Question

Determine the tangent to the curve 3y² = x³ at (3,3) and calculate the area of the triangle bounded by the tangent line, the x-axis and the line x=3.

Answer 1

Let us determine the tangent to the curve $3y^2 = x^3$ at $(3,3)$. Since $6yy'=3x^2,$ we conclude that $y'=\frac{x^2}{2y},$ and hence $y'(3)=\frac{3^2}{2\cdot 3}=\frac{3}{2}.$ It follows that the equation of the tangent to the curve is $y-3=\frac{3}{2}(x-3),$ which is equivalent to $y=\frac{3}{2}x-\frac{3}{2}.$

Let us calculate the area of the triangle bounded by the tangent line, the x-axis and the line $x=3$. If $y=0,$ then $\frac{3}{2}x-\frac{3}{2}=0,$ and hence $x=1.$ If $x=3,$ then $y=\frac{3}{2}\cdot 3-\frac{3}{2}=3.$ We conclude that the vertices of a triangle are $A(1,0),\ B(3,0)$ and $C(3,3).$ It follows thta this triangle is right with legs $3-1=2$ and $3$. Therefore, its area is $\frac{1}{2}\cdot2\cdot3=3.$