Consider the equation xy2 − x2y = 6.
(a) Find all points on the graph when x = 2.
(b) Find a formula for dy
dx in terms of x and y.
(c) Find the tangent lines for the points you found in the first part
a) 2(y^2-2)=6
solve for y.
y^2-2=6
y= +-sqrt8, x=2
b) y^2+2yy'x-2xy-y'x^2=0
y'(2xy-x^2)=2xy-y^2
y'= you finish it.
c. slope=y', put in the two points in a), and calculate
(a) x=2
2y^2-4y = 6
y^2-2y-3 = 0
(y-3)(y+1) = 0
So, the points are (2,3) and (2,-1)
(b)
xy^2-x^2y = 6
y^2 + 2xyy' - 2xy - x^2y' = 0
y'(2xy-x^2) = 2xy-y^2
y' = (2xy-y^2)/(2xy-x^2)
(c)
y'(2,3) = 3/8
y'(2,-1) = 5/8
check:
http://www.wolframalpha.com/input/?i=plot+xy%5E2-x%5E2y%3D6,+y%3D(3%2F8)(x-2)%2B3,+y%3D(5%2F8)(x-2)-1,x%3D2
To solve the given equation xy^2 - x^2y = 6, we will address each part of the question:
(a) Finding all points on the graph when x = 2:
To find the points on the graph when x = 2, substitute x = 2 into the equation. We get:
2y^2 - 4y = 6
Rearrange the equation to get it in quadratic form:
2y^2 - 4y - 6 = 0
Now we need to solve this quadratic equation. We can either factor it or use the quadratic formula. Let's use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -4, and c = -6. Plugging in these values, we find:
y = (-(-4) ± √((-4)^2 - 4*2*(-6))) / (2*2)
= (4 ± √(16 + 48)) / 4
= (4 ± √64) / 4
= (4 ± 8) / 4
Thus, we have two possible solutions:
y = (4 + 8) / 4 = 12 / 4 = 3
and
y = (4 - 8) / 4 = -4 / 4 = -1
So when x = 2, the points on the graph are (2, 3) and (2, -1).
(b) Finding a formula for dy/dx in terms of x and y:
To find the derivative dy/dx, we need to differentiate the given equation implicitly with respect to x. First, let's rearrange the equation:
xy^2 - x^2y = 6
Now, implicitly differentiate both sides with respect to x:
d(xy^2)/dx - d(x^2y)/dx = d(6)/dx
(y^2 + 2xy * dy/dx) - (2xy * dy/dx + x^2 * dy/dx) = 0
Simplifying this equation, we have:
y^2 + 2xy * dy/dx - 2xy * dy/dx - x^2 * dy/dx = 0
Combining like terms, we get:
y^2 - x^2 * dy/dx = 0
Rearranging this equation to make dy/dx the subject, we have:
dy/dx = y^2 / x^2
Therefore, the formula for dy/dx in terms of x and y is dy/dx = y^2 / x^2.
(c) Finding the tangent lines for the points found in part (a):
To find the tangent lines for the points (2, 3) and (2, -1), we need to substitute these points into the equation we found for dy/dx.
For the point (2, 3):
Substitute x = 2 and y = 3 into the equation dy/dx = y^2 / x^2:
dy/dx = 3^2 / 2^2
= 9 / 4
So the slope of the tangent line at the point (2, 3) is 9/4.
For the point (2, -1):
Substitute x = 2 and y = -1 into the equation dy/dx = y^2 / x^2:
dy/dx = (-1)^2 / 2^2
= 1 / 4
So the slope of the tangent line at the point (2, -1) is 1/4.
Now we can use the point-slope formula to find the equations of the tangent lines:
For the point (2, 3):
y - 3 = (9/4)(x - 2)
For the point (2, -1):
y + 1 = (1/4)(x - 2)
These are the equations of the tangent lines for the points (2, 3) and (2, -1).