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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
up vote 3 down vote accepted

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 $-60$.

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To make this clearer: the speed of car A is 60 km/h. When we talk about the velocity, we need to specify the car's speed and direction of travel. If we say going north is positive and south is negative, then since the car is traveling south we say the velocity of car A is 60 km/h south, i.e. -60 km/h. – Gyu Eun Lee Dec 11 '12 at 4:40

The distance between the car that is going south and point P is decreasing. For this reason, dx/dt is negative. It's not about North or South, it's about a change in distance compared to a change in time. You should always ask yourself if a value is increasing or decreasing when determining the sign on a rate of change. You will usually be able to determine this from the problem or, even better, from a sketch. You'll also notice that most "math people" will label horizontal distances as x and vertical distances as y. Your problem statement is opposite of this, which is ok, but makes some of these responses difficult to follow.

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