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Math Help - In desperate need of some help!!

  1. #1
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    In desperate need of some help!!

    Hey everyone...

    I am about to start university next week, where I will have a maths test just testing our mathematical ability. However, since I have been on a gap year, my maths skills have gone somewhat rusty, and this is me being polite to myself!

    So, I was wandering if you guys could help me with a few maths problems? Please only feel free to answer which ever ones you can, and I am terribly sorry I am asking such a huge favour without contributing anything first. I have got the answers to the questions, but just don't know how to arrive to them!


    1) Rearrange the following formula to give d in terms of D and f:

    (2 / D - d) + (2 / D + d) = (1/f)

    It does look simpler if you write it out!
    The answer is d = (D^2 - 4Df)^(1/2)

    2) Given that y = loge (Ae^(-Bx)), find x in terms of y, A and B.

    The answer is 1/B(loge A - y)

    Differentiation Questions

    3) Differentiate y = ln(3x^(2) +1) with respect to x.

    The answer is 6x/(3x^(2) +1)

    4) Differentiate y = (x^(2) + x)^(-2) with respect to x

    The answer is -(4x+2)(x^(2)+x)^(-3)

    5) By differentiating the relation x^2 + y^2 = e^k (where k is a constant) with respect to x, find dy/dx.

    The answer is -x/y

    I would really appreciate it you guys could help me out here, as I genuinely require some help.

    Thank you everyone in advance
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Singh_87 View Post
    1) Rearrange the following formula to give d in terms of D and f:

    (2 / D - d) + (2 / D + d) = (1/f)

    It does look simpler if you write it out!
    The answer is d = (D^2 - 4Df)^(1/2)
    \frac{2}{D - d} + \frac{2}{D + d} = \frac{1}{f}

    Multiply both sides by f(D - d)(D + d).

    2f(D + d) + 2f(D - d) = (D - d)(D + d)

    2fD + 2fd + 2fD - 2fd = D^2 - d^2

    4fD = D^2 - d^2

    d^2 = D^2 - 4fD

    d = \sqrt{D^2 - 4fD}

    -Dan
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  3. #3
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    1) Rearrange the following formula to give d in terms of D and f:

    (2 / D - d) + (2 / D + d) = (1/f)

    It does look simpler if you write it out!
    The answer is d = (D^2 - 4Df)^(1/2)


    The idea is to do anything so that d will be isolated by itselt in the Lefthand side (LHS).

    I think that should be
    [2/(D-d)] +[2/(D+d)] = 1/f
    Combine the two fractions in the LHS.
    [2(D+d) +2(D-d)] /[(D-d)(D+d)] = 1/f
    [2D +2d +2D -2d] /(D^2 -d^2) = 1/f
    [4D] /(D^2 -d^2) = 1/f
    Cross multiply,
    (4D)(f) = (D^2 -d^2)(1)
    4Df = D^2 -d^2
    d^2 = D^2 -4Df
    Take the square roots of both sides,
    d = sqrt[D^2 -4Df]
    Or,
    d = [D^2 -4Df]^(1/2)


    2) Given that y = loge (Ae^(-Bx)), find x in terms of y, A and B.

    The answer is 1/B(loge A - y)


    y = ln[A*e^(-Bx)]
    y = ln[A] +ln[e^(-Bx)]
    y = ln[A] +(-Bx)ln[e]
    y = ln[A] -(Bx)(1)
    y = ln[A] -Bx
    Bx = ln[A] -y
    x = (1/B)(ln[A] -y)

    We used:
    ---Log(ab) = Log(a) +Log(b)
    ---Log(a^b) = b*Log(a)
    ---Log(to the base a)[a] = 1

    Differentiation Questions

    3) Differentiate y = ln(3x^(2) +1) with respect to x.


    The answer is 6x/(3x^(2) +1)


    y = ln[3x^2 +1]
    dy/dx = (1/[3x^2 +1])*(6x)
    dy/dx = (6x) /(3x^2 +1)


    4) Differentiate y = (x^(2) + x)^(-2) with respect to x

    The answer is -(4x+2)(x^(2)+x)^(-3)


    y = (x^2 +x)^(-2)
    dy/dx = {(-2)[x^2 +x]^(-3)}*(2x +1)
    dy/dx = (-4x -2)[x^2 +x]^(-3)
    dy/dx = -(4x +2)[x^2 +x]^(-3)


    5) By differentiating the relation x^2 + y^2 = e^k (where k is a constant) with respect to x, find dy/dx.

    The answer is -x/y


    x^2 +y^2 = e^k
    Differentiate both sides,
    (2x)dx +(2y)dy = 0
    (x)dx +(y)dy = 0
    Divide both sides by dx,
    x +y(dy/dx) = 0
    y(dy/dx) = -x
    dy/dx = -x/y
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Singh_87 View Post
    2) Given that y = loge (Ae^(-Bx)), find x in terms of y, A and B.

    The answer is 1/B(loge A - y)
    The standard symbol (as I learned it, anyway) for log_e is ln. Thus
    y = ln(Ae^{-Bx}) <-- ln(ab) = ln(a) + ln(b)

    y = ln(A) + ln(e^{-Bx}) <-- ln(a^b) = b \cdot ln(a)

    y = ln(A) + -Bx \cdot ln(e) <-- ln(e) = 1

    y = ln(A) - Bx

    Bx = y - ln(A)

    x = \frac{y - ln(A)}{B}

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Singh_87 View Post
    3) Differentiate y = ln(3x^(2) +1) with respect to x.

    The answer is 6x/(3x^(2) +1)
    y = ln(3x^2 + 1)

    Recall that \frac{d}{dx}ln(x) = \frac{1}{x}. So applying the chain rule:
    \frac{dy}{dx} = \frac{1}{3x^2 + 1} \cdot (6x) = \frac{6x}{3x^2 + 1}

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Singh_87 View Post
    4) Differentiate y = (x^(2) + x)^(-2) with respect to x

    The answer is -(4x+2)(x^(2)+x)^(-3)
    Again, use the chain rule:
    y = (x^2 + x)^{-2}

    \frac{dy}{dx} = (-2) \cdot (x^2 + x)^{-3} \cdot (2x + 1) = -(4x + 2)(x^2 + x)^{-3}

    -Dan
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Singh_87 View Post
    5) By differentiating the relation x^2 + y^2 = e^k (where k is a constant) with respect to x, find dy/dx.

    The answer is -x/y
    This is an implicit differentiation problem. Note that when we use the chain rule to find the derivative of a function f(y) with respect to x we get \frac{df}{dy} \cdot \frac{dy}{dx}, so
    x^2 + y^2 = e^k <-- Notice that the right hand side is merely a constant!

    Taking the derivative of both sides gives:
    2x + 2y \cdot \frac{dy}{dx} = 0

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

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

    \frac{dy}{dx} = - \frac{x}{y}

    -Dan
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