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Math Help - Calc test tomorrow.. need help studying

  1. #1
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    Calc test tomorrow.. need help studying

    I have a calc test tomorrow, and as I was studying, there were some problems I couldn't figure out. Can anyone help me with these, please..?

    1) Find the equations of the two tangent lines to the graph of y=(x^2)+3 that pass through the point (-2,-9)

    2) Consider the curve defined by the equation y+siny=x+1 for 0 less than or equal to y less than or equal to 2pi

    a) find dy/dx in terms of y

    b) Write an equation for each vertical tangent to the curve.

    c) Find d^2y/dx^2 in terms of y


    thanks so much for the help! i really need it, hehe:]
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Adrienne View Post
    I have a calc test tomorrow, and as I was studying, there were some problems I couldn't figure out. Can anyone help me with these, please..?

    1) Find the equations of the two tangent lines to the graph of y=(x^2)+3 that pass through the point (-2,-9)

    2) Consider the curve defined by the equation y+siny=x+1 for 0 less than or equal to y less than or equal to 2pi

    a) find dy/dx in terms of y

    b) Write an equation for each vertical tangent to the curve.

    c) Find d^2y/dx^2 in terms of y


    thanks so much for the help! i really need it, hehe:]
    i already responded to this here. do not double posts. it wastes people's time. in fact, i think i see Soroban working on this right now when he could be helping someone else
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  3. #3
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    Hello, Adrienne!

    1) Find the equations of the two tangent lines to the graph of y\:=\:x^2+3
    that pass through the point P(-2,\,-9)

    Let the point of tangency be Q(p,\,p^2+3)

    The slope of the tangent is: . y \:=\:2x
    At Q(p,\,p^2+3), the slope is: . m \:=\:2p

    We want the slope of PQ to equal 2p.

    The slope of PQ is: . \frac{(p^2+3) - (-9)}{p - (-2)} \;=\;\frac{p^2+12}{p+2}

    Then we have: . \frac{p^2+12}{p+2} \:=\:2p\quad\Rightarrow\quad p^2 + 4p - 12\:=\:0

    . . which factors: . (p-2)(p+6) \:=\:0

    . . and has roots: . p \:=\:2,\,-6

    The corresponding y-values are: . 7,\:39


    We have two points of tangency: . Q(2,\:7),\;\;R(-6,\:39)

    . . At Q(2,\:7), the slope is: . m = 4

    . . At R(-6,\:39), the slope is: . m = -12

    You can write the equations of the tangent lines now . . .




    2) Consider the curve defined by the equation y+\sin y=\:x\:+1 for 0 \leq y \leq 2\pi

    a) Find \frac{dy}{dx} in terms of y

    b) Write an equation for each vertical tangent to the curve.

    c) Find \frac{d^2\!y}{dx^2} in terms of y

    a) Differentiate implicitly: . \frac{dy}{dx} + \cos y\cdot\frac{dy}{dx} \:=\:1

    Then: . \frac{dy}{dx}(1 + \cos y) \:=\:1\quad\Rightarrow\quad\boxed{\frac{dy}{dx} \:=\:\frac{1}{1 + \cos y}}



    b) For a vertical tangent, the slope is undefined; its denominator is zero.

    So we have: . 1 + \cos y \:=\:0\quad\Rightarrow\quad\cos y \:=\:-1\quad\Rightarrow\quad y \:=\:\pi

    Substitute into the equation: . \pi + \sin\pi \:=\:x + 1\quad\Rightarrow\quad\boxed{ x \:=\:\pi-1}



    c) We have: . \frac{dy}{dx}\;=\;(1 + \cos y)^{-1}

    Then: . \frac{d^2\!y}{dx^2} \;\;=\;\;-(1+\cos y)^{-2}(-\sin y)\cdot\frac{dy}{dx} \;\;=\;\;\frac{\sin y}{(1+\cos y)^2}\cdot\frac{dy}{dx}<br />

    Since \frac{dy}{dx}\,=\,\frac{1}{1+\cos y}, we have: . \frac{d^2\!y}{dx^2} \;\;=\;\;\frac{\sin y}{(1+\cos y)^2}\cdot\frac{1}{1+\cos y}

    Therefore: . \boxed{\frac{d^2\!y}{dx^2} \;\;=\;\;\frac{\sin y}{(1+\cos y)^3}}

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