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by **jaybird0827** on Wed May 27, 2009 12:01 am

This airplane is flying at 500 km/h in a straight direction at a constant 5000 m. $\theta$ is the angle of elevation from a fixed point of observation. What is the rate of change of $\theta$ when it is 30°? Required to answer in DMS/s and round to one decimal place.

Solution:

Drew the right triangle as follows: Vertical segment labeled 5000 m = 5 km, assigned variable hj. At the top, a stick figure of a jet plane. Horizontal segment labeled 5 $\sqrt3$ km, assigned variable b. Hypotenuse labeled 10 km, assigned variable c. The 5 $sqrt3$ and the 10 we get from the 30-60-90 relationship. Angle of elevation is angle $\theta$ , labeled $\theta$ = 30 deg (using degree symbol).

We also know that $\frac{db}{dt}$ = 500 km/h and $\frac{dh}{dt}$ = 0. So,

tan $\theta$ = h/b and

sec² d $\theta$ /dt = (b dh/dt - h db/dt)/h²

Plugging in,
sec² $\theta$ d $\theta$ /dt = (0 - 5*500)/25 = -100. But sec $\theta$ = 10/5sqrt3, then sec² $\theta$ = 4/3 and
4/3 d $\theta$ /dt = -100. Multiplying both sides by 3/4 gives d $\theta$ /dt = -75; i.e. -75 radians/h.

Using a calculator, I get -75 /h * 1 h / 60 min * 1 min / 60 s = approx -0.0208333333 / s which converts to -1.19+°, or (-1 deg 11 min 37.2 sec) /s , answer

**Sponsor**

by **stapel_eliz** on Wed May 27, 2009 1:13 am

I think you swapped "h" and "b" in the denominator when you took the derivative.

$\tan(\theta)\, =\, \frac{h}{b}$

$\sec^2(\theta)\,\frac{d\theta}{dt}\, =\, \frac{\frac{dh}{dt}\,b\, -\, h\,\frac{db}{dt}}{b^2}$

:wink:

by **jaybird0827** on Fri May 29, 2009 7:35 pm

Eliz,

Appreciate your attention to details. Hmm, let's see, "Low d high minus high d low, over the square of what's below") -

$tan(\theta)\, =\, \frac{h}{b}$ ,

And here's where I made my error - I swapped the "square of "what's below" with the square of what was above ... :oops:

Correction, should be

$\color{red}b^2$ , not $b$ , in the denominator,

$sec^2(\theta)\,\frac{d\theta}{dt}\, =\, \frac{\frac{dh}{dt}\,b\, -\, h\,\frac{db}{dt}}{b^2}\$

The rest of the fix should be easy.

$sec(\theta)\, =\, \frac{2}{\sqrt3}$

$\frac{4}{3}\,\frac{d\theta}{dt}\, =\, \frac{-5\,*\,500}{75} =\, \frac{-100}{3}$ , and

$\frac{4}{3}\,\frac{d\theta}{dt}\, =\,\frac{-100}{3}$

$\frac{d\theta}{dt}\, =\,\frac{-100}{3}\,*\,\frac{3}{4}$

$\frac{d\theta}{dt}\, =\,50\,\frac{radians}{hr}$

Using a calculator, I get

$\frac{-50}{hr}\,*\,\frac{180}{\pi}\,*\frac{hr}{60\,min}\,*\,\frac{min}{60\,s}$

$=\,approx\,-0.795774715\,\frac{\,deg}{s}\$ , or

^Answer: $\,approx\,(-0\,deg\,\,47\,min\,\,44.8\,sec)\,/s$

Does this make sense?

by **stapel_eliz** on Sat May 30, 2009 1:39 pm

jaybird0827 wrote: $\frac{d\theta}{dt}\, =\,\frac{-100}{3}\,*\,\frac{3}{4}$

$\frac{d\theta}{dt}\, =\,50\,\frac{radians}{hr}$

Why did the sign change? (The angle should be getting smaller, shouldn't it?)

Check your division: 100/4 = 25, not 50. :oops:

by **jaybird0827** on Mon Jun 01, 2009 2:44 pm

Wow, I must have been in "la la land". So, further corrections

$\frac{d\theta}{dt}\, =\,\frac{-100}{3}\,*\,\frac{3}{4}$

$\frac{d\theta}{dt}\, =\,-25\,\frac{radians}{hr}$

$\frac{-25}{hr}\,*\,\frac{180}{\pi}\,*\frac{hr}{60\,min}\,*\,\frac{min}{60\,s}$

$=\,approx\,-0.3978873577\,\frac{\,deg}{s}\$ , or

^Answer: $\,approx\,(-0\,deg\,\,23\,min\,\,52.4\,sec)\,/s$

I definitely agree it would have to be negative, and at that, a very small change per second.

Thanks for your help and your patience!

by **stapel_eliz** on Mon Jun 01, 2009 2:48 pm

Looks good to me! :thumb:

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