MHB Calculus FRQ: Implicit Curve, Slope & Vertical Tangents - Ethan

[MarkFL]

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]

Hello Ethan,

We are given the implicitly defined curve:

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

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

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

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

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

Factor the left side:

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

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

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

b.) The tangents are vertical where:

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

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

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

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

Arrange in standard polynomial form:

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

Applying the quadratic formula, we find:

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

Hence:

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

$$x=\frac{1}{2}\sqrt[3]{155\pm16\sqrt{91}}$$

Thus, the points having vertical tangents are:

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

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