# Related Rates problem involving pythagorean theorem.

I know the basics of Related Rates, but this practice problem I have seems to have an incorrect answer... Either that or I don't have as strong of a grasp as I thought I did...

Car A is being driven south toward point P at a speed of 60 km/h. Car B is being driven to the east away from point P. When car A is 0.6 km from point P, car B is 0.8 km from point P and the straight line distance between them is increasing at 20 km/h. What is the speed of car B?

Now... Given a right angle triangle, $x=\frac{6}{10}$, $y=\frac{8}{10}$ and $z=1$. Differentiating the Pythagorean theorem with respect to time gives $$2x\cdot \frac{dx}{dt} + 2y\cdot \frac{dy}{dt} = 2z\cdot \frac{dz}{dt}$$

This is where the solution seems wrong. Because car A is going south, the solution says that $x=\frac{6}{10}$ and $\frac{dx}{dt}=-60$. But the car can't have a negative velocity.

When I plug my values into the equation, I get $$2\cdot \frac{6}{10}\cdot 60+2\cdot \frac{8}{10}\cdot \frac{dy}{dt}=2\cdot 1\cdot 20$$

Solving for $\frac{dy}{dt}$ gives me $$\frac{dy}{dt}=\frac{(40-72)(10)}{8}=\frac{-320}{8}=-40$$ But as I stated, since the car can't have a negative velocity, I just take the absolute value to be $40$ km/hr. But the solution shows an answer of $70$ km/hr.

Did I mess up somewhere? I just can't see what's wrong with my algebra.

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That is a negative velocity - which can be used to signify direction. If you plug in the -60 as-is, you get the result desired. – Amzoti Dec 11 '12 at 4:41
Why not let $y$ be the distance we are north from $P$, and $x$ the distance we are east of $P$. Then $\dfrac{dy}{dt}$ is negative, sensible enough. – André Nicolas Dec 11 '12 at 6:20

A car can have a negative velocity, because velocity takes into account both speed and direction. A car cannot have a negative speed. Speed and velocity are two different things. So go ahead and plug in the $-60$.