Posted by Jessica Chaney on Thursday, March 1, 2012 at 1:13pm.
3x^2 + 3y^2 + 6xyy' + 3y^2 y' = 0
y' = -(x^2+y^2)/(2xy + y^2)
when x=0, y=1
y'(0) = -(0+1)/(0+1) = -1
tangent line at (0,1) is y=-x+1
slope of normal = 1
line is y=x+1
only place where tangent could be horizontal (y'=0) would be at (0,0) but that point is not on the graph.
Vertical tangents where 2xy+y^2 = 0
at y=0 x=1
y' = -1/0 -- vertical tangent
y(2x+y) = 0
y = -2x
y' = -(x^2 + 4x^2)/(2x(-2x) + 4x^2) = -5x^2/0 -- vertical tangent
x^3 + 3x(4x^2) - 8x^3 = 1
x^3 + 12x^3 - 8x^3 = 1
5x^3 = 1
x = 0.5848
so, vertical tangent at (.5848,-1.1696)
physics - A swimmer is capable of swimming 0.45 m/s in still water. (a) If she ...
physics - A batter hits a fly ball which leaves the bat 0.90 m above the ground ...
Please help for my test CALculus - using the chain rule find the min and max ...
English - 1. I like taking walks. 2. I like taking a walk. (Which one is OK? Are...
religious education - Re teachers - I need to write an essay about baptism in ...
AP Calculus AB - The sum of x and y is 8. What is the maximum value of xy-x^2? ...
ict year 8 - i`m struggling with my ict homework. i need to find the definitions...
URGENT!!!!9th grade tech - i have my ISUs due tomorrow and if i dont get them in...
Algebra - factor the expression x(a+2) - 2(a+2) i missed this day because i was ...
Just for I.AM..RUNNING.FROM.MYSELF, - I'm doing a poster in science and its ...
For Further Reading