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December 29, 2014

December 29, 2014

Posted by **Little** on Sunday, February 13, 2011 at 2:47pm.

a.find dy/dx

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 dont need the answers.

- calculus -
**Reiny**, Sunday, February 13, 2011 at 3:11pmdifferentiate implicitly.

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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