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?

Question

baru · Answer

|dw:1446351243748:dw|

baru · Answer

you understand how i got that diagram?

hpfan101 · Answer

Yeah I understand

baru · Answer

can you write 'd' in terms of 'h' ? 
hint: pythogoras theorem

hpfan101 · Answer

$$d^2=h^2+30^2$$
$$h=\sqrt{d^2-30^2}$$

baru · Answer

differentiate both sides with respect to 't'

hpfan101 · Answer

$$\frac{ dh }{ dt }=\frac{ 1 }{ 2 }(d^2-30^2)^{-1/2}\times2d$$

baru · Answer

you have missed out a $\frac{dd}{dt}$ term on the RHS

hpfan101 · Answer

Oh okay.

baru · Answer

its easier to work with$$d^2=h^2+30^2$$

baru · Answer

can you differentiate that on both sides with respect to 't'?

hpfan101 · Answer

Yeah $$2d \times \frac{ dd }{ dt }=2h \times \frac{ dh }{ dt }+0$$

baru · Answer

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 · Answer

Yeah d would be 50

baru · Answer

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 · Answer

$$2(50)\times100=2(40)\times \frac{ dh }{ dt }$$
$$\frac{ dh }{ dt }=125ft/\sec $$

baru · Answer

yep, that's right :)

hpfan101 · Answer

Cool! Thanks!