Course:MATH102/Question Challenge/2002 December Q6

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Question

Consider the curve

$\displaystyle x^{3}+y^{3}-3xy=0$

This curve intersects the line $y=x$ at a point $P$ in the first quadrant $(x>0,y>0)$. Find the slope of the tangent line to the curve at $P$. Show that this tangent line is perpendicular to the line $y=x$.

Hints

[show]Hint
Consider implicit differentiation, and don't forget to use the product rule on the term $xy$.

Solution

[show]Solution

This is a problem in which implicit differentiation is to be used. The equation is $x^{3}+y^{3}-3xy=0$ To solve for $dy/dx$ we derive each term with respect to $x$ and solve isolate for $dy/dx$, remembering to use the chair rule and product rules where needed. $3x^{2}+3y^{2}\cdot {\frac {dy}{dx}}-3x\cdot {\frac {dy}{dx}}-3y=0$ We know that the curve intersects the line $y=x$, so we can substitute $y$ for $x$: $3x^{2}+3x^{2}\cdot {\frac {dy}{dx}}-3x\cdot {\frac {dy}{dx}}-3y=0$ $x+x\cdot {\frac {dy}{dx}}-{\frac {dy}{dx}}-1=0$ $x\cdot {\frac {dy}{dx}}-{\frac {dy}{dx}}=1-x$ ${\frac {dy}{dx}}(x-1)=1-x$ ${\frac {dy}{dx}}={\frac {1-x}{x-1}}$ ${\frac {dy}{dx}}=-1$ The slope of the line $y=x$ is equal to $1$. By proving the curve has a slope of $-1$ at the intersection point, we have shown that the tangent line of the curve at the intersection point is perpendicular to the tangent of the line $y=x$.