hpfan101
  • hpfan101
A rocket rises straight up from point X on the ground. A tracking device is on the ground 30 ft from point X. The distance from the tracking device to the rocket is changing at a rate of 100 ft/sec. At what rate is the height of the rocket changing when the rocket if 40 ft high?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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baru
  • baru
|dw:1446351243748:dw|
baru
  • baru
you understand how i got that diagram?
hpfan101
  • hpfan101
Yeah I understand

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baru
  • baru
can you write 'd' in terms of 'h' ? hint: pythogoras theorem
hpfan101
  • hpfan101
\[d^2=h^2+30^2\] \[h=\sqrt{d^2-30^2}\]
baru
  • baru
differentiate both sides with respect to 't'
hpfan101
  • hpfan101
\[\frac{ dh }{ dt }=\frac{ 1 }{ 2 }(d^2-30^2)^{-1/2}\times2d\]
baru
  • baru
you have missed out a \(\frac{dd}{dt}\) term on the RHS
hpfan101
  • hpfan101
Oh okay.
baru
  • baru
its easier to work with\[d^2=h^2+30^2\]
baru
  • baru
can you differentiate that on both sides with respect to 't'?
hpfan101
  • hpfan101
Yeah \[2d \times \frac{ dd }{ dt }=2h \times \frac{ dh }{ dt }+0\]
baru
  • baru
yep you are given that the rocket is 40 ft high, so h=40 can you find what 'd' is when h=40 (Pythagoras again)
hpfan101
  • hpfan101
Yeah d would be 50
baru
  • baru
so now you have h=40 d=50 and dd/dt=100 (given in the question) substitute these values in the equation you just got after differentiating
hpfan101
  • hpfan101
\[2(50)\times100=2(40)\times \frac{ dh }{ dt }\] \[\frac{ dh }{ dt }=125ft/\sec \]
baru
  • baru
yep, that's right :)
hpfan101
  • hpfan101
Cool! Thanks!

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