# Spivak, Chapter 10, "Differentiation", problem 9: Rate of change of distance between moving particles. Is the solution manual solution incorrect?

Spivak, Chapter 10, "Differentiation", problem 9.

1. Particle A moves along the positive horizontal axis, and particle B along the graph of $$f(x)=-\sqrt{3}x, x \leq 0$$. At a certain time, A is at the point (5,0) and moving with speed 3 units/sec; and B is at a distance of 3 units from the origin and moving with speed 4 units/sec. At what rate is the distance between A and B changing?

The solution I came up with is similar to the solution manual solution, but it seems we disagree on certain signs which leads to different results. My result is that the distance is changing at a rate of $$-\frac{5}{14}$$ units/sec (ie it is decreasing), and the solution manual says the distance is increasing at $$\frac{83}{14}$$ units/sec.

Given that the vertical distance is certainly decreasing, and that since horizontally both particles are moving to the right the relative speed must be smaller than 3 units/second, it seems to me that the result $$\frac{83}{14} \approx 5.92$$ units/sec is incorrect.

Here is my solution

Everything in this problem is analyzed at a single point in time. The variables below are all functions of time, though I will omit the time parameter.

Initial Position of Particle B

$$D_{BO}^2=x_B^2+y_B^2$$

$$y_B=-\sqrt{3}x_B$$

$$D_{BO}^2=x_B^2+3x_B^2=4x_B^2\tag{1}$$

$$9=4x_B^2$$

$$x_B=\frac{3}{2}$$

$$\implies y_B=-\frac{3\sqrt{3}}{2}$$

Rates of Change of B's Coordinates

$$y_B(t)=-\sqrt{3}x_B(t)$$

$$y_B'=-\sqrt{3}x_B'$$

Here is a tricky part regarding signs. If we take the square root of $$(1)$$ then

$$D_{BO}=2x_B\tag{2}$$

Note that $$D_{BO}$$ is a distance, so it must be positive. $$x_B$$ on the other hand is actually negative. But $$(2)$$ is actually incorrect, it should be

$$D_{BO}=2|x_B|$$

$$=\begin{cases} -2x_B, x_B<0 \\ 2x_B, x_B \geq 0 \end{cases}$$

But in our single point in time we are in a situation where $$x_B<0$$. In this scenario, we have

$$D_{BO}'=-2x_B'$$

Note that $$D_{BO}'=-4$$ because particle B is moving towards the origin.

$$-4=-2x_B'$$

$$x_B'=2$$

$$\implies y_B=-2\sqrt{3}$$

These derivatives make sense: $$x_B$$ is increasing from a negative value towards $$0$$, and $$y_B$$ is decreasing from a positive value.

Rate of Change of Distance Between the Two Particles

$$D = \sqrt{y_B^2+(x_A-x_B)^2}$$

Note that since $$x_B$$ is negative, I am subtracting from $$x_A$$ to get difference in x-coordinate between the particles.

$$D'=\frac{y_By_B'+(x_A-x_B)(x_A'-x_B')}{\sqrt{y_B^2+(x_A-x_B)^2}}$$

We have the values of all these variables at the time we are considering. If we plug them in we get

$$D'=-\frac{5}{14}$$

Thus the distance between the particles is decreasing at a rate of $$\frac{5}{14}$$ units/sec.

Is this correct (ie is the solution manual wrong?)

• "Particle B moves along the graph, $f(x)=-\sqrt{3}x, x \leq 0$". I don't know why you chose the direction of $B$ to be towards the origin, but in my understanding, the initial position of $B$ is at the origin and it is moving away from the origin along the curve, $f$. So, your $D'_{BO}$ should be positive while $x'_{B}$ will be negative. Commented May 12, 2022 at 23:58
• You can save a great many steps by exploiting the fact that the line meets the $y$ axis at an angle of $\frac \pi 6$, so that cosine law for the triangle takes the form.. $$D^2 = a^2+b^2+ ab$$
– WW1
Commented May 13, 2022 at 1:23