Answer:
[tex]\frac{dc}{dt}\approx13.8146\text{ km/min}[/tex]
Step-by-step explanation:
We know that the plane travels at a constant speed of 14 km/min.
It passes over a radar station at a altitude of 11 km and climbs at an angle of 25°.
We want to find the rate at which the distance from the plane to the radar station is increasing 4 minutes later. In other words, if you will please refer to the figure, we want to find dc/dt.
First, let's find c, the distance. We can use the law of cosines:
[tex]c^2=a^2+b^2-2ab\cos(C)[/tex]
We know that the plane travels at a constant rate of 14 km/min. So, after 4 minutes, the plane would've traveled 14(4) or 56 km So, a is 56, b is a constant 11. C is 90+20 or 115°. Substitute:
[tex]c^2=(56)^2+(11)^2-2(56)(11)\cos(115)[/tex]
Evaluate:
[tex]c^2=3257-1232\cos(115)[/tex]
Take the square root of both sides:
[tex]c=\sqrt{3257-1232\cos(115)}[/tex]
Now, let's return to our law of cosines. We have:
[tex]c^2=a^2+b^2-2ab\cos(C)[/tex]
We want to find dc/dt. So, let's take the derivative of both sides with respect to t:
[tex]\frac{d}{dt}[c^2]=\frac{d}{dt}[a^2+b^2-2ab\cos(C)][/tex]
Since our b is constant at 11 km, we can substitute this in:
[tex]\frac{d}{dt}[c^2]=\frac{d}{dt}[a^2-(11)^2-2a(11)\cos(C)][/tex]
Evaluate:
[tex]\frac{d}{dt}[c^2]=\frac{d}{dt}[a^2-121-22a\cos(C)][/tex]
Implicitly differentiate:
[tex]2c\frac{dc}{dt}=2a\frac{da}{dt}-22\cos(115)\frac{da}{dt}[/tex]
Divide both sides by 2c:
[tex]\frac{dc}{dt}=\frac{2a\frac{da}{dt}-22\cos(115)\frac{da}{dt}}{2c}[/tex]
Solve for dc/dt. We already know that da/dt is 14 km/min. a is 56. We also know c. Substitute in these values:
[tex]\frac{dc}{dt}=\frac{2(56)(14)-22\cos(115)(14)}{2\sqrt{3257-1232\cos(115)}}[/tex]
Simplify:
[tex]\frac{dc}{dt}=\frac{1568-308\cos(115)}{2\sqrt{3257-1232\cos(115)}}[/tex]
Use a calculator. So:
[tex]\frac{dc}{dt}\approx13.8146\text{ km/min}[/tex]
And we're done!
