# Related Rates - Not getting answer in book

1. Mar 27, 2006

### GeoMike

The problem given in my book is:

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 = $$\frac{1000}{\sqrt{5}}$$

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-

Last edited: Mar 27, 2006
2. Mar 27, 2006

### 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

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

3. Mar 27, 2006

### 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.

4. Mar 27, 2006

### arildno

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

It is s that equals 2, not x!

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

5. Mar 27, 2006

### GeoMike

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

Thanks!
-GM-