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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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A kite is moving horizontally at a height of 151.5 meters If the speed of kite is 10m/s,how fast is the string being let out,when the kite is 250m away from the boy who is flying the kite?The height of boy is 1.5m

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  • 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
Given height at which the kite is flying is
$AC=151.5 m$
Height of the boy = $BC= 1.5m$
By applying pythagoras theorem in $\Delta \:ABE$,
$i.e.,\:\:\:h^2+x^2=y^2\: \: \: \: \: \: \: \: \: \: \: (1)$
when $x=BE=250\:m$,
$ \Rightarrow AE=y=291.5m$
Differentiating equation (1) w.r.t. $t$ we get
$ \Rightarrow x\large\frac{dx}{dt}=y \large\frac{dy}{dt}$
( Since height is always a constant )
Step 2
Given $\large\frac{dx}{dt}$$=10m/sec$ and $x=250\:m$
Now substituting the values for $ x, y$ and $ \large\frac{dx}{dt}$ we get
$ 10 \times 250 = 291.5 \times \large\frac{dy}{dt}$
$ \Rightarrow \large\frac{dy}{dt}= \large\frac{10 \times 250}{291.5}$
$ = 8 m/sec$
Hence the rate at which the string is released is $ 8m/ sec$
answered Jul 30, 2013 by thanvigandhi_1
edited Jan 18, 2014 by rvidyagovindarajan_1
incorrect answer
Jitender the correction has been done. Please  go through

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