# Rate of change of distance

A plane flying with a constant speed of $19 \,\text{km/min}$ passes over a ground radar station at an altitude of $10 \, \text{km}$ and climbs at an angle of $20^\circ$. At what rate is the distance from the plane to the radar station increasing $2$ minutes later?

So I drew up a triangle with a vertical height of $10\, \text{km}$ and an angle of elevation of $20^\circ$. But I'm not sure how to proceed after this. What equation do I have to set up so that I can implicitly differentiate it? How would I relate the triangle into it?

Any help?

• Carefully draw a diagram showing the radar station, the position of the plane as it passes over the station, and the position of the plane two minutes later. Do not assume that these three points form a right triangle. There are multiple ways to proceed from there, depending on whether you're more comfortable giving $x,y$ coordinates to everything, or using the law of cosines, or maybe something else. – David K Nov 3 '17 at 21:15

The altitude in the question should not be the altitude of a triangle (at least if you want to use the cosine law).

Let $x$ represent the distance of the plane to the radar station.

Let $y$ represent the distance of the plane from its initial position at $t=0$ (when it was 10 km above the radar station). Essentially $y$ is the distance of the plane after $t$ minutes.

We are given $\displaystyle \frac{dy}{dt} = \text{19 km/min}$

We are trying to find $\displaystyle \frac{dx}{dt}$ at $t=2\text{ min}$.

The climbing angle of $20^\circ$ is assumed to be the angle the plane makes with the horizontal. So sketching a diagram:

Notice we form an obtuse triangle with the sides $x$, $y$, and 10 km.

Directly beneath the $20^\circ$ is a $90^\circ$ right angle that can be formed, since the altitude is perpendicular to the horizontal.

That means that the obtuse angle in that triangle is $20^\circ + 90^\circ = 110^\circ$

Therefore, we have:

We can use the cosine law to relate these quantities together. (With cosine law, $a$ is assumed to be the length that is across from an angle $A$. $b$ and $c$ are the other sides of the triangle.)

\begin{align} a^2 &= b^2 + c^2 -2bc\cos A \\ x^2 &= 10^2 + y^2 - 2(10)(y)\cos 110^\circ \\ x^2 &= 10^2 + y^2 - 20\cos 110^\circ \cdot y \\ \end{align}

Note that $\cos 110^\circ$ is just number (a negative number actually), so I write it as a coefficient on $y$ along with the $20$.

Differentiate both sides with respect to time, $t$. Then isolate $\frac{dx}{dt}$.

\begin{align} \frac{d}{dt} \left[x^2\right] &= \frac{d}{dt} \left[ 10^2 + y^2 - 20\cos 110^\circ y \right] \\ 2x \frac{dx}{dt} &= 0 + 2y \frac{dy}{dt} - 20\cos 110^\circ \frac{dy}{dt} \\ 2x \frac{dx}{dt} &= 2y \frac{dy}{dt} - 20\cos 110^\circ \frac{dy}{dt} \\ \frac{dx}{dt} &= \frac{2y \frac{dy}{dt} - 20\cos 110^\circ \frac{dy}{dt}}{2x} \end{align}

In order to get $\displaystyle \left.\frac{dx}{dt}\right|_{t=2}$, we need to figure out the lengths of $x$ and $y$ at $t=2 \text{ minutes}$.

Since $\displaystyle \frac{dy}{dt} = \text{19 km/min}$, the length of $y$ is $19 \times 2 = 38\text{ km}$.

The cosine law can be used to find $x$ at $t=2 \text{ minutes}$.

\begin{align} a^2 &= b^2 + c^2 -2bc\cos A \\ a &= \sqrt{b^2 + c^2 -2bc\cos A} \\ x &= \sqrt{10^2 + 38^2 -2(10)(38)\cos 110^\circ } \\ x &\approx 42.47275 \end{align}

Using the above:

\begin{align} \frac{dx}{dt} &= \frac{2y \frac{dy}{dt} - 20\cos 110^\circ \frac{dy}{dt}}{2x} \\ \left.\frac{dx}{dt}\right|_{t=2}&= \frac{2(38)(19) - 20\cos 110^\circ (19)}{(2 )(42.47275)}\\ &= 18.529 \text{ km/min} \end{align}

The distance between plane and radar station increases at $18.529 \text{ km/min}$ two minutes after it passes the radar station.

I call the distance from R to D after 2 minutes of fly as a function $x(t)$

The horizontal distance can be described with this formula:

\begin{align}L(t)=Vt \cos\alpha &&(1)\end{align}

The vertical distance can be described with this formula: \begin{align}h(t)=H+Vt \sin\alpha &&(2)\end{align}

Therefore, by using Pythagorean theorem.

By using (1) and (2), (4) can be rewritten as

\begin{align} x(t) &=\sqrt{(H+Vt \sin\alpha)^2+ (Vt \cos\alpha)^2} \\ &= \sqrt{H^2+2H Vt\sin\alpha+V^2t^2 \sin^2{\alpha}+ V^2t^2 \cos^2{\alpha}} && (5) \end{align}

By using a trigonometric formula (the Pythagorean identity), $\sin^2{\alpha}+\cos^2{\alpha}=1$, (5) can be simplified to

\begin{align} x(t)&=\sqrt{(H^2+2H Vt\sin\alpha+V^2t^2 \sin^2{\alpha}+ V^2t^2 \cos^2{\alpha}} \\ &= \sqrt{H^2+2H Vt\sin\alpha+ V^2t^2} && (6) \end{align}

Now differentiate (6).

\begin{align}x^{'}(t) &= \left[\sqrt{H^2+2H Vt\sin\alpha+ V^2t^2}\right]^{'} \\ &= \frac{1}{2}\left[H^2+2H Vt\sin\alpha+ V^2t^2\right]^{\frac{1}{2}-1}\cdot \left[H^2+2H Vt\sin\alpha+ V^2t^2\right]^{'} \\ &= \frac{V^2t+HV\sin\alpha}{\sqrt{H^2+2H Vt\sin\alpha+ V^2t^2}} && (7) \end{align}

Now add values and calculate the result: \begin{align} x^{'}(t) &= \frac{V^2t+HV\sin\alpha}{\sqrt{H^2+2H Vt\sin\alpha+ V^2t^2}} \\ x^{'}(2) &= \frac{19^2\cdot2+10\cdot19 \sin 20^\circ}{\sqrt{10^2+2 \cdot 19 \cdot 10 \cdot 2\sin20^\circ+ 19^2 \cdot 2^2}} \\ &= 18.53 \frac{\mathrm{km}}{\mathrm{min}} \end{align}

Questions? Correct my calculations if you find errors.

You can write x and y as functions of time.

x = 5t cos 20◦

y = 5tsin20◦

Now the distance$D =\sqrt(x^{2} + (1 + y)^{2})$.

But it will be easier to work with D2 . $D^{2} = x^{2} + (1 + y)^{2}$

Then differentiating the above we get 2D ($dD/dt)$ = ...