the curve with equation 2(x^2+y^2)^2=9(x^2-y^2) is called a lemniscate. find the points on the lemniscate where the tangent is horizontal.
just plug and chug:
4(x^2 + y^2)(2x + 2yy') = 9(2x - 2yy')
Now, the tangent is horizontal when y' = 0
4(x^2 + y^2)(2x) = 9(2x)
8x^3 + 8xy^2 = 18x
2x(4x^2 + 4y^2 - 9) = 0
So, either x=0
substitute back into original equation:
2y^4 = -9y^2
y=0
or,
4x^2 + 4y^2 = 9
x^2 = (9 - 4y^2)/4
Substitute that back into the original equation
2((9 - 4y^2)/4 + y^2)^2 = 9((9 - 4y^2)/4 - y^2)
Expand the binomial and solve for y^2
Tan^3(xy^2+y)=x
Use derivative
Find .... Tan^3(xy^2+y)=x
Use derivative
To find the points on the lemniscate where the tangent is horizontal, we need to find the points where the derivative of the curve equation with respect to y is equal to zero. Let's start by differentiating the equation with respect to y.
Given curve equation: 2(x^2 + y^2)^2 = 9(x^2 - y^2)
Differentiating implicitly with respect to y:
d/dy [2(x^2 + y^2)^2] = d/dy [9(x^2 - y^2)]
To differentiate, we treat x as a constant:
4(x^2 + y^2)(2y) = 9(-2y)
Simplifying further:
8xy(x^2 + y^2) = -18y
Now, we need to find the values of x and y that satisfy this equation:
1. If y = 0, then the term -18y becomes 0, and we are left with:
8xy(x^2 + y^2) = 0
For this equation to be satisfied, either x = 0 or (x^2 + y^2) = 0.
2. If (x^2 + y^2) ≠ 0, then we can divide both sides of the equation by y:
8xy(x^2 + y^2)/y = -18y/y
8x(x^2 + y^2) = -18
Simplifying further, we get:
8x^3 + 8xy^2 = -18
Now, we eliminate y from the equation to find x. Substitute y^2 = 9(x^2 - y^2)/2 from the given equation into the above equation:
8x^3 + 8x(9(x^2 - y^2)/2) = -18
8x^3 + 4x(9x^2 - 9y^2) = -18
8x^3 + 36x^3 - 36xy^2 = -18
44x^3 - 36xy^2 = -18
Now, let's find the values of x that satisfy this equation:
1. If x = 0, then the term 44x^3 becomes 0, and we are left with:
-36xy^2 = -18
Again, either y = 0 or (x^2 + y^2) = 0.
2. If (x^2 + y^2) ≠ 0, then we can divide both sides of the equation by x:
(44x^3 - 36xy^2)/x = -18/x
44x^2 - 36y^2 = -18/x
Now, we can substitute y^2 = 9(x^2 - y^2)/2 from the given equation into this new equation:
44x^2 - 36(9(x^2 - y^2)/2) = -18/x
44x^2 - 36(9x^2 - 9y^2)/2 = -18/x
44x^2 - 36(9x^2 - 9(9(x^2 - y^2)/2))/2 = -18/x
Simplifying further, we get:
44x^2 - 18x^2 + 162y^2 = -18/x
26x^2 + 162y^2 = -18/x
This equation will help us find the values of x and y where the tangent to the curve is horizontal.