AP Calculus

Consider the curve given by x^2+4y^2=7+3xy

a) Show that dy/dx=(3y-2x)/(8y-3x)

b) Show that there is a point P with x-coordinate 3 at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P.

c) Find the value of d^2y/dx^2 (second derivative) at the point P found in part b). Does the curve have a local maximum, a local minimum, or neither at the point P? Justify your answer..

The only part that I did was a). I don't really know how to find Point P for b) and c)

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  1. a)
    2 x dx + 8 y dy = 3x dy + 3y dx

    dy(3x-8y)=dx(2x-3y)

    b)
    Where is the slope = 0?
    where 2x = 3y
    if x = 3
    9 + 4 y^2 = 7 + 9 y
    4 y^2 - 9 y + 2 = 0
    (4y-1)(y-2) = 0
    y = 1/4 or y = 2
    check both of those to find which works
    (3,2)
    3*2 - 2*3 = 0
    yes, (3,2) works
    (3,1/4)
    3/4 -6 = 0 no way so (3,2) is it

    c)
    dy/dx=(3y-2x)/(8y-3x)
    d^2y/dx^2 = [(8y-3x)(-2)-(3y-2x)(-3)] / (8y-3x)^2
    = [(16-9)(-2)]/(16-9)^2 = -2/7
    that is negative, so a maximum of the function

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  2. Thank you so much! It makes perfect sense and when I calculated it, it matched up.

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  3. I'm confused as to how to do part C), I don't remember how to find the second derivative of an implicit differentiation problem!

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