1
$\begingroup$

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 $\frac{dc}{dt}=\frac{2*2*500}{2*\sqrt5}=\frac{1000}{\sqrt5}$

$\endgroup$
1
$\begingroup$

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 $250/\sqrt3$)

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ – Max Power Mar 9 '19 at 23:26
1
$\begingroup$

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 $250\sqrt{3}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.