use implicit differentiation, ....
2y^3 + y^2 - y^5 = x^4 -2x^3 +x^2
6y^2 (dy/dx) + 2y (dy/dx) - 5y^4 (dy/dx) = 4x^3 - 6x^2 + 2x
dy/dx = (4x^3 - 6x^2 + 2x)/(6y^2 + 2y - 5y^4)
for a horizontal tangent, dy/dx = 0 ---> the numerator is zero
5x^3 - 6x^2 + 2x = 0
x(5x^2 - 6x + 2) = 0
the quadratic has no real solutions, so x = 0
then if x = 0
2y^3 + y^2 - y^5 = 0-0+0
y^2(2y + 1 - y^3) = 0
y = 0 , or y^3 - 2y - 1 = 0 ---> one point is (0,0)
y^3 - 2y - 1 = 0, easy to see that y = -1
using synthetic division by y+1 , I was left with
y^2 - y - 1 = 0 which solves to get
y = (1 ± √5)/2 ,..... mmmh, I see the golden ratio in there
So you got 3 points where the tangent is horizontal.
for vertical tangents, the denominator is zero
6y^2 + 2y - 5y^4 = 0
y(6y + 2 - 5y^3) = 0
y = 0 or 5y^3 - 6y - 2 = 0
so there is the (0,0) again, but 5y^3 - 6y - 2 = 0 is a mess to solve (unless I made an arithmetic error)
So I went with Wolfram:
http://www.wolframalpha.com/input/?i=5y%5E3+-+6y+-+2+%3D+0
to get 3 more points where the tangent is vertical
Might as well look at the original graph to see if this makes sense.
http://www.wolframalpha.com/input/?i=2y%5E3+%2B+y%5E2+-+y%5E5+%3D+x%5E4+-2x%5E3+%2Bx%5E2
Wow!
Can someone show how this question is solved.
Consider the curve given by the equation
2y^3 + y^2 - y^5 = x^4 -2x^3 +x^2
Find all points at which the tangent line to the curve is horizontal or vertical.
Thanks!
1 answer
1 answer