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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Two men A and B start with velocities v at the same time from the junction of two roads inclined at $45^\circ$ to each other.If they travel by different roads,find the rate at which they are being separated.

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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 $
  • $ \bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
The angle between the two roads is given as
$ \theta = 45^{\circ}$
Hence we can take the ratios as
$h=k$
$x=k\: and \: y=\sqrt2k$
According to pythagoras theorem we get,
$h^2+x^2=y^2$
Differentiating w.r.t $t$ we get
$2h.\large\frac{dh}{dt}+2x\large\frac{dx}{dt}=2y\large\frac{dy}{dt}$
$ \Rightarrow h.\large\frac{dh}{dt}+x.\large\frac{dx}{dt}=y\large\frac{dy}{dt}$
Step 2
Now substituting the values for $ h, x, \: and \: y$ and $ \large\frac{dx}{dt}=\large\frac{dy}{dt}=v$
$ k.\large\frac{dh}{dt}+k.v=\sqrt2k.v$
$ \Rightarrow \large\frac{dh}{dt}=\sqrt 2v-v$
$ = v ( \sqrt2-1)$
Hence the rate at which they seperate is
$v(\sqrt2-1)$
answered Jul 31, 2013 by thanvigandhi_1
 

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