Find the smallest value of x for which the following curve has vertical tangent lines: y= x^2 + xy^2 - x/ x^2
since it's kind of dumb just to have x/x^2, I assume you meant
y = (x^2 + xy^2 - x)/x^2
y = (x+y^2-1)/x
xy = x+y^2-1
xy - y^2 - x + 1 = 0
This is a degenerate conic -- just two straight lines:
y=1 and y=x-1
So, why don't use use some parentheses and tell us what you really meant ...
To find the smallest value of x for which the curve given by the equation y = x^2 + xy^2 - x / x^2 has vertical tangent lines, we need to find the value of x where the derivative of y with respect to x is undefined.
First, let's find the derivative of y with respect to x. We can use the Quotient Rule to differentiate the equation:
dy/dx = d/dx (x^2 + xy^2 - x / x^2)
= ( d/dx (x^2 + xy^2 - x) * x^2 - (x^2 + xy^2 - x) * d/dx (x^2) ) / ( x^2 )^2
Simplifying this expression, we get:
dy/dx = (2x + y^2 - 1) / x^2
For the derivative to be undefined, the denominator of this expression should be equal to zero. So we need to find the value of x that satisfies the equation x^2 = 0.
The value of x that satisfies this equation is x = 0.
Therefore, the smallest value of x for which the given curve has vertical tangent lines is x = 0.
To find the smallest value of x for which the given curve has vertical tangent lines, we need to find the derivative of y with respect to x and solve for the values of x that make the derivative undefined.
Let's start by finding the derivative of y with respect to x. The given equation is y = x^2 + xy^2 - x / x^2.
To better understand the equation, we can rewrite it as:
y = x^2 + xy^2 - (x / x^2)
Now, let's differentiate both sides of the equation implicitly with respect to x. We will use the quotient rule to differentiate the term (x / x^2):
dy/dx = d/dx (x^2) + d/dx(xy^2) - d/dx(x / x^2)
Differentiating each term:
The derivative of x^2 with respect to x is 2x.
The derivative of xy^2 with respect to x involves using both the product rule and chain rule:
d/dx(xy^2) = y^2 + 2xy(dy/dx)
To calculate dy/dx, we will solve the equation dx/dy = 0, indicating that the tangent line is vertical:
d/dx(x / x^2) = (1 * x^2 - x * 2x) / (x^2)^2 = (x^2 - 2x^2) / x^4 = -x^2 / x^4 = -1 / x^2
Now, substituting the derivatives back into the derivative of y:
dy/dx = 2x + y^2 + 2xy(dy/dx) - (-1 / x^2)
= 2x + y^2 + 2xy(dy/dx) + 1 / x^2
Since we want to find where the curve has vertical tangents, we set dy/dx equal to infinity:
2x + y^2 + 2xy(dy/dx) + 1 / x^2 = ∞
2x + y^2 = ∞- 2xy(dy/dx) - 1 / x^2
2x + y^2 = - 2xy(dy/dx) - 1 / x^2
2x + y^2 = - 2xy(∞) - 1 / x^2
As y is a function of x, y = y(x). Also, let's recall that ∞ is not a number but an indication of an undefined quantity, so the equation becomes:
2x + y^2 = - ∞ - 1 / x^2.
To find the smallest value of x for which the curve has vertical tangent lines, we need to find the x-values that make this equation hold. Since we have an undefined value (∞) on the right side and a finite term (2x + y^2) on the left side, the equation is only possible if the left side is also ∞ (undefined).
For 2x + y^2 to be ∞, y^2 should also be ∞, which means y should be ±∞.
Hence, to find the smallest value of x, we need to determine the x-values that make y ±∞.
To find these critical points, we substitute y = ±∞ into the equation and solve for x:
For y = ∞:
2x + (∞)^2 = - ∞ - 1 / x^2
2x + ∞ = - ∞ - 1 / x^2
Since we have ∞ on both sides, we cannot solve this equation.
For y = -∞:
2x + (-∞)^2 = - ∞ - 1 / x^2
2x + ∞ = - ∞
Again, we have ∞ on both sides, making it impossible to solve.
Therefore, the given curve does not have any x-values for which the tangent lines are vertical.