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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$\begingroup$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 $-60$.
$\endgroup$ 1 $\begingroup$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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