Related Rates Bikers

Question

I was given the question:

Bikes A and B are traveling on perpendicular roads. At the same time bike A is leaving the intersection at a rate of 2 feet per second and bike B is leaving the intersection at 3 feet per second. How fast in the distance, in feet per second, between them changing after 5 seconds?

A) -13/5 B) 13/5 C) sqrt(13) D) (13sqrt(5))/5 E) 5sqrt(13)

I know that dA/dt = 2 ft/sec, dB/dt = 3 ft/sec, and I am trying to find dD/dt, but I don't know where to start.

Answer 1

Hint:

There are two approaches:

$\underline{\text{Calculus}}:$

The distance as a function of time is just what Henry's comment says it is: $\sqrt{(2t)^2 + (3t)^2} = t\sqrt{13}.$ This is what you differentiate.

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$\underline{\text{Without Calculus}}:$

Actually, since this is a multiple choice problem, and since the choices offered have significant differences among them, the problem does not require Calculus.

Simply compute the distance between the two bikers after $5$ seconds, and then after $6$ seconds. The difference between the two distances gives you the approximate rate of change per second, at $5$ seconds. Then, plug each of the choices in, and see which one is close to your computation.

Answer 2

Their velocities are constant. So the rate of change of the distance between them is also constant. At time $t=0$ (and indeed at any subsequent time), they have perpendicular velocities of magnitudes $2$ and $3$. So their relative speed is $\sqrt{2^2+3^2}$.