# Use parameterization to deal with curves that have more than one tangent line at a point. Then...

## Question:

Use parameterization to deal with curves that have more than one tangent line at a point. Then you'll use implicit differentiation to relate two derivative functions, and solve for one using given information about the other.

Curve A: {eq}y^2 = x^2(x + 1) {/eq}

Curve B: {eq}y^2(1-\frac{1}{3}y) = x^2 {/eq}

(a) Use implicit differentiation to find all the points in Curve A with a horizontal tangent line.

(c) Try to find {eq}\frac{dy}{dx} {/eq} at the point {eq}(0,0) {/eq} on both graphs. What goes wrong?

## Implicit Differentiation

To find the derivative {eq}\displaystyle \frac{dy}{dx}, {/eq} from the implicit form {eq}\displaystyle f(x,y)=0, {/eq} we will use implicit differentiation,

{eq}\displaystyle \frac{d}{dx}(f(x,y))=0 \iff \frac{df}{dx}+\frac{df}{dy}\cdot \frac{dy}{dx}=0, {/eq}

and solve for {eq}\displaystyle \frac{dy}{dx}. {/eq}

When solving for the slope of a tangent line, we may need to divide by an expression that can be zero for some points on the graph.

Therefore, to find the slope of the tangent lines in those points, we need to use limits.

(a) To find the points on the curve {eq}\displaystyle A: y^2=x^3+x^2 {/eq} with horizontal tangent lines, we will obtain the first derivatives, {eq}\displaystyle y'(x) {/eq} and solve for zero slope: {eq}\displaystyle y'(x)=0. {/eq}

{eq}\displaystyle \begin{align} \frac{d}{dx}\left(y^2\right)&=\frac{d}{dx}\left(x^3+x^2 \right)\\ \implies 2yy'&=3x^2+2x\implies \text{ if } y\neq 0: y'(x)=\frac{3x^2+2x}{2y}\\ &\text{ so, the slope is zero when }3x^2+2x=0\iff x=0 \text{ or } x=-\frac{2}{3}, &\left[ \text{ but }x=0 \text{ corresponds to } y=0\text{ which is excluded from this analysis, when we solved for }y'\right]\\ \implies &\boxed{\text{ the points on curve A with horizontal tangent lines are }\left(-\frac{2}{3}, \frac{2}{3\sqrt{3}}\right), \left(-\frac{2}{3},- \frac{2}{3\sqrt{3}}\right)}, &\left[\text{ by solving for } y\text{ when } x=-\frac{2}{3} \text{ using }y^2=x^3+x^2\right]. \end{align} {/eq}

(b) To find the points on the curve {eq}\displaystyle B: y^2-\frac{y^3}{3} = x^2 {/eq} with horizontal tangent lines, we will obtain the first derivatives, {eq}\displaystyle y'(x) {/eq} and solve for zero slope: {eq}\displaystyle y'(x)=0. {/eq}

{eq}\displaystyle \begin{align} \frac{d}{dx}\left(y^2-\frac{y^3}{3} \right)&=\frac{d}{dx}\left(x^2 \right)\\ \implies (2y-y^2)y'&=2x\implies \text{ if } y\neq 0 \text{ and }y\neq 2: y'(x)=\frac{2x}{2y-y^2}\\ &\text{ so, the slope is zero when } x=0\\ &\text{ the point }x=0 \text{ corresponds to } y=0 \text{ (which is excluded from this analysis, when we solved for }y') \text{ or } y=3, &\left[\text{ by solving for } y\text{ when } x=0 \text{ using }y^2-\frac{y^3}{3} = x^2 \right] \\ \implies &\boxed{\text{ the point on curve B with horizontal tangent line is }\left(0, 3\right)}. \end{align} {/eq}

(c) When finding the slope of the tangent line {eq}\displaystyle \frac{dy}{dx}(0) {/eq} for curve A, {eq}\displaystyle \frac{dy}{dx}(0)=\frac{3x^2+2x}{2y} {/eq} {eq}\displaystyle \boxed{\text{ we obtain the indeterminate case of } \frac{0}{0}} {/eq}

so we need to find the slope using limits.

{eq}\displaystyle \begin{align} \frac{dy}{dx}(0) &=\lim_{x\to 0} \frac{3x^2+2x}{2y}\\ &=\lim_{x\to 0} \frac{3x^2+2x}{\pm 2\sqrt{x^3+x^2}}, &\left[\text{ using } y^2=x^3+x^2 \text{ and solving for }y \text{ in terms of } x\right]\\ &\overset{\frac{0}{0}}{=}\lim_{x\to 0} \frac{x(3x+2)}{\pm 2 |x|\sqrt{x+1}}\\ &= \lim_{x\to 0} \frac{3x+2}{\pm 2 \sqrt{x+1}}\\ &=\pm1. \end{align} {/eq}

Similarly, when finding the slope of the tangent line {eq}\displaystyle \frac{dy}{dx}(0) {/eq} for curve B, {eq}\displaystyle \frac{dy}{dx}(0)=\frac{2x}{2y-y^2} {/eq} {eq}\displaystyle \boxed{\text{ we obtain the indeterminate case of } \frac{0}{0}}. {/eq}