# Calculus AB

1) Find the equation of the line tangent to the given curve at the indicated point: 3y^3 + 2x^2 = 5 at a point in the first quadrant where y=1.

2) Show that there is no point on the graph of x^2 - 3xy + y^2 = 1 where the line tangent to the curve is horizontal.

3) Find all points on the curve xy^2 - x^3y = 6 where the tangent line is vertical. Explain your reasoning.

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1. 3y^3 + 2x^2 = 5
9y^2 dy/dx + 4x = 0
dy/dx = -4x/9y^2

when y = 1 , 3 + 2x^2 = 5
2x^2 = 2
x = Âą 1 , but in quadrant 1, x = 1

dy/dx = -4(1)/9 = -4/9
equation of tangent:
y-1 = (-4/9)(x-1)
9y-9 = -4x + 4
4x + 9y = 13

check:
http://www.wolframalpha.com/input/?i=plot+3y%5E3+%2B+2x%5E2+%3D+5,+4x+%2B+9y+%3D+13
notice the tangent at (1,1)

2.
x^2 - 3xy + y^2 = 1
2x -3x(dy/dx) - 3y + 2y dy/dx = 0
dy/dx(-3x + 2y) = 3y - 2x
dy/dx = (3y - 2x)/(2y - 3x)
to have a horizontal slope, dy/dx = 0 , that is,
3y - 2x = 0
y = 2x/3
sub that back into the original
x^2 - 3x(2x/3) + 4x^2/9 = 1
times 9
9x^2 - 18x^2 + 4x^2 = 1
-5x^2 = 1
There is no real solution to this, so there is no such tangent.

3.
find the derivative like I showed you in the first two.
Since the tangent is vertical, the denominator of the slope must be zero.
Show that this is not possible.

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posted by Reiny
2. Thank you so much! And I will

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2. đ 0
posted by Annie

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