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Thread: 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 $\displaystyle y=(x^2)+3$ that pass through the point (-2,-9)

    2) Consider the curve defined by the equation $\displaystyle y+siny=x+1$ 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 $\displaystyle y=(x^2)+3$ that pass through the point (-2,-9)

    2) Consider the curve defined by the equation $\displaystyle y+siny=x+1$ 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 $\displaystyle y\:=\:x^2+3$
    that pass through the point $\displaystyle P(-2,\,-9)$

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

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

    We want the slope of $\displaystyle PQ$ to equal $\displaystyle 2p$.

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

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

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

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

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


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

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

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

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




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

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

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

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

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

    Then: .$\displaystyle \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: .$\displaystyle 1 + \cos y \:=\:0\quad\Rightarrow\quad\cos y \:=\:-1\quad\Rightarrow\quad y \:=\:\pi$

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



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

    Then: .$\displaystyle \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}
    $

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

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

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