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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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A particle moves along the curve \( 6y = x^3 +2\). Find the points on the curve at which the \(y\)-coordinate is changing $8$ times as fast as the \(x\)-coordinate.

$\begin{array}{1 1} \big(-4,\frac{-31}{3}\big)\; and (4,11) \\ \big(-4,\frac{31}{3}\big)\; and\;(4,11) \\\big(-4,\frac{-31}{3}\big)\; and \;(-4,11) \\ \big(-4,\frac{-31}{3}\big)\; and \;(4,-11) \end{array} $

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Toolbox:
  • If $y=f(x)$,then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
  • $\big(\large\frac{dy}{dx}\big)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1:
Let the required point be $P(x,y)$.It is given that rate of change of $y$-coordinate =8(Rate of change of $x$-coordinate)
(i.e)$\large\frac{dy}{dt}$$=8.\large\frac{dx}{dt}$
It is given $6y=x^3+2$
On differentiating w.r.t $t$ on both sides,
$6\large\frac{dy}{dt}$$=3x^2\large\frac{dx}{dt}$$+2$
$\Rightarrow 6.\big(8\large\frac{dx}{dt}\big)=$$3x^2\large\frac{dx}{dt}$
$\Rightarrow 3x^2=48$
$\Rightarrow x^2=16$
$\Rightarrow x=\pm 4$
Step 2:
When $x=4$,then $6y=4^3+2$
$\Rightarrow 66$
Therefore $y=11$
When $x=-4$,then $6y=(-4)^3+2$
$\Rightarrow -62$
$y=\large\frac{-31}{3}$
Hence the required points are $\big(-4,\large\frac{-31}{3}\big)$ and $(4,11)$.
answered Jul 8, 2013 by sreemathi.v
 

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