# Ethan's question at Yahoo! Answers regarding an implicit curve, slope and vertical tangents

#### MarkFL

Staff member
Here is the question:

Calculus FRQ help!! Given the curve x^3+y^3=4xy+1?

a. Write the general expression for the slope of the curve.
b. Find the coordinates of the point on the curve where the tangents are vertical.
c. At the point (2,1) find the rate of change in the slope of the curve with respect to x.
I have posted a link there to this thread so the OP can view my work.

#### MarkFL

Staff member
Hello Ethan,

We are given the implicitly defined curve:

$$\displaystyle x^3+y^3=4xy+1$$

a.) Implicitly differentiating with respect to $x$, we obtain:

$$\displaystyle 3x^2+3y^2\frac{dy}{dx}=4\left(x\frac{dy}{dx}+y \right)+0$$

Arranging with all terms having $$\displaystyle \frac{dy}{dx}$$ as a factor on the left, and everything else on the right, we have:

$$\displaystyle 3y^2\frac{dy}{dx}-4x\frac{dy}{dx}=4y-3x^2$$

Factor the left side:

$$\displaystyle \left(3y^2-4x \right)\frac{dy}{dx}=4y-3x^2$$

Dividing through by $$\displaystyle 3y^2-4x$$ we find:

$$\displaystyle \frac{dy}{dx}=\frac{4y-3x^2}{3y^2-4x}$$

b.) The tangents are vertical where:

$$\displaystyle 3y^2-4x=0\implies x=\frac{3}{4}y^2$$

Substituting for $x$ into the original curve, there results:

$$\displaystyle \left(\frac{3}{4}y^2 \right)^3+y^3=4\left(\frac{3}{4}y^2 \right)y+1$$

$$\displaystyle \frac{27}{64}y^6+y^3=3y^3+1$$

Arrange in standard polynomial form:

$$\displaystyle 27y^6-128y^3-64=0$$

Applying the quadratic formula, we find:

$$\displaystyle y^3=\frac{8}{27}\left(8\pm\sqrt{91} \right)$$

Hence:

$$\displaystyle y=\frac{2}{3}\sqrt{8\pm\sqrt{91}}$$

$$\displaystyle x=\frac{1}{2}\sqrt{155\pm16\sqrt{91}}$$

Thus, the points having vertical tangents are:

$$\displaystyle (x,y)=\left(\frac{1}{2}\sqrt{155\pm16\sqrt{91}},\frac{2}{3}\sqrt{8\pm\sqrt{91}} \right)$$

c.) $$\displaystyle \left.\frac{dy}{dx} \right|_{(x,y)=(2,1}=\frac{4(1)-3(2)^2}{3(1)^2-4(2)}=\frac{8}{5}$$