# Related rates, my answer differs from the book, misprint or me?

Did my answer go wrong or does the book have a misprint?(there have been some inconstancies between the definitions used in the chapters and answer key, like two different authors, though only one is listed)

The problem: An airplane is flying 500 miles per hour horizontal one mile high over a radar station. Find the rate at which the distance is increasing when the plane is 2 miles from the station.

My answer is $$1000/\sqrt5$$ or $$200*\sqrt5$$

The book gives $$250*\sqrt3$$

My method: triangle abc, a=1, b=2, and c is the hypotenuse, $$db/dt=500$$, and $$c^2=1^2+b^2$$, and I want $$dc/dt$$ at b=2

I took the derivative: $$2c*\frac{dc}{dt}=0+2b*\frac{db}{dt}$$,

solved for $$dc/dt$$; $$dc/dt=\frac{2b*db/dt}{2c}$$

Substitute the variables; $$c=\sqrt{1+4}$$ and so $$\frac{dc}{dt}=\frac{2*2*500}{2*\sqrt5}=\frac{1000}{\sqrt5}$$

It may be a misprint, but I don't think your answer is right either. The question asks for $$dc/dt$$ when the distance from the station is 2 - this means when c=2 not when b=2. Other than that your answer is correct; plugging a=1, b=$$\sqrt 3$$, c=2 in, I get $$250\sqrt3$$ (not $$250/\sqrt3$$)
• Ah semantics. Probably not the last time they get the better of me. And yes $250/\sqrt3$, was a typo on my part, fixed now. – Max Power Mar 9 '19 at 23:26
You approach is good, you just mixed up $$b$$ and $$c$$. When plane is two miles from the station, $$c=2$$, $$b=\sqrt{3}$$. So I think the answer should be $$250\sqrt{3}$$.