Here is a question from a Cambridge practice paper. Year 12 standard, math B. It involves finding the time when the cars are at a minimum distance from each other (part a). I have solved part a below. However, I am stuck on part b. A solution for part c would also be very much appreciated. Thanks in advance.
Two straight roads cross at right angles at a point $X$. A red car travels at $v \: km/h$ along the first road away from $X$. A blue car travels at $w \: km/h$ along the other road towards $X$. At a particular time the red car is $a \: km$ from $X$ and the blue car is $b \: km$ from $X$. (Only consider positive values of $a$, $b$, $v$, $w$.)
a. If $a=b=100\:km$ and $w=100\:km/h$ and $v=80\:km/h$ determine the time, correct to the nearest minute when the cars are closest and determine their distance apart at this point, correct to the nearest meter.
b. Returning to the general situation: If $\frac{a}{b}<\frac{w}{v}$ show that the cars will be closest when their distances from $X$ are in the ratio $w:v$.
c. Given that $a=b=100\:km$ and that the cars are closest after 1 hour show that the pairs $(w, v)$ of possible values of $w$ and $v$ lie on an arc the circle $(v+50)^2+(w-50)^2=5000$ and verify $w=60$, $v=20$ is one such pair.
Part A solution
Here, I am using $d$ as referring to the distance to point $X$. $r$ is red and $b$ is blue.
It is known that speed is given by $s=\frac{d}{t}$
Thus distance with respect to time can be given for both cars:
$$d_r=a+vt$$
$$d_b=b-wt$$
Since the roads are a right angles, pythagoras can be used to solve for the total distance $$d_t=\sqrt{(a+vt)^2+(b-wt)^2}$$
In part a, values are given as follows: $a=b=100\:km$ and $w=100\:km/h$ and $v=80\:km/h$.
Therefore, $$d_t=\sqrt{(100+80t)^2+(100-100t)^2}$$
Now differentiate and solve for a minimum
$\frac{dd_t}{dt}=\frac{2\cdot 80(100+80t)-2\cdot 100(100-100t)}{2\sqrt{(100+80t)^2+(100-100t)^2}}$
$\frac{dd_t}{dt}=\frac{80(100+80t)-100(100-100t)}{\sqrt{(100+80t)^2+(100-100t)^2}}$
$0=80(100+80t)-100(100-100t),\:\:\:\: 0\neq (100+80t)^2+(100-100t)^2$
since the denominator cannot equal zero.
Thus:
$100(100-100t)=80(100+80t)$
$10000-10000t=8000+6400t$
$2000=16400t$
$t=0.1219512195$
$$\therefore t=7\:min, \:\:\:19.024\:sec$$
Substituting back in, that gives
$d_{t\: min}=\sqrt{(100+80(0.12195…))^2+(100-100(0.12195…))^2}$ (carried decimal in calculator)
$d_{t\: min}=140.556\:km$
Attempted Part B
Not too hard. But how do I do part B? This is what I've tried so far:
I figured that with the inequality: $\frac{a}{b}<\frac{w}{v}$, perhaps I could find the double derivative and find when that is positive, since $\frac{d^2d}{dt^2}>0$ for a local minimum. However, it gets messy and I don't think I'm on the right track. Anyway, see what you think.
Starting with the general equation: $$d_t=\sqrt{(a+vt)^2+(b-wt)^2}$$
Then differentiate it: $$\frac{dd_t}{dt}=\frac{2v(a+vt)-2w(b-wt)}{2\sqrt{(a+vt)^2+(b-wt)^2}}$$
Simplify: $$\frac{dd_t}{dt}=\frac{va+v^2t-wb+w^2t}{\sqrt{(a+vt)^2+(b-wt)^2}}$$
And differentiate again: $$\frac{d^2d_t}{dt^2}=\frac{\sqrt{(a+vt)^2+(b-wt)^2}\cdot (v^2+w^2)-\frac{2v(a+vt)-2w(b-wt)}{2\sqrt{(a+vt)^2+(b-wt)^2}}\cdot (va+v^2t-wb+w^2t)}{\sqrt{(a+vt)^2+(b-wt)^2}^2}$$
And now set it to be greater than zero: $$0<\frac{\sqrt{(a+vt)^2+(b-wt)^2}\cdot (v^2+w^2)-\frac{va+v^2t-wb+w^2t}{\sqrt{(a+vt)^2+(b-wt)^2}}\cdot (va+v^2t-wb+w^2t)}{(a+vt)^2+(b-wt)^2}$$
$$0<\frac{(v^2+w^2)\sqrt{(a+vt)^2+(b-wt)^2}-\frac{(va+v^2t-wb+w^2t)^2}{\sqrt{(a+vt)^2+(b-wt)^2}}}{(a+vt)^2+(b-wt)^2}$$
Then
$$0[(a+vt)^2+(b-wt)^2]<(v^2+w^2)\sqrt{(a+vt)^2+(b-wt)^2}-\frac{(va+v^2t-wb+w^2t)^2}{\sqrt{(a+vt)^2+(b-wt)^2}}$$
$$\frac{(va+v^2t-wb+w^2t)^2}{\sqrt{(a+vt)^2+(b-wt)^2}}<(v^2+w^2)\sqrt{(a+vt)^2+(b-wt)^2}$$
$$(va+v^2t-wb+w^2t)^2<(v^2+w^2)\sqrt{(a+vt)^2+(b-wt)^2}\cdot \sqrt{(a+vt)^2+(b-wt)^2}$$
$$(va+v^2t-wb+w^2t)^2<(v^2+w^2)[(a+vt)^2+(b-wt)^2]$$
$$a^2v^2-2abvw+2atv^3+2atvw^2+b^2w^2-2btv^2w-2btw^3+t^2v^4+2t^2v^2w^2+t^2w^4<a^2v^2+a^2w^2+2atv^3+2atvw^2+b^2v^2+b^2w^2-2btv^2w-2btw^3+t^2v^4+2t^2v^2w^2+t^2w^4$$
This simplifies to: $$-2abvw<a^2w^2+b^2v^2$$
$$\frac{-2abvw}{ab}<\frac{a^2w^2}{ab}+\frac{b^2v^2}{ab}$$
$$-2vw<\frac{aw^2}{b}+\frac{bv^2}{a}$$
Just added more. Any ideas?
Any help for part c would also be welcome.
I hope I've been clear enough, but let me know if you need further clarification on anything.
Thanks all.
Hugo.
Best Answer
Using the derivative likes what you did is overcomplicated. You can do the first and second parts it in a simple way:
a. $d_t=\sqrt{(100+80t)^2+(100-100t)^2}= \sqrt{16400t^2-2000t+20000}$. You can easily recognize that the square root gets the minimum value when the expression inside it gets the minimum value. So differentiating $16400t^2-2000t+20000$ now is much easier
b. $d_t=\sqrt{(a+vt)^2+(b-wt)^2}=\sqrt{(v^2+w^2)t^2+2(av-bw)t+a^2+b^2}$. Doing the same thing in part a, by differentiating the expression inside the square root, you will get $t=\dfrac {bw-av} {v^2+w^2}$. Now plug $t$ back in $a+vt$ and $b-wt$ and check the ratio