Problem:
At a certain moment, a ladybug is at position $x_0$ and moving with velocity vector $v$. At the moment, the angle $<ax_0b = \pi/2$, her velocity bisects the angle, and her speed is $5$ units/sec. At what rate is the sum of her distances from $a$ and $b$ decreasing at that moment?
My question is why my final answer is positive instead of negative.
Solution:
I will begin by defining two functions. let $p$ be the function that gives the position of the bug at time $t$ such that $p(t) = x_0 + tv$, and let $s$ be the function that gives the sum of the distances between the current position of the bug and the points $a$ and $b$ such that $s(x) = \rVert x - a\lVert + \rVert x - b\lVert$. Now I need the function which gives the sum of the distances given the time. This function which I will call $h$ is the composition of $s$ and $t$, so $h = s \circ p$.
The question is asking for the rate of decrease of $h$ at $t = 0$ so I need the derivative of $h$ at $t = 0$ which is $h'(0) = Ds(p(0))Dp(0)$.
$h'(0)$ is a scalar so it equals its transpose, so $h'(0) = (Ds(p(0))Dp(0))^\intercal = (Dp(0))^\intercal \nabla s(p(0))$.
Now $(Dp(0))^\intercal = v^\intercal$, and $\nabla s(p(0)) = \dfrac{p(0) - a}{\rVert p(0) - a\lVert} + \dfrac{p(0) - b}{\rVert p(0) - b\lVert}$.
So now $h'(0) = v^\intercal(\dfrac{p(0) - a}{\rVert p(0) - a\lVert} + \dfrac{p(0) - b}{\rVert p(0) - b\lVert}) = v \cdot(\dfrac{p(0) - a}{\rVert p(0) - a\lVert} + \dfrac{p(0) - b}{\rVert p(0) - b\lVert}) = (\dfrac{v \cdot (p(0) - a)}{\rVert p(0) - a\lVert} + \dfrac{v \cdot (p(0) - b)}{\rVert p(0) - b\lVert})$.
From the question we know that $\rVert v\lVert = 5$ and that the angle between $v$ and the two vectors is $\pi/4$, so $h'(0) = 5 \cos{45} + 5 \cos{45} = 5\sqrt{2}$ but that is a positive number which is wrong as the distance should be decreasing.
It seem to me from the picture as if the bug is moving in the direction of the smallest rate of change, so maybe somehow my mistake was in confusing $v$ with $-v$ which should be the direction of the greatest rate of change and the correct answer should be $-5\sqrt{2}$. Perhaps all my solution is just wrong.