Question

Find the points on the curve y = x³ where the slope of the tangent is equal to the x-coordinate of the point.

Answer 1

Let (x₁, y₁) be the required point.
x coordinate of the point is x₁.

$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},y_1={x_1}^3...\left(1\right) \\ \mathrm{Now},y=x^3 \\ \Rightarrow \frac{dy}{dx}=3x^2 \\ \text{Slope of tangent at}\left(x,y\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}3{x_1}^2 \\ \text{Given that} \\ \text{Slope of tangent at}\left(x_1,y_1\right)\text{=}x\mathit{}\text{co-ordinate of the point} \\ \Rightarrow 3{x_1}^2=x_1 \\ \Rightarrow x_1\left(3x_1-1\right)=0 \\ \Rightarrow x_1=0\text{or}{x}_1=\frac{1}{3} \\ \Rightarrow y_1=0^3\text{or}{y}_1={\left(\frac{1}{3}\right)}^3\text{(From (1))} \\ \Rightarrow y_1=0\text{or}{y}_1=\frac{1}{27} \\ \text{So, the points are}\left(x_1,y_1\right)\text{=}\left(0,0\right)\text{,}\left(\frac{1}{3},\frac{1}{27}\right) \end{gathered}$$