The rate of change of the distance from the plane to the radar station

Question

Problem statement:

A plane flying with constant speed of 4 km/min passes over a ground radar station at an altitude of 6 km and climbs at an angle of 35 degrees. At what rate, in km/min, is the distance from the plane to the radar station increasing 6 minutes later?

Hint: The law of cosines for a triangle is $c^2=a^2+b^2− 2ab \cos(\theta)$ where $\theta$ is the angle between the sides of length $a$ and $b$.

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Okay, so following the typical related rates algorithm:

Identify the problem at hand by drawing a picture. I did that. It appears to me that after the plane passes over the ground station, we get an angle between the measures a and b of 90 + 35 degrees.

Awesome.

Now, let's build a table of values: $$ a = 6\,\mathrm{km}\\ a' = 0 \quad \text{(duh)}\\ b = b'\cdot t = 4\cdot 6 = 24\\ b' = 4\,\mathrm{km}/\mathrm{min}\\ t' = 6min\\ c = \sqrt{a^2 + b^2} = \sqrt{36+24} = 2\sqrt{15} \approx 2.78316\\ c' = ? $$

Okay, let's differentiate the law of cosines, which is given to us in form of a hint:

$2cc' = 2aa' + 2bb' - \frac{du}{dt}$

$ u = 2ab\cos(\theta)$

$\frac{du}{dt} = \frac{ds}{dt} \cdot \cos(\theta) - \sin(\theta)\cdot(2ab)$

$\frac{ds}{dt} = 0\cdot ab + (a'b + b'a)$

And so now we rewind: $\frac{du}{dt} = \left[ \left(0 \cdot ab + \left(a'b + b'a \right) \right) \right] \cdot \cos(\theta) - \sin(\theta) \cdot 2ab $

And so:

$2cc' = 2aa' + 2bb' - \left[ \left(0 \cdot ab + \left(a'b + b'a \right) \right) \right] \cdot \cos(\theta) - \sin(\theta) \cdot 2ab $

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Boom, boom. Let's substutute what we know

$2\left(2 \sqrt{15} \right)(c') = 2(6)(0) + 2(24)(4) - \left[ \left(0\cdot (6)(24) + \left((0)(24) + (4)(6) \right) \right) \right] \cdot \cos(125^\circ) - \sin(125^\circ) \cdot 2(4)(24)$

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Now what? Well, evaluate. Not sure if that's going to yield the right answer. But I don't think that it will.

Please help! I don't know how to solve this kind of problem! It seems my solution/process is incorrect!

Answer 1

We use your notation. Let $c=c(t)$ be the distance from the plane to the radar station at time $t$. Let $b$ be the distance from the plane to the point where the climb began. Then $$c^2=36+b^2-12b\cos \theta=36+b^2+12b\sin\phi,\tag{1}$$ where $\theta=125^\circ$ and $\phi=35^\circ$. Differentiating, we get $$2c\frac{dc}{dt}=2b\frac{db}{dt}+12\sin\phi\frac{db}{dt}.\tag{2}$$ "Freeze" at time $6$ minutes after the climb began. We know everything on the right-hand side of (2), since at that instant $b=(6)(4)$.

We need to calculate $c$ at time $t=6$. Equation (1) does that, with $b=24$.

Answer 2

Your solution looks okay for the most part; you've just made some mistakes. A couple points:

1. $t'$ means the rate at which $t$ changes with respect to itself. This is of course just 1.
2. Keep track of what's constant before breaking out the product rule, which can cause your formulae to explode. In the expression $2 a b \cos(\theta)$, both $a$ and $\cos(\theta)$ are simply constants, so it's quite easy to differentiate.

Answer 3

At the end of your table of values, $c$ isn’t $\sqrt{a^2+b^2}$; the correct formula is

$$c=\sqrt{a^2+b^2-2ab\cos\theta}\;,$$

which in this case is

$$c=\sqrt{6^2+(4t)^2-2\cdot6\cdot4t\cdot\cos 125°}=\sqrt{36+16t^2-48t\cos125°}\;.$$

The differentiation is easier if we start with the square of that, and we get $$2cc'=32t-48\cos125°\;,$$ or $$c'=\frac{16t-24\cos 125°}c\;.$$ At $t=6$ we have $c\approx 27.87813$ km and hence $c'\approx 3.93734$ km/min.

Note that with this approach you’d have been less likely to get lost in the differentiation (e.g., forgetting that $a$ and $\cos\theta$ are constants).

Answer 4

This seems way too complicated. I must be missing something.

We have $x(t) = 4t \cos \theta$, $y(t) = 6+4t \sin \theta$. Hence the distance from the radar station is $d(t)^2 = x(t)^2 + y(t)^2 = 48\,t\,\sin\left( \theta\right) +16\,{t}^{2}+36$.

Taking the square root and differentiating gives $d'(t) = \frac{48\,{\sin}\left( \theta\right) +32\,t}{2\,\sqrt{48\,t\,\mathrm{sin}\left( \theta\right) +16\,{t}^{2}+36}}$.

Substituting $t=6$ gives: $d'(6) = \frac{48\,\sin\left( \theta\right) +192}{2\,\sqrt{288\,\sin\left( \theta\right) +612}}$.

This gives me, roughly, $d'(6) \approx 3.94$.

Answer 5

Do you really need to use calculus?

At the moment in question, the plane is downrange from the radar station $24\cos(35^\circ)$ km at an altitude of $(6+24\cdot\sin(35^\circ)) $ km. So, $\theta$, the angle of elevation of the plane as seen from the radar station is given by:$$\theta=\tan^{-1}\frac{6+24\cdot\sin(35^\circ)}{24\cos(35^\circ)}$$ $\theta$ is obviously greater than $35^\circ$.

The plane is still climbing at $35^\circ$ at 4 km/ min. The component of its velocity along the line of sight from the radar station is:$$V_{los}=4\cdot\cos(\theta-35^\circ)$$