dy/dx= (2x-2xy)/(x^2+2y)
Find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of those points.
To find all points on the curve where x = 2, we substitute x = 2 into the equation dy/dx = (2x - 2xy) / (x^2 + 2y):
dy/dx = (2(2) - 2(2)(y)) / (2^2 + 2y)
dy/dx = (4 - 4y) / (4 + 2y)
Now, we want to find the value of y that satisfies the equation when x = 2. We can set x = 2 in the original equation and solve for y:
dy/dx = (2(2) - 2(2)(y)) / (2^2 + 2y)
dy/dx = (4 - 4y) / (4 + 2y)
Now, let's solve for y:
dy/dx = (4 - 4y) / (4 + 2y)
Multiplying both sides by (4 + 2y):
(4 + 2y) * dy/dx = 4 - 4y
Expanding:
4(dy/dx) + 2y(dy/dx) = 4 - 4y
Substituting dy/dx = (4 - 4y) / (4 + 2y):
4[(4 - 4y) / (4 + 2y)] + 2y[(4 - 4y) / (4 + 2y)] = 4 - 4y
Simplifying:
(16 - 16y + 8y - 8y^2) / (4 + 2y) + (8y - 8y^2) / (4 + 2y) = 4 - 4y
Combining like terms:
(16 - 8y - 8y^2) / (4 + 2y) = 4 - 4y
Multiplying both sides by (4 + 2y):
16 - 8y - 8y^2 = (4 - 4y)(4 + 2y)
Expanding:
16 - 8y - 8y^2 = 16 - 8y - 16y + 8y^2
Rearranging the terms:
0 = 16y^2 - 8y - 8y^2
Combining like terms:
0 = 8y^2 - 8y
Factoring out 8y:
0 = 8y(y - 1)
Setting each factor equal to zero:
8y = 0 or y - 1 = 0
Solving for y:
y = 0 or y = 1
Therefore, the points on the curve where x = 2 are (2, 0) and (2, 1).
Now, to show that there is a horizontal tangent to the curve at one of those points, we need to find the derivative of y with respect to x:
dy/dx = (2x - 2xy) / (x^2 + 2y)
Substituting x = 2 and y = 0:
dy/dx = (2(2) - 2(2)(0)) / (2^2 + 2(0))
dy/dx = 4 / 4
dy/dx = 1
The derivative dy/dx at the point (2, 0) is 1. Since the derivative represents the rate of change or slope of the curve at a point, a derivative of 1 means the curve is increasing at that point. Therefore, the tangent line to the curve at (2, 0) is horizontal.
Hence, one of the points on the curve where x = 2 has a horizontal tangent.
To find the points on the curve where x = 2, we can substitute x = 2 into the given equation and solve for y. Let's do the math step-by-step:
1. Start with the given equation: dy/dx = (2x - 2xy) / (x^2 + 2y).
2. Substitute x = 2 into the equation: dy/dx = (2(2) - 2(2)y) / (2^2 + 2y).
3. Simplify the equation: dy/dx = (4 - 4y) / (4 + 2y).
Now, we need to find the value of y for which dy/dx will become undefined or approach infinity. This would indicate a horizontal tangent. To do this, we set the denominator of dy/dx to zero:
4 + 2y = 0.
Solving this equation, we find that y = -2.
So the point on the curve where x = 2 is (2, -2).
Lastly, we need to show that there is a horizontal tangent at this point. To do this, we need to find the derivative of the given equation and evaluate it at (2, -2). If the derivative is equal to zero at that point, then there is a horizontal tangent.
Taking the derivative of the equation dy/dx = (4 - 4y) / (4 + 2y):
d^2y/dx^2 = (-4(4 + 2y) - (4 - 4y)(2)) / (4 + 2y)^2.
Now we substitute x = 2 and y = -2 into this second derivative:
d^2y/dx^2 = (-4(4 + 2(-2)) - (4 - 4(-2))(2)) / (4 + 2(-2))^2
= (-4(4 - 4) - (4 + 8)(2)) / (4 - 4)^2
= (-4(0) - 12(2)) / 0.
We see that the second derivative is undefined at (2, -2) since it involves division by zero. Hence, there is a horizontal tangent at this point on the curve where x = 2.