Calculus airplane related rates problem ( cosine rule)

[jaychay]

A student has test his airplane and he is far from the airplane for 5 meter.He start to test his airplane by letting his airplane to move 60 degree from the horizontal plane with constant velocity for 120 meter per minute.Find the rate of distance between the student and the plane when the plane is moving 60 degree from the horizontal plane for 10 meter in the air ?[img][img][/img][/img]

[Guest]

Ok, so when the plane is 10 meters in the air, we can drop a perpendicular. This tells us that the plane has been flying [tex]\boxed{\frac{20\sqrt3}{3}}[/tex] meters, which is just our answer. No need for law of cosines.

[Baltuilhe]

Good night!

If I have understood:

b=5m
c=10m
a=?

[tex]a^2=b^2+c^2-2bc\cos\theta[/tex]
[tex]a^2=5^2+10^2-2\cdot 5\cdot 10\cdot \cos\; 120^{\circ}[/tex]
[tex]a^2=25+100-2\cdot 50\cdot\frac{-1}{2}[/tex]
[tex]a^2=25+100+50[/tex]
[tex]a^2=175[/tex]
[tex]a=\sqrt{175}=5\sqrt{7}[/tex]

Well, now we have to calculate the rate of distance:
Again:
[tex]a^2=b^2+c^2-2bc\cos\theta[/tex]
The derivative:
[tex]\frac{d(a^2)}{dt}=\frac{d(b^2)}{dt}+\frac{d(c^2)}{dt}-\frac{d(2bc\cos\theta)}{dt}[/tex]
[tex]2a\cdot\frac{da}{dt}=0+2c\cdot\frac{dc}{dt}-2b\cos\theta[/tex]

Now, with the known values:
[tex]2\cdot 5\sqrt{7}\cdot\frac{da}{dt}=2\cdot 10\cdot\frac{dc}{dt}-2\cdot 5\cdot\frac{-1}{2}[/tex]
[tex]2\cdot 5\sqrt{7}\cdot\frac{da}{dt}=2\cdot 10\cdot 120m/min-2\cdot 5\cdot\frac{-1}{2}[/tex]
[tex]10\sqrt{7}\cdot\frac{da}{dt}=2\,400+5=2\,405[/tex]
[tex]\frac{da}{dt}=\frac{2\,405}{10\sqrt{7}}=\frac{481}{2\sqrt{7}}=\frac{481\sqrt{7}}{14}\approx 90,9m/min[/tex]

Hope to have helped!