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

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The discussion focuses on solving a calculus problem involving the implicitly defined curve x^3 + y^3 = 4xy + 1. The general expression for the slope of the curve is derived as dy/dx = (4y - 3x^2) / (3y^2 - 4x). Vertical tangents occur where 3y^2 - 4x = 0, leading to the coordinates of points with vertical tangents being expressed in terms of cubic roots. At the specific point (2,1), the rate of change in the slope of the curve is calculated to be 8/5. The solution provides a comprehensive approach to implicit differentiation and the analysis of tangent behavior on the curve.
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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.
 
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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.) $$\left.\frac{dy}{dx} \right|_{(x,y)=(2,1}=\frac{4(1)-3(2)^2}{3(1)^2-4(2)}=\frac{8}{5}$$
 
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