Calculus Spotlight Angle Question

[akbarali]

A plane drifts over some area of land. The men on the ground train a spotlight on the airship, which is traveling at 90 km/hour, and at a constant altitude of 1 km. The beam of the spotlight makes an angle θ with the ground.

1. Draw a diagram.

2. When the airship is 3 kilometers from the spotlight, how fast is θ changing?

 

[MarkFL]

Hello and welcome to MHB, akbarali!(Wave)

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Have you drawn a diagram? You should have a right triangle, with an angle $\theta$ at the spotlight representing the angle of inclination of the beam, a side opposite the angle $\theta$ which is the altitude of the airship, and a hypotenuse (I call it $h$) representing the distance of the airship from the spotlight. You should also have a side adjacent to the spotlight (I call it $x$).

Once you draw this, can you think of a way to relate these quantities using trigonometry?

 

[akbarali]

View attachment 788

That's what I've got so far!

 

[MarkFL]

Okay, good! (Cool)

We need to let the hypotenuse be variable as it is changing...only the opposite side is remaining constant. So let the hypotenuse be $h$. How can we relate $x$, $h$, and $\theta$?

 

[akbarali]

I'd assume they're related via some trig function, but I'm not sure how to proceed at all, to be honest. I'm thinking we need to use the tan function?

 

[Nono713]

Yes, what I would do as well is work out the rate of change of the hypotenuse $h$ in the diagram, and express this result as a rate of change of the angle $\theta$ using trigonometric relations ;)

A diagram helps a lot in this situation, indeed.

If you look at the diagram, you have the relation $\tan{\theta} = \frac{\text{altitude}}{\text{horizontal distance}}$. Now the rate of change of $\text{horizontal distance}$ is known, and $\text{altitude}$ is a constant, so what can you deduce about the rate of change of $\tan{\theta}$? What about $\theta$?​

 

[MarkFL]

You are thinking along the right lines...we do want a trig. function. :D

The tangent function relates the angle to the opposite side and the adjacent side, but we want the trig. function that relates the angle to the adjacent side and the hypotenuse. Think of how the trig. functions are defined...which one has the adjacent side and the hypotenuse in its definition?

 

[akbarali]

Cosine, I believe.

 

[MarkFL]

Yes, that's correct, so can you state the relationship using the cosine function?

 

[MarkFL]

Yes, what I would do as well is work out the rate of change of the hypotenuse $h$ in the diagram, and express this result as a rate of change of the angle $\theta$ using trigonometric relations ;)

A diagram helps a lot in this situation, indeed.

If you look at the diagram, you have the relation $\tan{\theta} = \frac{\text{altitude}}{\text{horizontal distance}}$. Now the rate of change of $\text{horizontal distance}$ is known, and $\text{altitude}$ is a constant, so what can you deduce about the rate of change of $\tan{\theta}$? What about $\theta$?​

This is actually computationally a much simpler way to go. I suggest we go this route.

 

[akbarali]

View attachment 790

After watching a video or two, and checking the book, I tried it this way, but not confident at all. =/

 

[MarkFL]

I agree with:

\(\displaystyle \tan(\theta)=\frac{1}{x}\)

Implicitly differentiating with respect to time $t$:

\(\displaystyle \sec^2(\theta)\frac{d\theta}{dt}=-\frac{1}{x^2}\cdot\frac{dx}{dt}\)

Now, we know:

\(\displaystyle \sec(\theta)=\frac{h}{x}\)

and so we may write:

\(\displaystyle \left(\frac{h}{x} \right)^2\frac{d\theta}{dt}=-\frac{1}{x^2}\cdot\frac{dx}{dt}\)

\(\displaystyle \frac{d\theta}{dt}=-\frac{1}{h^2}\cdot\frac{dx}{dt}\)

Now, we are not told whether the airship is approaching the soptlight or moving away from it, but we see that the sign of \(\displaystyle \frac{d\theta}{dt}\) is the opposite of the sign of \(\displaystyle \frac{dx}{dt}\), which means $\theta$ is increasing when $x$ is decreasing and $\theta$ is decreasing when $x$ is increasing. We should just concern ourselves with the magnitude.

