by Baltuilhe » Sun Sep 06, 2020 6:59 pm
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!