- 187.237 km/hr fast is θ changing 12 min after the plane passes over the radar station
Explanation:
Let the distance x and angle θ be defined as in the figure below. Then
[tex]\tan \theta=\frac{6}{x}[/tex]
Now, differentiate with respect to t, we get
[tex]\sec ^{2} \theta \frac{d \theta}{d t}=-\left(\frac{6}{x^{2}}\right) \frac{d x}{d t}[/tex]
Now, calculate the travel distance from radar station to plane after 12min
Distance, [tex]x=800 \times \frac{12}{60}=160[/tex]
Substituting ‘x’ value, we get
[tex]\tan \theta=\frac{6}{160}=\frac{3}{80}[/tex]
Find the rate of change of theta after 12 min,
[tex]\frac{d \theta}{d t}=-\frac{1}{\sec ^{2} \theta} \times \frac{6}{x^{2}} \times \frac{d x}{d t}[/tex]
We know, the formula for,
[tex]\sec ^{2} \theta=1+\tan ^{2} \theta=1+\frac{3^{2}}{80^{2}}[/tex]
So, then, [tex]\frac{d x}{d t}=800 \mathrm{km} / \mathrm{hr}(\text { let assume })[/tex]
[tex]\frac{d \theta}{d t}=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800[/tex]
[tex]=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800[/tex]
[tex]=-\frac{1}{1.0014} \times \frac{6}{25600} \times 800[/tex]
[tex]=-\frac{4800}{25635.84}=-0.187237 \mathrm{rad} / \mathrm{hr}[/tex]
When express the value in km/he, we get, the change in theta as
[tex]=-187.237 \mathrm{km} / \mathrm{hr}[/tex]
The rate of change of the radar station 12 minutes after the plane passes over the radar station is approximately [tex]-4.982\times 10^{-3}[/tex] radians per minute.
The geometric diagram of the situation of the plane is shown below. The angle of the radar station ([tex]\theta[/tex]), in radians, is represented by the following trigonometric expression:
[tex]\tan \theta = \frac{y}{x}[/tex] (1)
Where:
Since we know that [tex]h[/tex] is constant and the plane travels at constant speed, then we have the following expression for the rate of change of the radar station ([tex]\dot \theta[/tex]), in radians per minute:
[tex]\sec^{2}\theta\cdot \dot \theta = -\frac{y\cdot\dot x}{x^{2}}[/tex]
[tex](\tan^{2}\theta +1)\cdot \dot \theta = -\frac{y\cdot \dot x}{(\dot x\cdot t)^{2}}[/tex]
[tex]\left(\frac{y^{2}}{x^{2}}+1 \right) \cdot \dot \theta = -\frac{y}{\dot x\cdot t^{2}}[/tex]
[tex]\left(\frac{y^{2}}{\dot x^{2}\cdot t^{2}} +1\right)\cdot \dot \theta = -\frac{y}{\dot x\cdot t^{2}}[/tex]
[tex]\left(\frac{y^{2}+\dot x^{2}\cdot t^{2}}{\dot x^{2}\cdot t^{2}} \right)\cdot \dot \theta = -\frac{y}{\dot x\cdot t^{2}}[/tex]
[tex](y^{2}+\dot x^{2}\cdot t^{2})\cdot \dot \theta = -y\cdot \dot x[/tex]
[tex]\dot \theta = -\frac{y\cdot \dot x}{y^{2}+\dot x^{2}\cdot t^{2}}[/tex] (2)
Where [tex]t[/tex] is the time, in minutes.
If we know that [tex]y = 6\,km[/tex], [tex]\dot x = 8.333\,\frac{km}{min}[/tex] and [tex]t = 12\,min[/tex], then the rate of change of the angle of the radar station is:
[tex]\dot \theta = -\frac{(6\,km)\cdot \left(8.333\,\frac{km}{min} \right)}{(6\,km)^{2}+\left(8.333\,\frac{km}{min} \right)^{2}\cdot (12\,min)^{2}}[/tex]
[tex]\dot \theta \approx -4.982\times 10^{-3}\,\frac{rad}{min}[/tex]
The rate of change of the radar station 12 minutes after the plane passes over the radar station is approximately [tex]-4.982\times 10^{-3}[/tex] radians per minute. [tex]\blacksquare[/tex]
The plane speed is missing (500 kilometers per hour).
To learn more on rates of change, we kindly invite to check this verified question: https://brainly.com/question/11606037
