Calculus problem

[Howitzer]

I have a calculus problem and I am not sure what I am doing wrong.
A plane is flying along 2000m above the ground at constant speed toward someone, and the angle of elevation (A) of the observer to the plane is increasing as it gets closer to them. We are told that when the angle is at 30 degrees the angle increasing at 2 degrees per second.

So the way I though you would say you need to find dX/dt, where X is the horizontal distance of the plane from the person. To do this I would find dX/dA and dA/dt.
X = 2000 / tan A or 2000*cot A
so dX/dA = -2000cosec^2 A or -2000/(sin A)^2
And then I know dA/dT = 2 when the angle is 30, so I substitute 30 into the equation and I get -2000/0.5^2 or -8000
So (dA/dT) * (dX /dA) is -16000m/s
but it should be ~277

 

[ApostateAbe]

What value are you trying to find, what is your initial angle, what is your horizontal distance from the man, what is your initial speed, what is the initial rate of angular increase, and is there anything else relevant you left out?

 

[Howitzer]

We have to find the speed of the plane, which is constant. That is all the information given.

 

[ApostateAbe]

We have to find the speed of the plane, which is constant. That is all the information given.

Is there anything else that is constant? Does the elevation of the plane increase at a constant rate? I would like to know what the function of the rate of angular increase might be. Maybe you should write the problem exactly as it is given.

 

[Howitzer]

"As a plane travelling at a constant speed flies twards an observer its apparent angle of elevation increases. The plane will pass 2000m directly above the observer. When the angle of elevation is 30 degress, the angle of elevation is increasing at 2 degrees per second. Estimate the speed of the plane to the nearest 10km/h"
And there is a visual aid which is just a right angle triangle like
............x
......|===Plane
2km| ... /
......| . /
......|/
.....guy

 

[ApostateAbe]

"As a plane travelling at a constant speed flies twards an observer its apparent angle of elevation increases. The plane will pass 2000m directly above the observer. When the angle of elevation is 30 degress, the angle of elevation is increasing at 2 degrees per second. Estimate the speed of the plane to the nearest 10km/h"
And there is a visual aid which is just a right angle triangle like
............x
......|===Plane
2km| ... /
......| . /
......|/
.....guy

That still is not enough information, I'm afraid, and that may be your final answer in the end. The plane could be going at any speed and all of the given information would hold. Are you sure that no more numerical information can be discerned from that illustration? Like, maybe a horizontal distance where the plane starts out level? I did a Google search, I found a similar problem (not quite the same problem) solved here:

http://www.sosmath.com/CBB/viewtopic.php?t=4981&sid=d35a9f112ca0648e97cb8dbfde8c55d9

and here:

http://www.freemathhelp.com/forum/viewtopic.php?t=37472

 

[ApostateAbe]

Seems like the problem was adapted from that previous more-popular math problem, and the editor did not think it through all of the way. The right triangle would be misleading, because of course the plane would not be traveling in a linear path but it would be following a curved path upward. If the plane is traveling fast, then the curved path at the top is stretched out long horizontally. If the plane is traveling slowly, then the curved path at the top will be short in its horizontal dimension. If we have a fixed horizontal distance or other fixed related value, then there can be only one constant speed, and the problem can be solved. Otherwise, not enough info.

 

[Howitzer]

I think you are misunderstanding the problem. The angle is the angle of elevation from the person to the plane (I wrote it the other way round originally sorry). So the plane is not changing angle and is flying exactly level, and its velocity is constant.
e.g. You can just approximate the speed by substituting 30 degrees and 30.01 or whatever degrees into the equation for X and use the given rate of the angle changing to solve it for the correct speed that it has in the answers.

But it is the last question in the related rates of change section and we are supposed to use chain rules and such to solve it. And I am doing it with all my current differentiation knowledge which makes sense to me until the answer comes out incorrectly.

I checked out the links and the way the solve it is basically the same kind of route. But I am sure my working is correct it just doesn't work.

 

[ApostateAbe]

I think you are misunderstanding the problem. The angle is the angle of elevation from the person to the plane (I wrote it the other way round originally sorry). So the plane is not changing angle and is flying exactly level, and its velocity is constant.
e.g. You can just approximate the speed by substituting 30 degrees and 30.01 or whatever degrees into the equation for X and use the given rate of the angle changing to solve it for the correct speed that it has in the answers.

But it is the last question in the related rates of change section and we are supposed to use chain rules and such to solve it. And I am doing it with all my current differentiation knowledge which makes sense to me until the answer comes out incorrectly.

