Find the points on the curve
Question:

Find the points on the curve $y=x^{3}$ where the slope of the tangent is equal to the $x$-coordinate of the point.

Solution:

Let $\left(x_{1}, y_{1}\right)$ be the required point.

$x$ coordinate of the point is $x_{1}$.

Since, the point lies on the curve.

Hence, $y_{1}=x_{1}{ }^{3}$       ….(1)

Now, $y=x^{3}$

$\Rightarrow \frac{d y}{d x}=3 x^{2}$

Slope of tangent at $(x, y)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=3 x_{1}^{2}$

Given that

Slope of tangent at $\left(x_{1}, y_{1}\right)=x$ co-ordinate of the point

$\Rightarrow 3 x_{1}^{2}=x_{1}$

$\Rightarrow x_{1}\left(3 x_{1}-1\right)=0$

$\Rightarrow x_{1}=0$ or $x_{1}=\frac{1}{3}$

$\Rightarrow y_{1}=0^{3}$ or $y_{1}=\left(\frac{1}{3}\right)^{3}($ From $(1))$

$\Rightarrow y_{1}=0$ or $y_{1}=\frac{1}{27}$

So, the points are $\left(x_{1}, y_{1}\right)=(0,0),\left(\frac{1}{3}, \frac{1}{27}\right)$

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