(x^2+y^2)^2=16y
Find all values of x where the tangent line is vertical
I found the derivative and looked for where it was undefined but it didn't work
using implicit differentiation,
2(x^2+y^2)(2x+2yy') = 16y'
x(x^2+y^2) + y(x^2+y^2)y' - 4y' = 0
y' = -x(x^2+y^2)/(x^2y + y^3 - 4)
So, the slope is vertical where
x^2y-y^3 -4 = 0
y(x^2-y^2) = 4
Not easy to solve in general, but a little playing around gives
x = ±√3, y=1
Can you show some more steps in between I can't get how to solve for y and x with only 1 equation.
Also isn't there an infinite number of solutions to this equation? How are you supposed to solve it then?
To find the values of x where the tangent line is vertical for the given equation, you need to identify the points on the curve where the derivative is undefined. However, sometimes finding the derivative alone may not be sufficient in this case.
Let's go through the process step by step:
Step 1: Start by rewriting the equation in a manageable form:
(x^2 + y^2)^2 = 16y
Step 2: Differentiate both sides of the equation implicitly with respect to x. This will help us find the derivative dy/dx:
d/dx[(x^2 + y^2)^2] = d/dx[16y]
Step 3: Apply the chain rule to the left side of the equation:
2(x^2 + y^2)(2x + 2y(dy/dx)) = 16(dy/dx)
Step 4: Simplify the equation:
2(x^2 + y^2)(x + y(dy/dx)) = 8(dy/dx)
Step 5: Rearrange the equation to solve for dy/dx:
(dy/dx)(2(x^2 + y^2) - 8) = -2(x^2 + y^2)
Step 6: Simplify further to get the derivative dy/dx:
dy/dx = -2(x^2 + y^2) / (2(x^2 + y^2) - 8)
Step 7: Now, to find the values of x where the tangent line is vertical, we need to find the points (x, y) where dy/dx is undefined. This occurs when the denominator of dy/dx becomes zero.
Therefore, set the denominator equal to zero and solve for x:
2(x^2 + y^2) - 8 = 0
Simplifying this equation:
x^2 + y^2 = 4
Step 8: Now, you have an equation for a circle with a radius of 2 centered at the origin (0, 0). All the points on this circle will have vertical tangent lines.
Step 9: To find the x-values for which the tangent line is vertical, you can solve the equation x^2 + y^2 = 4 for x. This can be done by substituting y^2 with (4 - x^2) and solving for x:
x^2 + (4 - x^2) = 4
4 = 4
Since the equation becomes an identity where both sides are equal, it means that any x-value satisfying the equation x^2 + y^2 = 4 will have a vertical tangent line.
Thus, all values of x at the points on the circle x^2 + y^2 = 4 will give you the x-values where the tangent line is vertical for the given equation.