find the equation of all horizontal tangents to the curve y^2 = (x^2+4)/x, if any exist
this is what I have so far:
xy^2= x^2+4
2xydy/dx = 2x-y^2
dy/dx = (2x-y^2)/2xy
(2x-Y^2)/2xy = 0
2x- Y^2 = 0
I don't know what to do after this
you know that y^2 = (x^2+4)/x, so just plug that in and solve for x.
wouldn't everthing cancel?
2x - (x^2+4)/x = 0
2x^2 - x^2 - 4 = 0
x^2 - 4 = 0
x = -2 or 2
Now, find y at those values, and those are the horizontal tangents.
2x - ( x^2 + 4)/x = 0
times x
2x^2 -(x^2 + 4) = 0
x^2 = 4
x = ± 2
if x = 2, y^2 = 4, and dy/dx = 0 , of course !
a horizontal line through (2,4) is y = 4
if x = -2 ....... (you do it)
verification of result at
http://www.wolframalpha.com/input/?i=plot+y%5E2+%3D+%28x%5E2%2B4%29%2Fx
To find the equation of all horizontal tangents to the curve, we need to find the points where the derivative dy/dx is equal to zero. In other words, we need to solve the equation:
dy/dx = (2x - y^2) / (2xy) = 0
Let's break down the steps to solve this equation:
1. Start with the equation: dy/dx = (2x - y^2) / (2xy) = 0.
2. To make the equation equal to zero, the numerator (2x - y^2) must be equal to zero.
Set (2x - y^2) = 0.
3. Solve for y by isolating y in terms of x.
Subtract 2x from both sides of the equation:
-y^2 = -2x.
Divide both sides by -1:
y^2 = 2x.
Take the square root of both sides:
y = ± √(2x).
4. Now we have the possible values of y for each x. To find the horizontal tangents, we need to substitute y back into the original equation and solve for x.
Substitute y = √(2x):
y^2 = (x^2 + 4) / x.
(√(2x))^2 = (x^2 + 4) / x.
2x = (x^2 + 4) / x.
Multiply both sides by x to remove the fraction:
2x^2 = x^2 + 4.
Subtract x^2 from both sides:
x^2 = 4.
Take the square root of both sides:
x = ± 2.
Therefore, when y = √(2x), the corresponding x values are ±2.
Repeat step 4 for y = -√(2x):
Similarly, we obtain x = ±2.
5. The x-values obtained in step 4 correspond to the points where the graph of the curve has horizontal tangents. Therefore, the equation of all horizontal tangents is x = ±2.
In summary, there exist horizontal tangents to the curve y^2 = (x^2 + 4) / x, and their equation is x = ±2.