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Math Help - AP style Calculus Questions

  1. #1
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    AP style Calculus Questions

    1. Consider the differential equation dy/dx = 3(x^2) / e^2y

    (a) Find a solution for y = f(x) to the differential equation satisfying f(0) = .

    (b) Find the domain and range of the function f found in part (a). Justify your answer.



    2. Suppose that the function f has a continuous second derivative for all x, and that f(0) = 2, f (0) = -3, and f (0) = 0. Let g be a function whose derivative is given by g'(x) = e^-2x(3f(x) + 2f'(x)) for all x.

    (a) Write an equation of the line tangent to the graph of f at the point where x = 0.

    (c) Given that g(0) = 4, write an equation of the line tangent to the graph of g at the point where x = 0.

    (d) Show that g''(x) = e^-2x(-6f(x) 2f''(x)) At x = 0, is g (x) = 0 and g (x) < 0? Justify your answer.


    I apologize for not thanking whoever it was that helped me with the last questions i had. I ended up checking my answers with my friend, and kind of forgot to check here.
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  2. #2
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    Quote Originally Posted by Blue Griffin View Post
    2. Suppose that the function f has a continuous second derivative for all x, and that f(0) = 2, f (0) = -3, and f (0) = 0. Let g be a function whose derivative is given by g'(x) = e^-2x(3f(x) + 2f'(x)) for all x.

    (a) Write an equation of the line tangent to the graph of f at the point where x = 0.

    (c) Given that g(0) = 4, write an equation of the line tangent to the graph of g at the point where x = 0.

    (d) Show that g''(x) = e^-2x(-6f(x) 2f''(x)) At x = 0, is g (x) = 0 and g (x) < 0? Justify your answer.
    For A with the info given f(0)=2 \iff (0,2) and m=f'(0)=-3 so we get y-2=-3(x-0) \iff y=-3x+2

    for c) using the equation

    g'(x)=e^{-2x}(3f(x)+2f'(x)) \mbox{ and } g(0)=4 \iff (0,4)

    so g'(0)=e^{0}(3f(0)+2f'(0))=1(3 \cdot 2+2 \cdot(-3))=1(6-6)=0

    So the equation of the tangent line is y=4

    for the last one take the derivative of

    g'(x)=e^{-2x}(3f(x)+2f'(x))

    g''(x)=(3f(x)+2f'(x))\frac{d}{dx}e^{-2x}+e^{-2x}\frac{d}{dx}(3f(x)+2f'(x))=

    -2e^{-2x}(3f(x)+2f'(x))+e^{-2x}(3f'(x)+2f''(x))=

    e^{-2x}(-6f(x)-4f'(x)+3f'(x)+2f''(x))

    g''(x)=e^{-2x}(-6f(x)-f'(x)+2f''(x))

    I hope this helps.
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  3. #3
    Behold, the power of SARDINES!
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    1. Consider the differential equation dy/dx = 3(x^2) / e^2y

    (a) Find a solution for y = f(x) to the differential equation satisfying f(0) = .

    (b) Find the domain and range of the function f found in part (a). Justify your answer.
    The equation is seperable so

    \frac{dy}{dx}=\frac{3x^2}{e^{2y}} \iff e^{2y}dy=3x^2dx

    So we integrate both sides...

    \int e^{2y}dy= \int 3x^2dx \iff \frac{1}{2}e^{2y}=x^3+c

    Solving for y we get...

    e^{2y}=2x^3+C taking the ln

    2y = ln(2x^3+c) \iff y=\frac{1}{2}ln(2x^3+c)

    using the inital condition we get

    \frac{1}{2}=\frac{1}{2}ln(c) \therefore c= e

    so we get y=\frac{1}{2}ln(2x^3+e)

    Good luck.
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  4. #4
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    I'm sorry, i posted number 2 wrong. What you got was what it was supposed to be TheEmpty Set. Lol, thanks!
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