Posted by Little on Sunday, February 13, 2011 at 2:47pm.
Consider the curve given by the equation y^3+3x^2y+13=0
b. Write an equation for the line tangent to the curve at the point (2,-1)
c. Find the minimum y-coordinate of any point on the curve.
the work for these would be appreciated i don't need the answers.
- calculus - Reiny, Sunday, February 13, 2011 at 3:11pm
3y^2 dy/dx + 3x^2 dy/dx + 6xy + 0 = 0
dy/dx(3y^2 + 3x^2) = -6xy
dy/dx/ = -6xy/(3y^2 + 3x^2)
b) sub in x = 2, to get the slope, then use the grade 9 way to find the equation of a line given the slope and a point.
c) set dy/dx = 0
so -6xy = 0
x = 0 or y = 0
when x = 0, y^3 +1 = -13
y = (-13)^(1/3) = appr. - 2.35
when y = 0, 0 + 0 + 13 = 0 , not possible.
so we have one turning point at (0,-2.35)
what about when x becomes large either + or -
investigate for x = 100
y^3 + 30000y = -13
y = -13/30000 which approaches zero.
the same thing is true if x = -100
so the minimum value of y is -2.35 when x = 0
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