Now all you need to do is plug in the given values for $h$ and \(\displaystyle \frac{dx}{dt}\).

 

[akbarali]

If h = 3, and dx/dt = 90, dθ/dt = -10?

 

[MarkFL]

Using the method I initially suggested, we have:

\(\displaystyle \cos(\theta)=\frac{x}{h}\)

Implicitly differentiating with respect to time $t$:

\(\displaystyle -\sin(\theta)\frac{d\theta}{dt}=\frac{h\frac{dx}{dt}-x\frac{dh}{dt}}{h^2}\)

Now, by Pythagoras we know:

\(\displaystyle x^2+1=h^2\)

Implicitly differentiating with respect to time $t$:

\(\displaystyle 2x\frac{dx}{dt}=2h\frac{dh}{dt}\)

\(\displaystyle \frac{dh}{dt}=\frac{x}{h}\cdot\frac{dx}{dt}\)

Now substituting for \(\displaystyle \frac{dh}{dt}\) we have:

\(\displaystyle -\sin(\theta)\frac{d\theta}{dt}=\frac{h\frac{dx}{dt}-x\cdot\frac{x}{h}\cdot\frac{dx}{dt}}{h^2}\)

\(\displaystyle -\sin(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}\left(h-\frac{x^2}{h} \right)}{h^2}\)

\(\displaystyle -\sin(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}\left(h^2-x^2 \right)}{h^3}\)

Now \(\displaystyle \sin(\theta)=\frac{1}{h}\) and $h^2-x^2=1$ so this yields

\(\displaystyle -\frac{1}{h}\frac{d\theta}{dt}=\frac{1}{h^3}\cdot \frac{dx}{dt}\)

\(\displaystyle \frac{d\theta}{dt}=-\frac{1}{h^2}\cdot\frac{dx}{dt}\)

As you can see, we do obtain the correct result, but with many more steps than with the suggestion given by Bacterius.

 

[MarkFL]

If h = 3, and dx/dt = 90, dθ/dt = -10?

Yes, assuming the airship is moving away, I would just use the magnitude of 10...what are the units associated with this number?

I would also observe that as $h$ increases, the magnitude of \(\displaystyle \frac{d\theta}{dt}\) decreases...does this agree with intuition?

 

[akbarali]

km/h ?

So, if it was moving towards the spotlight, the answer would be +10, correct?

Yes, that does seem to agree.

 

[MarkFL]

The unit of $\theta$ is the radian (rad), while the unit of $t$ is the hour (hr), so the units of \(\displaystyle \frac{d\theta}{dt}\) is \(\displaystyle \frac{\text{rad}}{\text{hr}}\).

Typically we aren't expected to be as strict with our units in a calculus course as we are in a calculus based physics course. If you look at the result:

\(\displaystyle \frac{d\theta}{dt}=-\frac{1\text{ km}}{(h\text{ km})^2}\cdot\frac{dx}{dt}\,\frac{\text{km}}{\text{hr}}\)

We see through dividing out units we have:

\(\displaystyle \frac{d\theta}{dt}=-\frac{1}{h^2}\cdot\frac{dx}{dt}\,\frac{1}{\text{hr}}\)

Now we see we must attach the dimensionless unit of the radian to be consistent with the units:

\(\displaystyle \frac{d\theta}{dt}=-\frac{1}{h^2}\cdot\frac{dx}{dt}\,\frac{\text{rad}}{\text{hr}}\)

 

[akbarali]

Thanks for all the help!

 

[MarkFL]

Glad to be of assistance! Feel free to post any other questions you have as well! (Cool)

 

[akbarali]

View attachment 794

My tutor sent me a (belated) response to this question. His method yields -9, as opposed to -10. Can you see why one answer should be correct over the other?

 

[MarkFL]

Your tutor assumed a horizontal distance of 3 km from the spotlight ($x=3\text{ km}$), whereas I assume the true distance from the spotlight, i.e., along the hypotenuse ($x^2+1^2=3^2$).

Your tutor also uses a negated version of the quotient rule, that I found somewhat confusing, but the difference in numbers comes from the difference in interpretation of the problem.

 

[akbarali]

I see it now. Yes, I believe the distance should definitely be along the hypotenuse as well.