you'll get f(1)=\frac{1}{\sqrt{2-1}}=1

    Thus the equation of the tangent is: y = 1

    I've attached a diagram.

    EB
    Attached Thumbnails Attached Thumbnails high school / calculus-gebr_wurzelfkt.gif  
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  3. #3
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    thank you very much!

    for the 2nd question do i take the derivative of the function then plug in x into the equation to find the equation of the line that intercepts at that point ?
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  4. #4
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    Quote Originally Posted by philipsk View Post
    thank you very much!

    for the 2nd question do i take the derivative of the function then plug in x into the equation to find the equation of the line that intercepts at that point ?
    The equation of the tangent will be of the general form of any straight line: y=m\,x+c, where m is the slope of the line.

    The derivative of the given graph will give you the slope of the tangent at the given point. you then have to find c such that the line goes through the given point.

    Your function is:

    f(x)=\sqrt{2x-1}

    so:

    \frac{df}{dx}=\frac{1}{2}\,\frac{1}{\sqrt{2x-1}}\,2=\frac{1}{\sqrt{2x-1}}.

    Now the point in question has x coordinate 5, so the slope of the tangent at the point in question is:

    m=\frac{1}{\sqrt{2\times 5-1}}=1/3

    So the equation of the tangent is:

    y=x/3+c

    and passes through (5,3), so we must have:

    3=5/3+c,

    so c=4/3 and the equation of the tangent is:

    y=x/3+4/3.

    RonL
    Attached Thumbnails Attached Thumbnails high school / calculus-gash.jpg  
    Last edited by CaptainBlack; February 1st 2007 at 11:10 PM.
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  5. #5
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    4.

    a)Use the Chain rule
    y = sin(x^2 + 1)^3
    y' = 3(sin(x^2 + 1)^2)(cos(x^2+1)(2x)
    y' = 6xsin(x^2+1)^2cos(x^2+1)

    b)Use the Product and Chain rules.
    f(x) = ((2x-3)^3)((x^2+1)^2)
    f'(x) = (3)(2x-3)^2(2)(x^2+1)^2 + (2x-3)^3(2)(x^2+1)(2x)
    f'(x) = 6(2x-3)^2(x^2+1)^2 + 4x(2x-3)^3(x^2+1)
    f'(x) = 2(x^2+1)(2x-3)^2(3(x^2+1)+2x(2x-3))
    f'(x) = 2(x^2+1)(2x-3)^2(7x^2-6x+3)

    c)Use the Product and Chain rules.
    f(x) = x(sinx)^2
    f'(x) = (2x)(sin x)(cos x) + (sinx)^2

    d)Use the Product and Chain rules
    f(x) = x^2(x^2 + 4)^(1/2)
    f'(x) = 2x(x^2+4)^(1/2) + x^2(2x)(1/2)(x^2+4)^(-1/2)
    f'(x) = 2x(x^2+4)^(1/2) + x^3/(x^2+4)^(1/2)
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  6. #6
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    Quote Originally Posted by philipsk View Post
    ...but now he needs ... help again ...
    Hello,

    to #4:

    with this problem I'm going to use implicite derivation:

    I use f(x,y)=x^2 y^2 - 9x^2 - 4y^2. The derivation is calculated by

    f'(x,y)= - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}. That means here:

    f'(x,y)= - \frac{2x y^2-18x}{2x^2 y - 8y}. Now plug in the values of x and y:

    f'(-4, 2\sqrt{3})= - \frac{-8 \cdot 12 + 72}{64 \sqrt{3} - 16 \sqrt{3}}=\frac{24}{48\sqrt{3}}=\frac{1}{6}\sqrt{3  }

    Now you can use the point-slope-formula of a straight line to calculate the equation of the tangent:

    \frac{y - 2\sqrt{3}}{x+4}=\frac{1}{6}\sqrt{3} \Longleftrightarrow y =\left(\frac{1}{6}\sqrt{3} \right) x+\frac{8}{3}\sqrt{3}

    EB
    Attached Thumbnails Attached Thumbnails high school / calculus-cruciform_graph.gif  
    Last edited by earboth; February 2nd 2007 at 03:20 AM.
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  7. #7
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    Quote Originally Posted by philipsk View Post
    ...but now he needs my help again and he's giving me a few problems to show him how to solve so once again i'm seeking your help...
    Hello,

    it's me again.
    If you are not familiar with implcite derivation I can now show you a different way to do #4.

    First transform the given equation:

    x^2 y^2-9x^2-4y^2=0 \Longleftrightarrow y^2(x^2-4)=9x^2 \Longleftrightarrow y^2 = \frac{9x^2}{x^2-4}  \Longrightarrow y = \pm \sqrt{\frac{9x^2}{x^2-4}}

    The point in question lays above the x-axis (it's y-value is positive), therefore you have to use the positive square-root:

    y=f(x)= \sqrt{\frac{9x^2}{x^2-4}}

    Now calculate the 1rst derivative of this function (use chain rule, quotient rule):

    f'(x)=\frac{1}{2}\left( \frac{9x^2}{x^2-4} \right)^{-\frac{1}{2}}\cdot \frac{(x^2-4) \cdot 18x-9x^2 \cdot 2x}{(x^2-4)^2} =\left( \frac{9x^2}{x^2-4} \right)^{-\frac{1}{2}}\cdot \frac{(x^2-4) \cdot 9x-9x^2\cdot x}{(x^2-4)^2} finally:

    f'(x)=\frac{-36x}{(x^2-4)^2 \cdot \sqrt{\frac{9x^2}{x^2-4}}}

    Now plug in the value x = -4 :

    f'(-4)=\frac{-36 \cdot (-4)}{((-4)^2-4)^2 \cdot \sqrt{\frac{9 \cdot (-4)^2}{(-4)^2-4}}} =\frac{144}{144 \cdot \sqrt{12}}=\frac{1}{\sqrt{12}}=\frac{1}{2\sqrt{3}}  =\frac{1}{6}\sqrt{3}

    To calculate the equation of the tangent have a look at my previous post.

    EB
    Last edited by earboth; February 2nd 2007 at 09:44 AM.
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  8. #8
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    Play more.

    If you like to play more on the (x^2)(y^2) -9x^2 -4y^2 = 0, here is another way.

    (x^2)(y^2) -9x^2 -4y^2 = 0
    (x^2)[2yy'] +(y^2)[2x] -18x -8yy' = 0
    Divide all by 2,
    (x^2)yy' +xy^2 -9x -4yy' = 0
    yy'(x^2 -4) +x(y^2 -9) = 0
    yy'(x^2 -4) = -x(y^2 -9)
    y' = -x(y^2 -9) / y(x^2 -4) ---------slope of tangent line anywhere on the curve.

    At point (-4, 2sqrt(3)),
    y' = -(-4)[(2sqrt(3))^2 -9] / 2sqrt(3)*[(-4)^2 -4]
    y' = 4[3] / 2sqrt(3)*[12]
    y' = 1 / 2sqrt(3)
    y' = 1*sqrt(3) / 2*(sqrt(3))^2
    y' = (sqrt(3)) /6 -------------------***

    For the equation of the tangent line, using the point-slope form,
    y -y1 = m(x -x1)
    y - 2sqrt(3) = [sqrt(3) /6](x -(-4))
    y -2sqrt(3) = [sqrt(3) /6]x +4[sqrt(3) /6]
    y = [sqrt(3) /6]x +(2/3)sqrt(3) +2sqrt(3)
    y = [sqrt(3) /6]x +(8/3)sqrt(3) -------------------the tangent line.
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