I checked out the links and the way the solve it is basically the same kind of route. But I am sure my working is correct it just doesn't work.

Yes, I did misunderstand, sorry. I know the problem now, and I am working on it.

 

[ApostateAbe]

I found the answer. I basically gave up and used your math, only I did the math correctly. I will copy and paste what you wrote and place changes in bold.

So the way I though you would say you need to find dX/dt, where X is the horizontal distance of the plane from the person. To do this I would find dX/dA and dA/dt.
X = 2000 / tan A or 2000*cot A
so dX/dA = -2000cosec^2 A or -2000/(sin A)^2
And then I know dA/dT = 2 deg/sec = (pi/90) rad/sec when the angle is 30, so I substitute 30 into the equation and I get -2000/0.5^2 or -8000 m
So (dA/dT) * (dX /dA) is (pi/90) rad/sec * -8000 m= -279.25 m/sec
but it should be ~277 and it is, yay.​

Remember to convert it to km/hr and round. The lesson is to keep writing down your units.

 

[Howitzer]

I didn't know that we had to change things to radians. Thank you. I looked up the reasoning now too and everything fits together a lot better in my head

 

[ApostateAbe]

Yeah, it is a very common mistake to calculate using degrees and getting the units all jacked up. I have wondered why "radians" are basically the same as "unitless," and I figure it is just because it is an expression of a ratio of two distances, like the circumference of a circle over the diameter of a circle.

 

[Oblivious]

2 pi is the circumference of the unit circle of radius 1.

 

[Alice?]

Yeah, it is a very common mistake to calculate using degrees and getting the units all jacked up. I have wondered why "radians" are basically the same as "unitless," and I figure it is just because it is an expression of a ratio of two distances, like the circumference of a circle over the diameter of a circle.

Radians are the measure of an angle of a certain revolution around a circle. Say you take the unit circle, and wish to move counterclockwise around it 90 degrees starting from 0. This angle be the same as pi/2. The conversions between the two systems are very simple, the general formula for converting from degrees to radians is to multiply the degrees by pi/180.

So each 360 degree rotation around the circle is equal to 2pi. It's generally a lot simple to deal in radians when dealing with trigonometric functions rather than having to convert back and forth between the two.

 

[ApostateAbe]

There is a very fundamental formula that goes: s=r*theta, where s is the arc length, r is the radius, and theta is the angle in radians, and I think that is the expression of how radians are defined. theta = s / r = arc length over radius, which means you have a ratio of two distances and a unitless quantity.

 

[Oblivious]

Hard to get more fundamental then 2*pi*r, where r is 1.

 

[Howitzer]

Now I'm all confused trying to reconcile pi with radians with arc lengths with sin with degrees. Wish we had time to deal with these fundemental ideas in math D:

 

[Oblivious]

It's never too late to go back and learn your trigonometry. I find that from time to time I go back just to reaffirm concepts that I might have forgotten or missed the first time.

Geometry is a very important field to engineering and science. It's worth the effort to become proficient in.

 

[Agent Intellect]

Unit circle is a good place to start:

Spoiler

Although if you take a trig class or a precalculus class, they will probably show you graphically why this works (it has to do with the repeating nature of sine and cosine).

I would recommend not taking calculus II until you have trig mastered (especially trig identities).

 

[ProxyAmenRa]

Before you start to solve a question it is always good to draw a diagram.

 

[James Black]

Adding to AI's post about the unit circle: its really easy to remember if you just focus on the first quadrant and learn a system to it instead of just the numbers. After that, you can do some mirroring or a bit more thinking to come up with the rest.

Drawing 1/4th of the circle (the top right portion), you should find it easy to find (0,1) and (1,0)
Make a point to remember that 0 = (sqrt(0)/2) and 1 = (sqrt(4)/2) if you need to, in order to help you with the next part: With the angles 0, 30, 45, 60, and 90, each point is a square root over 2, and the number within the square root goes up by one, from 0 to 4, or down by one from 4 to 0.

Converting from degrees to radians isn't too difficult, as long as you remember 360 degrees (or... a circle) = 2pi radians. Theres a few ways to remember the rest from there.

Half a circle = pi. A fourth a circle (90 degree mark) = pi/2.
pi * (90 degrees / 180 degrees) [ or 2 pi * (90 degrees / 360 degrees) ] = 1/2 pi = pi/2.

 

[number676766]

Only on an INTP forum could you ask for calc help and get it.

 

[ApostateAbe]

Only on an INTP forum could you ask for calc help and get it.

Welcome to the forum.