Rate of change word problem

[flakine]

A rocket, rising vertically, is tracked by a radar station that is on the ground 5 miles from the launch pad. At what rate is the elevation angle changing when the rocket is 3000 feet up and rising vertically at 500 ft/sec?

First: C^2=B^2+A^2, A=the path of the rocket=3000'. B=the distance from the lanchpad of the radar station=5 miles or 26400' so, C=26569.9'

Now, I take the derivaitive of both sides:

2C dC/dt=2B dB/dt + 2A dA/dt
2(26569.9) dC/dt=2(26400) dB/dt + 2(3000) dA/dt
53139.8 dC/dt=52800 dB/dt + 6000 dA/dt

53139.8 dC/dt= 52800(0)(no rate change for B) + 6000(500)
dC/dt= 3000000/53139.8
dC/dt=56.46

sin(dA/dt)/(dC/dt)=sin(500/56.46) =.537 Rad

What am I doing wrong?????


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[Daniel_Feldman]

Try this:

tan(theta)=y/x

Differentiate both sides with respect to time. Remember that x is constant, so dx/dt is 0.

 

[galactus]

Yes indeed, \(\displaystyle tan({\theta})\) is a good place to start:

You need \(\displaystyle \frac{d{\theta}}{dt}\) given that \(\displaystyle \frac{dy}{dt}=500\ ft/sec\)

\(\displaystyle tan{\theta}=\frac{y}{26400}\)

Differentiate both sides because they're both functions of t:

\(\displaystyle sec^{2}{\theta}\frac{d{\theta}}{dt}=\frac{1}{26400}\frac{dy}{dt}\)

Solve for \(\displaystyle \frac{d{\theta}}{dt}\).

You know your hypoteneuse length, use it to find sec.

Then plug them all in and you're done.