Related Rates: Solving Radar Station Angle Change Problem

[asz304]

Homework Statement

At a certain moment, a plane passes directly above a radar station at an altitude of 3 miles at a speed of 300 mph. Suppose that the line through the radar station and the plane makes an angle θ with the horizontal. How fast is θ changing 50 min after the plane passes over the radar station?

Homework Equations

x^2 + y^2 = c^2?


How do I solve the problem without knowing the x-component of the triangle? I tried using trigonometry but it didn't work out well, because I'm only given with only y-component with the speeds which can't fit in the triangle..not sure...

 

[Mark44]

Homework Statement

At a certain moment, a plane passes directly above a radar station at an altitude of 3 miles at a speed of 300 mph. Suppose that the line through the radar station and the plane makes an angle θ with the horizontal. How fast is θ changing 50 min after the plane passes over the radar station?

Homework Equations

x^2 + y^2 = c^2?


How do I solve the problem without knowing the x-component of the triangle? I tried using trigonometry but it didn't work out well, because I'm only given with only y-component with the speeds which can't fit in the triangle..not sure...

The vertical side of the triangle is 3 (mi). Let's call the horizontal side x, which represents the horizontal distance the plane has traveled in t hours. Use one of the trig functions to get a relationship between the angle θ and the two sides of the right triangle, then differentiate to get a relationship betwee dθ/dt and dx/dt.

 

[asz304]

So tan(theta) = y/x, then the derivative is sec^2(theta)=-(x^-2)y + x^-1. I don't know y. Do I need the use 300mph and 50 min(0.83333h)? If I'll use them, where should I plug the numbers?

Just to make things clearer:

Is dy/dt=3 miles?

Thanks

 

[acddklr06]

First, you need to write the given down so you know what you have.

y is 3 mi. dy/dt = 0. The reason dy/dt = 0 is that the altitude does not change in this problem.
We are, also, given that dx/dt = 300 mph. You can find x at 50 mins by converting it to hours and then multiplying by 300.

You will, also, need to draw a diagram. Hint: The top of the triangle should be where the radar station is.

Hint 2: You will need to use an inverse trig function to solve this problem.

 

[Mark44]

First, you need to write the given down so you know what you have.

y is 3 mi. dy/dt = 0. The reason dy/dt = 0 is that the altitude does not change in this problem.
We are, also, given that dx/dt = 300 mph. You can find x at 50 mins by converting it to hours and then multiplying by 300.

You will, also, need to draw a diagram. Hint: The top of the triangle should be where the radar station is.

IMO this is not a good hint. The radar station is on the ground and the airplane is above it, in the air. In my drawing, the right angle is at the top left, the vertical side is 3 (no need to give it a name, since it doesn't change), and the horizontal side of the triangle, x, is along the top edge.

Hint 2: You will need to use an inverse trig function to solve this problem.

This is a misleading hint. Inverse trig functions are not needed. You don't need to solve for theta, but only need to find d(theta)/dt, which you can do by implicit differentiation.

 

[acddklr06]

IMO this is not a good hint. The radar station is on the ground and the airplane is above it, in the air. In my drawing, the right angle is at the top left, the vertical side is 3 (no need to give it a name, since it doesn't change), and the horizontal side of the triangle, x, is along the top edge.

This is what I meant for my hint.

This is a misleading hint. Inverse trig functions are not needed. You don't need to solve for theta, but only need to find d(theta)/dt, which you can do by implicit differentiation.

I know three different ways to solve this problem. Of the three, they all required the inverse trig function to find theta at that instant. However, only one of the ways require using implicit differentiation with a inverse trig function. If he knows the derivative of the inverse trig function, it will be easier than using implicit differentiation on what he provided since the algebra can get nasty, which results in a bigger chance of getting the problem incorrect.

Edit: I should say that two of the methods require finding the theta value at that instant. The method of differentiating inverse trig. function does not require you to find the theta value at that instant.

