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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: /0 on June 19, 2008, 06:37:02 pm

Title: Related Rates, looking at an aeroplane
Post by: /0 on June 19, 2008, 06:37:02 pm: An aeroplane is flying horizontally at a constant height of 1000m. At a certain instant the angle of elevation is $30^{\circ}$ and decreasing and the speed of the aeroplane is 480 km/h.

a) How fast is $\theta$ decreasing at this instant? (Answer in degrees/s.)

b) How fast is the distance between the aeroplane and the observation point changing at this instant?

I can't get A) :( I keep getting the angle in terms of a trig expression, not an inverse trig expression or etc. Thanks
Title: Re: Related Rates, looking at an aeroplane
Post by: dcc on June 19, 2008, 06:47:16 pm: Let h = horizontal displacement, so $\dfrac{dh}{dt} = \dfrac{400}{3} \; ms^{-1}$ .

We are looking for $\dfrac{d \theta}{dt} = \dfrac{dh}{dt} \cdot \dfrac{d \theta}{dh}$

Now lets connect h and theta:

$\tan \theta = \dfrac{1000}{h} \implies h = \dfrac{1000}{\tan \theta}$

$\implies \dfrac{dh}{d \theta} = \dfrac{-1000 \sec^2 \theta}{\tan^2 \theta} = -1000 \csc^2 \theta$

When theta is 30 degrees:

$\dfrac{d \theta}{dt} = \dfrac{400}{3} \cdot - \dfrac{\sin^2 \; 30^{\circ}}{1000} = -\dfrac{1}{30} \mbox{ radians / second}$
Title: Re: Related Rates, looking at an aeroplane
Post by: /0 on June 19, 2008, 06:52:29 pm: Thanks dcc! I looks like I forgot to convert to m/s :S

Still though, if you put in $\frac{\pi}{6}$ radians, wouldn't the result still give the answer in degrees/second?
Title: Re: Related Rates, looking at an aeroplane
Post by: dcc on June 19, 2008, 06:53:51 pm: If you are interested, there was a question almost exactly like this last years Specialist Maths exam two (except I think the height was changing, rather than the horizontal displacement).
Title: Re: Related Rates, looking at an aeroplane
Post by: dcc on June 19, 2008, 07:01:33 pm: Quote from: DivideBy0 on June 19, 2008, 06:52:29 pm
Thanks dcc! I looks like I forgot to convert to m/s :S

Still though, if you put in $\frac{\pi}{6}$ radians, wouldn't the result still give the answer in degrees/second?

Indeed, this complicates the matter slightly, so the answer is gave is incorrect, as you will have to convert from -1/30 radians/second to degrees/second (as the derivative of trig functions for an angle in degrees is different to that of an angle in radians).

$- \dfrac{1}{30} \mbox{ radians / second} = -\dfrac{1}{30} \cdot \dfrac{180}{\pi} = -\dfrac{6}{\pi} \mbox{ degrees / second}$

I think this should be the answer, though I am not 100% sure.
Title: Re: Related Rates, looking at an aeroplane
Post by: /0 on June 19, 2008, 07:04:02 pm: thanks, I just confirmed with the book that $-\frac{6}{\pi} degrees/second$ is correct