Calculating the Speed of Airplane Above an Observer

[jmed]

Homework Statement

An airplane flying horizontally at an altitude of 3 miles and at a speed of 480 miles per hour passes directly above an observer on the ground. How fast is the distance from the observer to the airplane increasing 30 seconds later?

Homework Equations

The Attempt at a Solution

When I attempted to draw the picture for this I got a 3, 4, 5 right triangle because the altitude is 3, and if you divide 480 by 60 you get 8 and half of that is 4. So this would create a 3, 4, 5 right triangle? Now would I use the distance formula with respect to distance?

 

[Mark44]

Homework Statement

An airplane flying horizontally at an altitude of 3 miles and at a speed of 480 miles per hour passes directly above an observer on the ground. How fast is the distance from the observer to the airplane increasing 30 seconds later?

Homework Equations

The Attempt at a Solution

When I attempted to draw the picture for this I got a 3, 4, 5 right triangle because the altitude is 3, and if you divide 480 by 60 you get 8 and half of that is 4. So this would create a 3, 4, 5 right triangle? Now would I use the distance formula with respect to distance?

At the time in question (1/2 minute after the plane was directly overhead), the plane is 5 miles from the observer, but that's not what the problem is after. It wants the rate of change of distance, with respect to time.

This is a related rates problem. In your right triangle, the only thing that isn't changing is the altitude (plane or triangle). You should label the other side of your triangle as x and the hypotenuse as D. Now can you think of a relationship between x and D? Can you use that relationship to get a relationship between the two rates of interest here?

 

[jmed]

Would law of cosine come into play? d^2 = 1^2 + x^2 - 2*1*x cos90. Then from here differentiate?

 

[Mark44]

It's a right triangle, right? You don't need the Law of Cosines for a right triangle. You can use it, but it's overkill. Also, what's the plane's altitude?

 

[jmed]

Ok so it is a 45-45-90? Then I can use law of sine? Also, the altitude is 3km. So if I use law of sine, I get d = 4.2426...i'm not sure how I can differentiate from here tho...

 

[Mark44]

Ok so it is a 45-45-90? Then I can use law of sine? Also, the altitude is 3km. So if I use law of sine, I get d = 4.2426...i'm not sure how I can differentiate from here tho...

How do you get it being a 45-45-90 right triangle? There is something much simpler than the Law of Cosines or the Law of Sines.

The altitude is 3 miles.

Take a look again at what I said in post #2.

 

[jmed]

It is 3 miles. I am confused on what to use? I thought 45-45-90 because I calculated x to be 4 which would make the hypotnuse 5...?

 

[jmed]

a^2 + b^2 = c^2?? but then how do I differentiate?

 

[Char. Limit]

It's not a 45-45-90.

Start with this:

x^2+y^2=r^2

Where x is the horizontal distance, y is the altitude, and r is the total distance.

Then differentiate with respect to time. d(x^2)/dt + d(y^2)/dt = d(r^2)/dt

 

[Mark44]

a^2 + b^2 = c^2?? but then how do I differentiate?

You know the altitude, so your formula should involve only two variables. And I would suggest using better letters than a, b, and c - letters that make it easier for you to understand what they represent.

 

[jmed]

I still need a rate tho. I have x*dx/dt= r*dr/dt. I am not sure how to use the 480mph and the 30sec to find a rate. Is dx/dt = 30sec?

 

[Mark44]

OK, that's the equation that shows how the rates are related. At any time t, x* dx/dt = r * dr/dt. Before getting to the answer, what does each of these symbols represent?

x is what?
dx/dt is what? (it's not 30 sec.)
r is what?
dr/dt is what?

 

[jmed]

x = distance plane has traveled...dx/dt is equal to rate at which plane is traveling...r is distance from observer to plane 30 sec. later. dr/dt is equal to rate of distance from observer to airplane 30 sec later. dx/dt would equal 480mph. dr/dt is what I am trying to find.

 

[Mark44]

x = distance plane has traveled...dx/dt is equal to rate at which plane is traveling

In other words, the plane's speed, which is known.

...r is distance from observer to plane 30 sec. later.

No, r is the distance from observer to plane. Period.

dr/dt is equal to rate of distance from observer to airplane 30 sec later.

No dr/dt is just the rate at which the distance from observer to plane is changing. Period. r and dr/dt don't have anything to do with the 30 seconds.

dx/dt would equal 480mph.

Yes.

dr/dt is what I am trying to find.

Yes.

OK, at any time t, you have 480*x = r*dr/dt (replacing dx/dt with 480 mi/hr).
Can you solve for dr/dt in terms of the other quantities? You won't get a number just yet.

At 30 seconds after the plane was directly overhead, what is x? r? Now you should be able to get dr/dt.

Notice that this value of dr/dt is only at the time in question (30 seconds after plane was overhead). At later times, dr/dt will have different values, even though dx/dt remains constant.

 

[tt2348]

If you have any knowledge of vectors and velocities, try considering the angle formed by the plane at the give time by its height, and the distance it traveled from the point it was over head of the observer , to the present location (d=480t). 480 mph would be your V(horizontal), so if you can think of a relation using the angle, to describe ratio of the V(horizontal) to V(away), using the triangle, you can do it without calculus.

 

[jmed]

so then x = 4 after 30 seconds and r is then = to 5. This equates to (480*4)/5 = dr/dt. Which equals 384 mph?

 

[Mark44]

Right.

 

[jmed]

awesome! thanks again!