Course:MATH102/Question Challenge/2002 December Q6

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Question

Consider the curve

$x^3+y^3-3xy=0$

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

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

Solution

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 $y=x$.