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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.

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  • $\begingroup$ $D=\sqrt{A^2+B^2}$ 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

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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.


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

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.

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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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