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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the points on the curve \(y = x^3\) at which the slope of the tangent is equal to the \(y\) - coordinate of the point.

$\begin{array}{1 1} (A)\;(0,0)(9,27) \\ (B)\;(0,0)(3,27)\\(C)\;(0,3)(0,27) \\(D)\;(0,-3)(9,27) \end{array} $

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Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Let $P(x_1,y_1)$ be the required point.
The given curve is $y=x^3$----(1)
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}$$=3x^2$
The slope of the tangent at $P(x_1,y_1)$ is
$\large\frac{dy}{dx}_{(x_1,y_1)}$$=3x^2$------(2)
Step 2:
It is given that the slope of the tangent is equal to the $y$ coordinate of the point.
Hence $\large\frac{dy}{dx}$$=y_1$------(3)
Equating equ(2) and equ(3) we get,
$3x_1^2=y_1$-----(4)
Also $(x_1,y_1)$ lies on equ(1)
Hence $y_1=x_1^3$----(5)
Step 3:
Now equating equ(4) and equ(5)
$3x_1^2=x_1^3$
$\Rightarrow 3x_1^2-x_1^3=0$
$x_1^2(3-x_1)=0$
Therefore $x_1=0$ and $x_1=3$
Step 4:
When $x_1=0,y_1=(0)^3=0$
When $x_1=3,y_1=(3)^3=27$
Hence the required points are $(0,0)$ and $(3,27)$
answered Jul 11, 2013 by sreemathi.v
 

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