the curve: (x)(y^2)-(x^3)(y)=6
(dy/dx)=(3(x^2)y-(y^2))/(2xy-(x^3))
a) find all points on the curve whose x-coordinate is 1 and write an equation for the tangent line of each of these points
b)find the x-coordinate of each point on the curve where the tangent line is vertical
a) If x = 1, y^2 -y -6 = 0
(y-3)(y+2) - 0
y = 3 or -2
b) Find the points where dy/dx = infinity
Those would be the places where the denominator of dy/dx = 0. x=0 is one such point. Any point where x^2 = 2y would be another, if there is such a point on the curve.
To find the points on the curve whose x-coordinate is 1, we need to substitute x = 1 into the equation of the curve and solve for y.
Given curve: (x)(y^2) - (x^3)(y) = 6
Substitute x = 1:
(1)(y^2) - (1^3)(y) = 6
y^2 - y = 6
y^2 - y - 6 = 0
Now we can solve this quadratic equation for y by factoring:
(y - 3)(y + 2) = 0
From this, we find two possible values for y: y = 3 or y = -2.
So the points on the curve whose x-coordinate is 1 are (1, 3) and (1, -2).
a) To find the equation for the tangent line at each of these points, we need the slope of the tangent line, which is given by the derivative of the curve.
Given derivative: (dy/dx) = (3(x^2)y - (y^2))/(2xy - (x^3))
For the point (1, 3):
Substitute x = 1 and y = 3 into the derivative:
(dy/dx) = (3(1^2)(3) - (3^2))/(2(1)(3) - (1^3))
(dy/dx) = (9 - 9)/(6 - 1)
(dy/dx) = 0/5
(dy/dx) = 0
Since the derivative is 0, this means that the tangent line is horizontal.
The equation for the tangent line at the point (1, 3) is y = 3.
For the point (1, -2):
Substitute x = 1 and y = -2 into the derivative:
(dy/dx) = (3(1^2)(-2) - ((-2)^2))/(2(1)(-2) - (1^3))
(dy/dx) = (-6 - 4)/(-4 - 1)
(dy/dx) = -10/-5
(dy/dx) = 2
The equation for the tangent line at the point (1, -2) is y = 2x - 4.
b) To find the x-coordinate of each point on the curve where the tangent line is vertical, we need to find the values of x where the derivative becomes undefined. In this case, it happens when the denominator of the derivative becomes 0.
For the derivative (dy/dx):
2xy - x^3 = 0
Factor out an x:
x(2y - x^2) = 0
From this equation, we can see that either x = 0 or 2y - x^2 = 0.
If x = 0, then the tangent line is vertical. However, this does not give us a point on the curve since it would be at the origin.
For 2y - x^2 = 0:
2y = x^2
y = (x^2)/2
So, the x-coordinate of each point on the curve where the tangent line is vertical is given by x = sqrt(2y).
Note that this only gives us the x-coordinate, while the y-coordinate can be found by substituting this x into the equation of the curve.