Rate of change of distance

1,625

drawing 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.

Share:
1,625

Related videos on Youtube

sktsasus
Author by

sktsasus

Updated on November 19, 2020

Comments

  • sktsasus
    sktsasus almost 3 years

    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?