Question

A rocket travels vertically at a speed of 1000 km/hr. The rocket is tracked through a telescope by an observer located 15 km from the launching pad. Find the rate at which the angle between the telescope and the ground is increasing 3 min after the lift-off.

Answer 1

Answer:

... a speed of 1200 km/h. The rocket is tracked through a telescope by an observer located 16. ... Of 1200 Km/h. The Rocket Is Tracked Through A Telescope By An Observer Located 16 Km From The Launching Pad. Find The Rate At Which The Angle Between The Telescope And The Ground Is Increasing 3 Min After Lift-off.

Answer 2

The rate at which the angle between the telescope and the ground is increasing 3 minutes after the lift-off is approximately 19.157 radians per hour.

First, we create a geometric diagram with the following variables:

• $x$ - Horizontal distance between the rocket and the telescope, in kilometers.
• $y$ - Vertical distance between the telescope and the rocket, in kilometers.
• $r$ - Straight line distance between the telescope and the rocket, in kilometers.
• $\theta$ - Angle of the telescope, in radians.

By trigonometric ratios we have the following expression for the angle of the telescope:

$\tan \theta = \frac{y}{x}$ (1)

Then, we derive an expression for the rate of change of the angle of the telescope in time ($\dot \theta$), in radians per hour, by differential calculus, trigonometric ratios and the Pythagorean theorem:

$\sec^{2}\theta \,\dot{\theta} = \frac{\dot {y}\cdot x - y \cdot \dot {x}}{x^{2}}$

$\frac{r\cdot \dot{\theta}}{x} = \frac{\dot y\cdot x - y\cdot \dot x}{x^{2}}$

$\dot {\theta} = \frac{x^{2}\cdot \dot {y}- x\cdot y\cdot \dot x}{r\cdot x^{2}}$

$\dot {\theta} = \frac{x^{2}\cdot \dot {y}-x\cdot y \cdot \dot x}{x^{2}\cdot \sqrt{x^{2}+y^{2}}}$ (2)

As of the rocket is travelling vertically, (2) is reduced into this form:

$\dot \theta = \frac{\dot y}{\sqrt{x^{2}+y^{2}}}$ (2b)

If we know that $x = 15\,km$, $y = 50\,km$ and $\dot y = 1000\,\frac{km}{h}$, then the rate at which the angle between the telescope and the ground is increasing is:

$\dot \theta = \frac{1000\,\frac{km}{h} }{\sqrt{(15\,km)^{2}+(50\,km)^{2}}}$

$\dot \theta \approx 19.157\,\frac{rad}{h}$

The rate at which the angle between the telescope and the ground is increasing 3 minutes after the lift-off is approximately 19.157 radians per hour.

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