Question

if $-xy + y^{4}=-2$ then find the equations of all tangent lines to the curve when $y = -1$.

Explanation:

Step1: Find the value of \( x \) when \( y = -1 \)

Substitute \( y = -1 \) into the equation \( -xy + y^4 = -2 \):

$$ -x(-1)+(-1)^4=-2 $$

Simplify the left - hand side:

$$ x + 1=-2 $$

Subtract 1 from both sides:

$$ x=-2 - 1=-3 $$

So the point of tangency is \( (-3,-1) \).

Step2: Implicitly differentiate the equation \( -xy + y^4=-2 \) with respect to \( x \)

Use the product rule for \( -xy \) (the product rule states that if \( u = -x \) and \( v = y \), then \( \frac{d(uv)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx} \)) and the chain rule for \( y^4 \) (if \( u = y \), then \( \frac{d(u^4)}{dx}=4u^3\frac{du}{dx}=4y^3\frac{dy}{dx} \)).

Differentiate each term:

• The derivative of \( -xy \) with respect to \( x \) is \( -y - x\frac{dy}{dx} \) (using the product rule: \( \frac{d(-x)}{dx}=-1 \), so \( -x\frac{dy}{dx}+y(-1)=-y - x\frac{dy}{dx} \))
• The derivative of \( y^4 \) with respect to \( x \) is \( 4y^3\frac{dy}{dx} \) (by the chain rule)
• The derivative of the constant \( -2 \) with respect to \( x \) is \( 0 \)

So, differentiating the entire equation \( -xy + y^4=-2 \) gives:

$$ -y - x\frac{dy}{dx}+4y^3\frac{dy}{dx}=0 $$

Step3: Solve for \( \frac{dy}{dx} \) (the slope of the tangent line)

Rearrange the equation to solve for \( \frac{dy}{dx} \):

$$ -x\frac{dy}{dx}+4y^3\frac{dy}{dx}=y $$

Factor out \( \frac{dy}{dx} \):

$$ \frac{dy}{dx}(-x + 4y^3)=y $$

Then,

$$ \frac{dy}{dx}=\frac{y}{4y^3 - x} $$

Step4: Substitute \( x = -3 \) and \( y=-1 \) into the formula for \( \frac{dy}{dx} \)

Substitute \( x=-3 \) and \( y = -1 \) into \( \frac{dy}{dx}=\frac{y}{4y^3 - x} \):

$$ \frac{dy}{dx}=\frac{-1}{4(-1)^3-(-3)} $$

Simplify the denominator:

$$ 4(-1)^3-(-3)=-4 + 3=-1 $$

So, \( \frac{dy}{dx}=\frac{-1}{-1}=1 \)

Step5: Use the point - slope form of a line to find the equation of the tangent line

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(-3,-1) \) and \( m = 1 \) (the slope we found).

Substitute the values into the point - slope form:

$$ y-(-1)=1(x - (-3)) $$

Simplify the equation:

$$ y + 1=x + 3 $$

Which can be rewritten as:

$$ y=x + 2 $$

Answer:

The equation of the tangent line is \( y=x + 2 \)