Find the Horizontal Tangent Line x^3+y^3=6xy

Problem

x3+y3=6*x*y

Solution

1. Differentiate implicitly with respect to x to find the slope of the tangent line.

d(x3)/d(x)+d(y3)/d(x)=(d(6)*x*y)/d(x)

3*x2+3*y2d(y)/d(x)=6*y+6*xd(y)/d(x)

2. Solve for the derivative d(y)/d(x) by grouping terms containing the derivative on one side.

3*y2d(y)/d(x)−6*xd(y)/d(x)=6*y−3*x2

d(y)/d(x)*(3*y2−6*x)=6*y−3*x2

d(y)/d(x)=(6*y−3*x2)/(3*y2−6*x)

d(y)/d(x)=(2*y−x2)/(y2−2*x)

3. Set the derivative to zero to find the condition for a horizontal tangent line.

(2*y−x2)/(y2−2*x)=0

2*y−x2=0

y=(x2)/2

4. Substitute the condition back into the original equation to find the specific x values.

x3+((x2)/2)3=6*x((x2)/2)

x3+(x6)/8=3*x3

(x6)/8−2*x3=0

x3*((x3)/8−2)=0

5. Solve for x by setting each factor to zero.

x3=0⇒x=0

(x3)/8−2=0⇒x3=16⇒x=√(3,16)=2√(3,2)

6. Find the y-coordinates for each x value using y=(x2)/2

For *x=0,y=0/2=0

For *x=2√(3,2),y=((2√(3,2))2)/2=(4√(3,4))/2=2√(3,4)

7. Verify the points in the original derivative's denominator. At (0,0) the denominator y2−2*x is 0 making the derivative undefined (a cusp or node). At (2√(3,2),2√(3,4)) the denominator is (2√(3,4))2−2*(2√(3,2))=4√(3,16)−4√(3,2)=8√(3,2)−4√(3,2)≠0

Final Answer

y=2√(3,4)

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