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.

  • $\begingroup$ $D=\sqrt{A^2+B^2}$ so differentiate. You also know $A=2t$ and $B=3t$ $\endgroup$
    – Henry
    Commented May 22, 2021 at 16:28
  • $\begingroup$ Let $A$ and $B$ be the distances of the bikers to the intersection. Then $D^2=A^2+B^2$. Differentiate both sides of this with respect to time, plug in what you know, and solve for what you don't. $\endgroup$ Commented May 22, 2021 at 16:29
  • $\begingroup$ When I differentiate I get 2(dD/dt) = 2(dA/dt) + 2(dB/dt). Would I then plug in to get 2(dD/dt) = 2(2) + 2(3)? (Did I do that correctly?) $\endgroup$ Commented May 22, 2021 at 16:31
  • $\begingroup$ Not quite:$2D\cdot { dD\over dt}=2A\cdot { dA\over dt}+2B\cdot { dB\over dt}.$ (Note you have to figure out what $A, B, D$ are when $t=5$) $\endgroup$ Commented May 22, 2021 at 16:34
  • $\begingroup$ If I plug in everything to what you wrote above, I got D(dD/dt) = 65. Here I am stuck again though because the problem never told me the value of D, it just asked me to find the rate of it with respect to time. $\endgroup$ Commented May 22, 2021 at 16:45

2 Answers 2



There are two approaches:


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.

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

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.


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 $\sqrt{2^2+3^2}$.


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