Help With High School Maths Problem
 

I am a senior maths tutor, however I am stumped on one question. I know that I must be doing something small wrong because I can do all the other questions in the exercise.

The question is on related rates of change

ie dy/dx = dy/du * du/dx .... sort of like the chain rule

The question is...

An aircraft at an altitude of 10000m is flying at a constant speed on a line that will take it directly over an observer on the ground. If the observer notes that the angle of elevation of the aircraft is 60 degrees and is increasing at a rate of 1 degree/second, find the speed on the plane in m/s.

Ok. This is how far that I have got.

ds/dt = dx/dt * ds/dx

s= distance
t= time... therefore ds/dt will give speed
x= angle of elevation

and we know that from the question that dx/dt = 1 when x=60

So now all that I need is an equation for distance travelled in terms of angle of elevation. Then, find the derivative to get ds/dx.

This is where I am stuck. I can develop equations for this. I have left them out as theygive quite embarassinging wrong answers. I can get the right answer (233m/s) by using approximations, but I can't get the answer properly by using the formula above (which my students must do to get full marks on an exam).

If you want to know what I am by approximations, I find out how far the plane has moved horisontally from angle of elevation of 60 degrees to 61 degrees. I know that it takes one second to move one degree (at 60 degrees) therefore this must be the distance moved in one second, hense it is the speed in m/s. This gives 230m/s, which is fairly close, but I know that my students would only recieve about half marks by working it out this way.

Can someone please give me some assistance??

Draw a diagram. Fortunately, we can keep this in two dimensions since the path of the plane will pass directly over the observer.

From the observer, extend a vertical line until it intersects the line representing the path of the plane. Since the horizontal distance of the plane from the observer and the elevation angle will change in time, give them algebraic values (eg. X and THETA). Label any other known values.

Use trig to express the relationship between X and THETA. Take the derivative wrt TIME. Plug in the known numbers. (Be sure that your units are correct) I get 232.71057 m/sec.

I suspect that this is the desired answer. However, the real answer is slightly different because the path of the plane is not a straight line, but rather a circle of constant elevation. I leave inclusion of the necessary adjustments as an exercise for the reader.

I under stand what you have written, because I have done hundreds of these questions. However, I was hoping for a litte more numbers because I am trying to work out where I am going wrong. I am sure that it is just something small that I am doing.

I will now use THETA for angle of elevation instead of my "x".
I will also use x for horisontal distance from plane to observer instead of my "s".

Therefore, I get horisontal distance from plane to observer as

x= 10000/tan(theta)

therefore dx/d(theta) = -10000/(sin(theta))^2

Sub in my values...

dx/dt = dx/d(theta) * d(theta)/dt

= -10000/(sin(60))^2 * 1
= -13333m/sec

Which I know is wrong...

I know that you said to take the derivate wrt time, however, I don't know how to do this seeing as though the function is only in x and theta, unless you mean what I have done by using related rates of change.

A little more help please?

[QUOTE=Wacky] (Be sure that your units are correct) [/QUOTE]

You are almost there. What are the units for theta?

[QUOTE]
dx/d(theta) = -10000/(sin(theta))^2

dx/dt = dx/d(theta) * d(theta)/dt

= -10000/(sin(60))^2 * 1
[/QUOTE]

Ah... So I have to convert from degrees to radians

(I have rounded numbers to try to keep fractions simple)

dx/d(theta) = -10000/(sin(theta))^2

dx/dt = dx/d(theta) * d(theta)/dt

= -10000/(sin(1.05))^2 * 0.017

= 232m/s

Thanks for your help. It is very much appreciated.

Could you just answer one last questions please??

I thought that it didn't matter whether you had your formula in degrees or radians, just as long as you kept it the same throughout the entire questions. Could you give me a simple explanation as to why it is important to use radians in this question.

I assume that it is because of the "sin". If the angle has either sin, cos or tan working on it, it doesn't matter, however seeing as d(theta)/dt didn't have any trig function working on it, it was neccassary to change to radians.

Once again, thanks again for you help.

You have to look at the definition of sin(angle).
Note that sin(1 degree) is very close to pi/180, not 1.
Only when the angle is expressed in radians will its derivative be cos(angle).

Thank you Wacky for all of your help. I really have appreciated it.

The role of the [i]radian[/i] in trig somewhat resembles the role of [i]e[/i], the natural base of logarithms, in that they both represent units that allow equations to be written without base-conversion constants (or, rather, with those conversion constants = 1).