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Math Help - Limit and tangent problem

  1. #1
    Junior Member
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    Limit and tangent problem

    I'm unsure how to go about either problem. Can I get some direction? Thanks.

    1. Find the values of the constants a and b such that \lim_{x\to0} \frac{\sqrt[3]{ax +b} -2}{x} =\frac{5}{12}

    My attempt:
    f'(0) = \lim_{x\to0} f(x) = \lim_{x\to0} \frac{f(x) - f(a)}{x-a}

    f(x) = \sqrt[3]{ax +b}
    f'(x) = \frac{1}{3}a(ax+b)^{-2/3}
    f'(0) = \frac{1}{3}ab^{-2/3}

    \frac{1}{3}ab^{-2/3} = \lim_{x\to0} \frac{\sqrt[3]{ax +b} -2}{x} =\frac{5}{12}

    \frac{1}{3}ab^{-2/3} = \frac{5}{12}

    ab^{-2/3} = \frac{5}{4}

    a = \frac{5b^{2/3}}{4}

    2. Show that the length of the portion of a-ny tangent line to the asteroid x^{2/3} + y^{2/3} = a^{2/3} cut off by the coordinate axes is constant.



    My attempt:

    x^{2/3} + y^{2/3} = a^{2/3}

    \frac{2}{3}x^{-1/3} +\frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0

    x^{-1/3} +y^{-1/3}\frac{dy}{dx} = 0

    \frac{dy}{dx} = \frac{-x^{-1/3}}{y^{-1/3}}

    \frac{dy}{dx} = \frac{-x^{-1/3}}{[(a^{2/3}-x^{2/3})^{3/2}]^{1/3}} = \frac{-x^{-1/3}}{(a^{2/3}-x^{2/3})^2}
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Cursed View Post
    I'm unsure how to go about either problem. Can I get some direction? Thanks.

    1. Find the values of the constants a and b such that \lim_{x\to0} \frac{\sqrt[3]{ax +b} -2}{x} =\frac{5}{12}

    My attempt:
    f'(0) = \lim_{x\to0} f(x) = \lim_{x\to0} \frac{f(x) - f(a)}{x-a}

    f(x) = \sqrt[3]{ax +b}
    f'(x) = \frac{1}{3}a(ax+b)^{-2/3}
    f'(0) = \frac{1}{3}ab^{-2/3}

    \frac{1}{3}ab^{-2/3} = \lim_{x\to0} \frac{\sqrt[3]{ax +b} -2}{x} =\frac{5}{12}

    \frac{1}{3}ab^{-2/3} = \frac{5}{12}

    ab^{-2/3} = \frac{5}{4}

    a = \frac{5b^{2/3}}{4}
    b must equal 8, so that the numerator approaches zero as the denominator does. Otherwise, the function will blow up as x\to0.

    Now use L'Hopital's Rule.

    Spoiler:
    \lim_{x\to0}\frac{\sqrt[3]{ax+8}-2}{x}=\lim_{x\to0}\frac{a}{3}(ax+8)^{-2/3}.

    Plugging in x=0 gives:

    \frac{a}{3}\cdot8^{-2/3}=\frac{a}{12}

    So a=5, b=8.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Cursed View Post
    I'm unsure how to go about either problem. Can I get some direction? Thanks.

    2. Show that the length of the portion of a-ny tangent line to the asteroid x^{2/3} + y^{2/3} = a^{2/3} cut off by the coordinate axes is constant.



    My attempt:

    x^{2/3} + y^{2/3} = a^{2/3}

    \frac{2}{3}x^{-1/3} +\frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0

    x^{-1/3} +y^{-1/3}\frac{dy}{dx} = 0

    \frac{dy}{dx} = \frac{-x^{-1/3}}{y^{-1/3}}

    \frac{dy}{dx} = \frac{-x^{-1/3}}{[(a^{2/3}-x^{2/3})^{3/2}]^{1/3}} {\color{red}~\neq~} \frac{-x^{-1/3}}{(a^{2/3}-x^{2/3})^2}
    Not sure I understand what the problem is asking, but there is a small error in your last calculation.

    \frac{dy}{dx} = \frac{-x^{-1/3}}{y^{-1/3}}=-\frac{y^{1/3}}{x^{1/3}}=-\frac{[(a^{2/3}-x^{2/3})^{3/2}]^{1/3}}{x^{1/3}}=-\frac{\sqrt{a^{2/3}-x^{2/3}}}{x^{1/3}} =-\frac{\sqrt{a^{2/3}-x^{2/3}}}{\sqrt{x^{2/3}}}=-\sqrt{\frac{a^{2/3}}{x^{2/3}}-1}=-\sqrt{(a/x)^{2/3}-1}

    Whatever calculation you need to do, the derivative is probably easiest to work with in that form.
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