# Calculus airplane related rates problem ( cosine rule)

### Calculus airplane related rates problem ( cosine rule)

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]
Attachments
cal.png (239.36 KiB) Viewed 180 times
jaychay

Posts: 5
Joined: Tue Sep 01, 2020 6:58 pm
Reputation: 0

### Re: Calculus airplane related rates problem ( cosine rule)

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 $$\boxed{\frac{20\sqrt3}{3}}$$ meters, which is just our answer. No need for law of cosines.
Guest

### Re: Calculus airplane related rates problem ( cosine rule)

Good night!

If I have understood:

b=5m
c=10m
a=?

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

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

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

Hope to have helped!

Baltuilhe

Posts: 91
Joined: Fri Dec 14, 2018 3:55 pm
Reputation: 56