 

[Mark44]

This is what I meant for my hint.

What you said was the opposite of what you meant, which is why I said it was misleading.
You said, "Hint: The top of the triangle should be where the radar station is."

The radar station should be on the ground, which in my drawing is at the bottom of the vertical leg.

I know three different ways to solve this problem. Of the three, they all required the inverse trig function to find theta at that instant. However, only one of the ways require using implicit differentiation with a inverse trig function. If he knows the derivative of the inverse trig function, it will be easier than using implicit differentiation on what he provided since the algebra can get nasty, which results in a bigger chance of getting the problem incorrect.

Edit: I should say that two of the methods require finding the theta value at that instant. The method of differentiating inverse trig. function does not require you to find the theta value at that instant.

It isn't necessary to find θ in this problem, so there is no need for inverse trig functions. The OP is on the right track where he has
tan θ = y/x

Even better would be tan θ = 3/x, since the plane's altitude doesn't change. A somewhat simpler relationship between θ and x is
cot θ = x/3

Now differentiate with respect to t to get the rates relationship, and solve for dθ/dt. The algebra is straightforward.

 

[asz304]

Wow...

So I get -1/sin^2(theta) = 1/3, then I get sin^2(theta) = -3? I can't do this on my calculator for some reason . Am I supposed to get the derivative of both sides? or just sin^2(theta)? My other trial was sin^2(theta) = -3/250.Another question:

A baseball player runs from home plate towards first base at 20 ft/s. How fast is the player's distance from second base changing when the player is quarter of the way to the first base? (One side of a baseball diamond is 90 ft.

I am finding dy/dt ,so the formula goes like this x^2 + y^2 = 90^2, and is x = 1/4 ? If so I will find y by plugging in 1/4 and do some algebra to get 89.9996 ( which is y ). Then have 2x ( dx/dt) + 2y( dy / dt ) = 0 which is [ 10 + 179.9993056 dy/dt = 0 ] and solving for dy/dt...which makes me end up with an unreasonable value of 0.055...Is my work right?

 

[Mark44]

Wow...

So I get -1/sin^2(theta) = 1/3

No. You need to use the chain rule.

Starting with cot(θ) = x/3, and differentiating both sides with respect to t, you should get
-csc²(θ) * dθ/dt = 1/3 * dx/dt

, then I get sin^2(theta) = -3? I can't do this on my calculator for some reason . Am I supposed to get the derivative of both sides? or just sin^2(theta)? My other trial was sin^2(theta) = -3/250.


Another question:

A baseball player runs from home plate towards first base at 20 ft/s. How fast is the player's distance from second base changing when the player is quarter of the way to the first base? (One side of a baseball diamond is 90 ft.

I am finding dy/dt ,so the formula goes like this x^2 + y^2 = 90^2, and is x = 1/4 ? If so I will find y by plugging in 1/4 and do some algebra to get 89.9996 ( which is y ). Then have 2x ( dx/dt) + 2y( dy / dt ) = 0 which is [ 10 + 179.9993056 dy/dt = 0 ] and solving for dy/dt...which makes me end up with an unreasonable value of 0.055...Is my work right?

You say you are finding dy/dt, but you don't say what either x or y represents. If you haven't drawn a sketch for this problem, you really need to do this.

"one quarter of the way to first base" does NOT mean that x = 1/4. How far is it from home plate to first base? How far is it from first base to second base?

 

[asz304]

So dx/dt = 300 or 250? And how do I get the exact numerical answer if theta is involved with -csc^2(theta). Could it just be d(theta)/dt = 3/250?

 

[Mark44]

So dx/dt = 300 or 250?

dx/dt = 300 mi/hr, which is given information.

And how do I get the exact numerical answer if theta is involved with -csc^2(theta). Could it just be d(theta)/dt = 3/250?

No, d(theta)/dt is not 3/250. How did you get that?