Related Rates - Not getting answer in book

[GeoMike]

The problem given in my book is:

A plane flying horizontally at an altitude of 1mi and a speed of 500 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2mi away from the station.

I set up and solved the problem this way:
dx/dt = 500 mi/h
x = 2mi
y = 1mi (constant)

Distance:
s² = y² + x²

s² = 1 + x²

s = (1 + x²)^1/2

d/dt = d/dt[sqrt(1 + x²)^1/2]

ds/dt = (1/2)(1 + x²)^-1/2 * (2x(dx/dt))

ds/dt = (1 + x²)^-1/2 * (x(dx/dt))

ds/dt = \frac{x(dx/dt)}{(1 + x^2)^{1/2}}

Substiting:
ds/dt = \frac{(2)(500)}{(1 + (2)^2)^{1/2}}

ds/dt = \frac{1000}{\sqrt{5}}

The final answer:
ds/dt = 200\sqrt{5} mi/h

However the back of my book has: 250\sqrt{3} mi/h

What am I doing wrong?
Thank you,
-GM-

 

[benorin]

don't solve for s, just leave it at s^2=1+x^2 and differentiate implicitly to get

2s\frac{ds}{dt}=2x\frac{dx}{dt}

solve for ds/dt

\frac{ds}{dt}=\frac{x}{s}\frac{dx}{dt}

 

[benorin]

You work was correct that far, however, since s=\sqrt{x^2+1} anyhow. The error is the sqrt(5) should be sqrt(3): 2 mi away is straight line distance.

 

[arildno]

You have misinterpreted the meaning of "2 mi away from the station".

It is s that equals 2, not x!

Thus, you have:
2^{2}=y^{2}+x^{2}\to{x}=\sqrt{3}

 

[GeoMike]

AH! Thank you!
I understand now, I was taking 2mi as horizontal distance.

Thanks!
-GM-