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Thread: differentiation

  1. #1
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    differentiation

    1. the curve y=ax^2+bx passes through the point (2,4) with gradient 8.

    find a and b

    2. show that if y=2x-x^2 then

    y 2^2y - 2 dy +2y = 4(x+1)
    dx^2 dx

    3. A child's height h cm at age 8 years can be modelled by the equation

    h= _ a^4
    500 + (a-1/2)^2 +55

    for ages 11 ≤ a ≤ 16

    Find the child's annual growth rate at age 12, and at age 15

    I am completely stuck with what to do, so a little guidance with how to go about solving would be great. Thanks

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  2. #2
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    Hello, lra11!

    1. The curve $\displaystyle y\:=\:ax^2+bx$ passes through the point $\displaystyle (2,4)$ with gradient $\displaystyle 8.$

    Find $\displaystyle a$ and $\displaystyle b.$

    We have: .$\displaystyle y \:=\:ax^2+b$

    Since (2,4) is on the curve: .$\displaystyle 4 \:=\:a\!\cdot\!2^2 + b\quad\Rightarrow\quad 4a + b \:=\:4$ .[1]

    The gradient is given by: .$\displaystyle y' \:=\:2ax$
    . . At (2,4), we have: .$\displaystyle 8 \:=\:2a(2)\quad\Rightarrow\quad\boxed{a \,=\,2}$

    Substitute into [1]: .$\displaystyle 4(2) + b \:=\:4\quad\Rightarrow\quad\boxed{b \:=\:-4}$




    2. Show that if $\displaystyle y\:=\:2x-x^2$, then

    y 2^2y - 2 dy +2y = 4(x+1) . . . . . impossible to read this!
    dx^2 dx
    If I were to guess what it said . . . maybe: .$\displaystyle y\!\cdot\!\frac{d^2y}{dx^2} - 2\!\cdot\!\frac{dy}{dx} + 2y \;=\;4(x+1)$
    . . but it is not true!

    We have: .$\displaystyle \begin{Bmatrix}y & = & 2x-x^2 \\ \\ \dfrac{dt}{dx} & = & 2-2x \\ \\ \dfrac{d^2y}{dx^2} &=& -2 \end{Bmatrix}$


    Then: . $\displaystyle \underbrace{y}_{\downarrow}\;\;\cdot\;\;\underbrac e{\frac{d^2y}{dx^2}}_{\downarrow} \:- \:2 \cdot \underbrace{\frac{dy}{dx}}_{\downarrow} \;\;\;+\;\;\;2\;\cdot\;\underbrace{y}_{\downarrow} $
    . . . . . $\displaystyle (2x-x^2)(-2) - 2(2-2x) + 2(2x-x^2) $

    . . . . $\displaystyle -4x + 2x^2 - 4 + 4x + 4x - 2x^2$

    . . . $\displaystyle = \;4x - 4 \;=\;{\color{blue}4(x-1)}$



    3. A child's height $\displaystyle h$ cm at age $\displaystyle a$ years can be modelled by the equation

    . . $\displaystyle h \;=\;-\frac{a^4}{500} + \left(a - \frac{1}{2}\right)^2 + 55$ . for ages $\displaystyle 11 \leq a \leq 16$ . . . . another guess

    Find the child's annual growth rate at age 12, and at age 15.

    Differentiate: .$\displaystyle \frac{dh}{da} \;=\;-\frac{4a^3}{500} + 2\left(a - \frac{1}{2}\right) \;=\;-\frac{a^3}{125} + 2a - 1$


    When $\displaystyle a = 12\!:\;\;\frac{dh}{da} \;=\;-\frac{12^3}{125} + 2(12) - 1 \;=\;9.176\text{ cm/year} $

    When $\displaystyle a = 15\!:\;\;\frac{dh}{dt} \;=\;-\frac{15^3}{125} + 2(15) - 1 \;=\;2\text{ cm/year}$

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