determine the point(s)at which the graph of y^4 = y^2 -x^2 has a horizontal tangent.
y^4 = y^2 - x^2
4y^3 y' = 2y y' - 2x
y' = -2x/(4y^3-2y)
Now, assuming that y ≠ 0 or ±1/√2,
y'=0 when x=0
so, y^4 = y^2, and y = ±1, so
(0,1) and (0,-1)
extra credit: what happens at y = 0 or ±1/√2?
how did you know it was + or - l? I got what x equals but I was confused on how to find the y values?
if x=0,
y^4 = y^2
y^2(y^2-1) = 0
y = 0 or ±1
but y=0 is not a possibility.
where did you get y^2 -1 from? Did you subtract y^2 from the right side? if so where did the -1 come from?
Hello? This is calculus. Have you forgotten you algebra I?
y^4 = y^2
y^4 - y^2 = 0
y^2(y^2-1) = 0
...
Well, determining the point(s) at which the graph of y^4 = y^2 - x^2 has a horizontal tangent might require some tangential thinking. So, let's dive in!
To find the points where the graph has a horizontal tangent, we need to find where the derivative of y with respect to x equals zero. So, let's start by finding the derivative.
Taking the derivative of both sides of the equation y^4 = y^2 - x^2, we get:
4y^3 * dy/dx = 2y * dy/dx - 2x
Now, let's isolate dy/dx to find where it equals zero:
dy/dx(4y^3 - 2y) = 2x
dy/dx = 2x / (4y^3 - 2y)
Now, for the graph to have a horizontal tangent, dy/dx needs to equal zero. So, we set 2x / (4y^3 - 2y) = 0.
For dy/dx = 0, either 2x = 0 or (4y^3 - 2y) = 0.
Let's start with 2x = 0. It implies that x = 0. So, one point where y^4 = y^2 - x^2 has a horizontal tangent is when x = 0.
Now, let's move on to (4y^3 - 2y) = 0:
2y(2y^2 - 1) = 0
This equation gives two possibilities:
1) 2y = 0, which means y = 0
2) 2y^2 - 1 = 0, which gives us two more potential points: y = -1/sqrt(2) and y = 1/sqrt(2).
So, in total, the points where the graph has a horizontal tangent are (0, 0), (0, -1/sqrt(2)), and (0, 1/sqrt(2)).
Remember, though, my approach might be comically mathematical, so feel free to double-check my work!
To determine the point(s) at which the graph of the equation y^4 = y^2 - x^2 has a horizontal tangent, we need to find the x and y coordinates of those points where the slope of the graph is zero.
Step 1: Differentiate the equation with respect to x.
Differentiating both sides of the equation y^4 = y^2 - x^2 with respect to x, we get:
4y^3(dy/dx) = 2y(dy/dx) - 2x
Step 2: Find the value of dy/dx when the slope is zero.
We set the derivative equal to zero since we are looking for the points where the slope is zero:
4y^3(dy/dx) - 2y(dy/dx) + 2x = 0
Factoring out dy/dx, we have:
(dy/dx)(4y^3 - 2y) + 2x = 0
For the slope to be zero, dy/dx must be zero. Therefore, we have:
(4y^3 - 2y) + 2x = 0
Step 3: Solve for y in terms of x.
Rearranging the equation, we get:
4y^3 - 2y = -2x
Now, solve for y in terms of x:
4y^3 - 2y + 2x = 0
2y(2y^2 - 1) = -2x
y(2y^2 - 1) = -x
Step 4: Substitute the value of y into the original equation.
Substitute y(2y^2 - 1) for x in the original equation y^4 = y^2 - x^2:
y^4 = y^2 - (y(2y^2 - 1))^2
Simplify the equation:
y^4 = y^2 - (2y^3 - y)^2
y^4 = y^2 - (4y^6 - 4y^3 + y^2)
Combine like terms:
y^4 - y^2 = -4y^6 + 4y^3 - y^2
Rearrange the equation:
4y^6 - 4y^3 + y^2 - y^4 = 0
Step 5: Solve for y.
Solve the equation 4y^6 - 4y^3 + y^2 - y^4 = 0 for the values of y. To solve this equation, you can use factoring or numerical methods such as graphical or numerical approximation.
Once you find the values for y, substitute them back into the equation y^4 = y^2 - x^2 from Step 4 to find the corresponding x values.
By following these steps, you will be able to find the point(s) at which the graph of y^4 = y^2 - x^2 has a horizontal tangent.