# Calculus Spotlight Angle Question

#### akbarali

##### New member
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

Staff member
Hello and welcome to MHB, akbarali! When I and others bring questions from other sites here, we provide a full solution, as a means of increasing our knowledge base and to demonstrate to guests the type of expertise available here at MHB.

As a registered member, you will now be encouraged to show what you have tried so that we may help you be a part of the learning process by taking part in getting to the solution. It would actually be lazy of us to provide full solutions to everyone, and would not meet our goal of teaching rather than simply providing answers, which is of minimal benefit to students.

We will provide suggestions/hints, and then expect you to either give feedback on your progress, or to ask for further clarification. We will continue until you have solved the problem, and you will have learned much more and will gain a sense of accomplishment in having actually taken part in finding the solution.

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?

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#### MarkFL

Staff member
Okay, good! 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

##### New member
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?

#### Bacterius

##### Well-known member
MHB Math Helper
[JUSTIFY]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 $\theta$?[/JUSTIFY]

#### MarkFL

Staff member
You are thinking along the right lines...we do want a trig. function. 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

##### New member
Cosine, I believe.

#### MarkFL

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

#### MarkFL

Staff member
[JUSTIFY]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 $\theta$?[/JUSTIFY]
This is actually computationally a much simpler way to go. I suggest we go this route.

#### MarkFL

Staff member
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 $$\displaystyle \frac{dx}{dt}$$.

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#### akbarali

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

#### MarkFL

Staff member
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

Staff member
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

##### New member
km/h ?

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

Yes, that does seem to agree.

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#### MarkFL

Staff member
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

##### New member
Thanks for all the help!

#### MarkFL

Staff member
Glad to be of assistance! Feel free to post any other questions you have as well! #### 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 ($x^2+1^2=3^2$).