AP Calculus AB Exam Question:

Consider the equation of a curve defined implicitly by the equation:

1. Find $\frac{{dy}}{{dx}}$ using implicit differentiation.
2. Determine the slope of the tangent line to the curve at the point $(1, -1)$.
3. Find the equation of the tangent line at that point.

Answer:

1. Find $\frac{{dy}}{{dx}}$ using implicit differentiation:

To find $\frac{{dy}}{{dx}}$, we will differentiate both sides of the equation with respect to $x$ while treating $y$ as an implicitly defined function of $x$. Let's differentiate each term on the left-hand side separately:

Differentiating $x^2$ with respect to $x$ gives $\frac{{d}}{{dx}}(x^2) = 2x$.

Differentiating $y^2$ with respect to $x$ gives $\frac{{d}}{{dx}}(y^2) = 2y \cdot \frac{{dy}}{{dx}}$.

Using the product rule, differentiating $xy$ with respect to $x$ gives $\frac{{d}}{{dx}}(xy) = x \cdot \frac{{dy}}{{dx}} + y$ (the derivative of $x$ with respect to $x$ is simply 1).

Bringing it all together, the derivative of the left-hand side is:

Since the derivative of the right-hand side is 0 (since $4$ is a constant), we have:

Now, let's isolate $\frac{{dy}}{{dx}}$ by combining like terms:

Factoring out $\frac{{dy}}{{dx}}$:

Finally, we can solve for $\frac{{dy}}{{dx}}$ by dividing both sides by $(2y + x)$:

2. Determine the slope of the tangent line to the curve at the point $(1, -1)$:

To find the slope of the tangent line at the point $(1, -1)$, we substitute $x = 1$ and $y = -1$ into the expression we obtained for $\frac{{dy}}{{dx}}$ in part 1.

Plugging in these values, we have:

Therefore, the slope of the tangent line at the point $(1, -1)$ is $-\frac{{1}}{{3}}$.

3. Find the equation of the tangent line at that point:

To find the equation of the tangent line at the point $(1, -1)$, we'll use the point-slope form of a line:

where $(x_1, y_1)$ is a point on the line and $m$ is the slope.

Substituting the given point $(1, -1)$ and the slope $-\frac{{1}}{{3}}$, we have:

Simplifying:

Rearranging the equation:

Therefore, the equation of the tangent line to the curve at the point $(1, -1)$ is $y = -\frac{{1}}{{3}}x - \frac{{2}}{{3}}$.