Rate of change of distance

Question

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?

Answer 1

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 ($dD/dt)$ = ...

Answer 2

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.

Answer 3

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.