logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

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

Can you answer this question?
 
 

1 Answer

0 votes
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
Given height at which the kite is flying is
$AC=151.5 m$
Height of the boy = $BC= 1.5m$
$\Rightarrow\:AB=h=151.5-1.5=150$
By applying pythagoras theorem in $\Delta \:ABE$,
$AB^2+BE^2=AE^2$
$i.e.,\:\:\:h^2+x^2=y^2\: \: \: \: \: \: \: \: \: \: \: (1)$
when $x=BE=250\:m$,
$150^2+250^2=AE^2$
$ \Rightarrow AE=y=291.5m$
Differentiating equation (1) w.r.t. $t$ we get
$2x.\large\frac{dx}{dt}$$=2y.\large\frac{dy}{dt}$
$ \Rightarrow x\large\frac{dx}{dt}=y \large\frac{dy}{dt}$
( Since height is always a constant )
Step 2
Given $\large\frac{dx}{dt}$$=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
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...