I am stuck at this question for hours. Can someone advise me how to solve the following problem?
Question: A thief is $10$ meters away ($8$ meters ahead of you, across a street $6$ meters wide). The thief runs on that side at $7$meters/second, and you run at $9$meters/second.
How fast are you approaching if:
(a) you follow on your side;
(b) you run toward the thief;
(c) you run away on your side?
I have completely no idea how to solve (b), because I can't think of a function relating the thief and me.
For (a) and (c), I kinda feel that the thief lies on the function $(8+7t,6)$ and my coordinate is along $(9t,0)$.
I tried to applied the distance formula and I ended up getting $2$m/s for (a) which is not the case because the answer is $1.6$m/s, $9$m/s and $12.8$m/s for (a),(b),(c), respectively.
For part (a), you are correct that the coordinates of Me are $(9t,0)$ and of the thief are $(8+7t,6)$. The distance between us at time $t$ is therefore
$$\begin{align} d(t) &= \sqrt{[(8+7t)-9t)]^2+[6-0]^2} \\ &= \sqrt{(8-2t)^2+36} \\ &= \sqrt{64-32t+4t^2+36} \\ &= \sqrt{4t^2-32t+100} \\ \end{align}$$
The rate of change of that distance is
$$\begin{align} d'(t) &= \frac 12(4t^2-32t+100)^{-1/2}(8t-32) \end{align}$$
and the rate of change at zero is
$$\begin{align} d'(0) &= \frac 12(100)^{-1/2}(-32) \\ &= \frac 12\cdot\frac1{10}\cdot -32 \\ &= -1.6 \end{align}$$
That is the rate that the distance increases, so the rate of approach is the negative of that, namely
Part (c) is very similar, except now my position is $(-9t,0)$ (note the negative sign now). So we get
$$\begin{align} d(t) &= \sqrt{[(8+7t)-(-9t)]^2+[6-0]^2} \\ &= \sqrt{(8+16t)^2+36} \\ &= \sqrt{64+256t+256t^2+36} \\ &= \sqrt{256t^2+256t+100} \\ \end{align}$$
The rate of change of that distance is
$$\begin{align} d'(t) &= \frac 12(256t^2+256t+100)^{-1/2}(512t+256) \end{align}$$
and the rate of change at zero is
$$\begin{align} d'(0) &= \frac 12(100)^{-1/2}(256) \\ &= \frac 12\cdot\frac1{10}\cdot 256 \\ &= 12.8 \end{align}$$
That is the rate that the distance increases, so the rate of approach is the negative of that, namely
Note that this is the negative of what you said the answer is.
For part (b) I now am running at an angle, up and to the right, toward the thief. At least this is true at time $t=0$ and is approximately true at times near zero. My velocity vector length is 9 m/s and the angle I am running has the cosine of $\frac 8{10}$ and sine of $\frac 6{10}$ due to the distances given. My position is thus
$$(9t\cos\theta, 9t\sin\theta) = (9t\cdot\frac 8{10}, 9t\cdot\frac 6{10}) = (7.2t, 5.4t)$$
The distance between us is now
$$\begin{align} d(t) &= \sqrt{[(8+7t)-7.2t)]^2+[6-5.4t]^2} \\ &= \sqrt{(8-0.2t)^2+(6-5.4t)^2} \\ &= \sqrt{64-3.2t+0.04t^2+36-64.8t+29.16t^2} \\ &= \sqrt{29.2t^2-68t+100} \\ \end{align}$$
The rate of change of that distance is
$$\begin{align} d'(t) &= \frac 12(29.2t^2-68t+100)^{-1/2}(58.4t-68) \end{align}$$
and the rate of change at zero is
$$\begin{align} d'(0) &= \frac 12(100)^{-1/2}(-68) \\ &= \frac 12\cdot\frac1{10}\cdot -68 \\ &= -3.4 \end{align}$$
That is the rate that the distance increases, so the rate of approach is the negative of that, namely
This differs from your answer but makes intuitive sense. If the thief is running at 7 m/s not directly away from you and you are running toward him at 9 m/s, there is no way you are making 9 m/s directly toward him--that ignores his own speed partly away from you. Clearly you are making something less than 9 m/s but greater than $9-7$, as in my answer.
I tested all three answers in Geogebra, and they all check. The drawings are sloppy and not worthy of publication, not to mention that some of the distances are so small that they cannot be seen, so I do not show my drawings here.